Calc Iii Lecture Notes

MATH 275: Calculus III

Lecture Notes by Angel V. Kumchev

Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lecture 1. Lecture 2. Lecture 3. Lecture 4. Lecture 5. Lecture 6. Lecture 7. Lecture 8. Lecture 9. Lecture 10. Lecture 11. Lecture 12. Lecture 13. Lecture 14. Lecture 15. Lecture 16. Lecture 17. Lecture 18. Lecture 19. Lecture 20. Lecture 21. Lecture 22. Lecture 23. Three-Dimensional Coordinate Systems . . . . . . . . . . . . . . . . . . . . . iii 1

Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Multivariable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Directional Derivatives and Gradients . . . . . . . . . . . . . . . . . . . . . . . 77 Tangent Planes and Normal Vectors . . . . . . . . . . . . . . . . . . . . . . . 81 Extremal Values of Multivariable Functions . . . . . . . . . . . . . . . . . . . 91 Lagrange Multipliers* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Double Integrals over Rectangles . . . . . . . . . . . . . . . . . . . . . . . . 105 Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . 111 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 121 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . 133 Applications of Double and Triple Integrals . . . . . . . . . . . . . . . . . . . 139 Change of Variables in Multiple Integrals* . . . . . . . . . . . . . . . . . . . 149 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 The Fundamental Theorem for Line Integrals . . . . . . . . . . . . . . . . . . 175 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 i Lecture 24. Lecture 25. Lecture 26.

Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Gauss’ Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

Appendix A. Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Appendix B. Deﬁnite Integrals of Single-Variable Functions . . . . . . . . . . . . . . . . 213 Appendix C. Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Appendix D. Introduction to Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Appendix E. Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 239

ii

Preface
These lecture notes originated from a set of class notes that I used to suplement my lectures for M408D Sequences, Series, and Multivariable Calculus at UT-Austin, and later for MATH 275 here, at TU. About 2006, the TU mathematics department set a goal to include in MATH 275 some topics that had previously been excluded “for lack of class time.” I found that diﬃcult to achieve within the constraints imposed by the organization of the standard calculus text that we were using at the time, yet quite manageable if I moved some material around and kept the applications more focused on geometry. Those changes were signiﬁcant enough to diminish the value of the textbook as a primary reference for the course—it became a (rather expensive) source of homework assignments and pretty pictures. That motivated me to add some exercises and illustrations to the class notes and to turn them in a self-contained resource—these lecture notes. I have not written these notes in vacuum, so there are several people whose help and suggestions I gratefully acknowledge. The earliest version of my UT-Austin class notes beneﬁted greatly from numerous exchanges with Michael Tehranchi and Vrej Zarikian, and my good friend and colleague Alexei Kolesnikov helped me with the addition of some new material to the current version. My colleagues John Chollet and Tatyana Sorokina have used these notes in their own teaching and in the process have not only helped me ﬁnd various typos, but also have provided me with insights regarding possible improvements—as have many of their and my own former students. The present version of these notes diﬀers from earlier versions in two main ways. First, I added a couple of new topics: the method of Lagrange multipliers (Lecture #12) and the general theory of change of variables in multiple integrals (Lecture #19). While those topics are not oﬃcially a part of the MATH 275 curriculum, I was convinced over time that they are too important not to be mentioned. The second major addition to this version is the inclusion of more visual illustrations and of several Mathematica tutorials. Among the latter, I want to mention speciﬁcally Appendix D, which is a version of the introduction to Mathematica that is part of the computer laboratories used in calculus courses at TU. To that end, I thank my colleagues Raouf Boules, Geoﬀrey Goodson, Ohoe Kim and Michael O’Leary for allowing me to use their work.

iii

LECTURE 1

Three-Dimensional Coordinate Systems
You are familiar with coordinate systems in the plane from single-variable calculus. The ﬁrst coordinate system in the plane that you have seen is most likely the Cartesian system, which consists of two perpendicular axes. These axes are usually positioned so that one of them, called the x-axis, is horizontal and oriented from left to right and the other, called the y-axis, is vertical and oriented upward (see Figure 1.1). The intersection point of the two axes is called the origin and is commonly denoted by O. y

O

x

Figure 1.1. Two-dimensional Cartesian coordinate system 1.1. Cartesian coordinates in space The three-dimensional Cartesian coordinate system consists of a point O in space, called the origin, and of three mutually perpendicular axes through O, called the coordinate axes and labeled the x-axis, the y-axis, and the z-axis. The labeling of the axes is usually done according to the right-hand rule: if one places one’s right hand in space so that the origin lies on one’s palm, the y-axis points toward one’s arm, and the x-axis points toward one’s curled ﬁngers, then the z-axis points toward one’s thumb. There are several standards for drawing two-dimensional models of three-dimensional objects. In particular, there are three somewhat standard ways to draw a three-dimensional coordinate system. These are shown on Figure 1.2. Model (c) is the one commonly used in manual sketches. Model (a) is a version of model (c) that is commonly used in textbooks and will be the one we shall use in these notes; unlike model (c), it is based on real mathematics that transforms threedimensional coordinates to two-dimensional ones. Model (b) is the default model preferred by computer algebra systems such as Mathematica. The coordinate axes determine several notable objects in space. First, we have the coordinate planes: these are the unique planes determined by the three diﬀerent pairs of coordinate axes. The coordinate planes determined by the x- and y-axes, by the x- and z-axes, and by the y- and z-axes are called the xy-plane, the xz-plane, and the yz-plane, respectively. The three coordinate planes divide space into eight “equal” parts, called octants; the octants are similar to the four quadrants
1

z

z y y (a) (b) x

z

y x (c)

x

Figure 1.2. Three-dimensional Cartesian coordinate systems that the two coordinate axes divide the xy-plane into. The octants are labeled I through VIII, so that octants I through IV lie above the respective quadrants in the xy-plane and octants V through VIII lie below those quadrants. Next, we deﬁne the Cartesian coordinates of a point P in space. We start by ﬁnding its orthogonal projections onto the yz-, the xz-, and the xy-planes. Let us denote those points by P x , Py , and Pz , respectively (see Figure 1.3). Let a be the directed distance from P x to P: that is, a= |PP x | if the direction from P x to P is the same as that of the x-axis, −|PP x | if the direction from P x to P is opposite to that of the x-axis.

Further, let b be the directed distance from Py to P, and let c be the directed distance from Pz to P. The three-dimensional Cartesian coordinates of P are (a, b, c) and we write P(a, b, c). Example 1.1. Plot the points A(4, 3, 0), B(2, −2, 2), and C(2, 2, −2). Example 1.2. Determine the sets of points described by the equations: z = 1; x = −2; and y = 3, x = −1. Answer. The equation z = 1 represents a plane parallel to the xy-plane. The equation x = −2 represents a plane parallel to the yz-plane. The equations y = 3, x = −1 represent a vertical line perpendicular to the xy-plane and passing through the point (3, −1, 0). z Py b a P c y x Pz Px

Figure 1.3. Cartesian coordinates of a point in space
2

1.2. Distance in space Let P1 = (x1 , y1 , z1 ) and P2 = (x2 , y2 , z2 ) be two points in space. Then the distance between P1 and P2 is given by |P1 P2 | = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 . (1.1) Example 1.3. The distance between the points A(4, 3, 0) and B(2, −2, 2) is √ |AB| = (4 − 2)2 + (3 + 2)2 + (0 − 2)2 = 33. Example 1.4. Find the equation of the sphere of radius 3 centered at C(1, −2, 0). Solution. The sphere of radius 3 centered at C(1, −2, 0) is the set of points P(x, y, z) such that |PC| = (x − 1)2 + (y + 2)2 + (z − 0)2 = 3, (x − 1)2 + (y + 2)2 + z2 = 9. By generalizing this example, we obtain the following proposition. Proposition 1.1. The sphere with center C(a, b, c) and radius r > 0 has an equation (x − a)2 + (y − b)2 + (z − c)2 = r2 . (1.2) Conversely, the set of points (x, y, z) satisfying (1.2) is a sphere in space with center (a, b, c) and radius r. Example 1.5. The equation x2 + y2 + z2 − 4x + 6y + z = 7 describes a sphere. Find its center and radius. Solution. We can rewrite the above equation as x2 − 4x + 4 + y2 + 6y + 9 + z2 + z +
1 4

that is, the points P(x, y, z), whose coordinates satisfy the equation

=7+4+9+
81 . 4

1 4

1 (x − 2)2 + (y + 3)2 + (z + 2 )2 =

The last equation is of the form (1.2) with a = 2, b = −3, c = − 1 , and r = 9 . Hence, the given 2 2 equation describes a sphere with center (2, −3, − 1 ) and radius 9 . 2 2 1.3. Cylindrical coordinates You should be familiar with the polar coordinate system in the plane from single-variable calculus (see also Appendix C). In the remainder of this lecture, we deﬁne two sets of coordinates in space that extend the polar coordinates in the plane to the three-dimensional setting. We start with the so-called “cylindrical coordinates”. The basic idea of the cylindrical coordinate system is to replace the Cartesian coordinates x, y in the xy-plane by the corresponding polar coordinates and to keep the Cartesian z-coordinate. Let P be a point in space with Cartesian coordinates (x, y, z). The cylindrical coordinates of P are the triple (r, θ, z), where (r, θ) are the polar coordinates of the point P0 (x, y) in the xy-plane (see Figure 1.4). In other words, we pass between Cartesian and cylindrical coordinates in space
3

z P(x, y, z) z x θ r P0 (x, y) y

Figure 1.4. Cylindrical coordinates of a point in space by passing between Cartesian and polar coordinates in the xy-plane and keeping the z-coordinate unchanged. The formulas for conversion from cylindrical to Cartesian coordinates are x = r cos θ, r= x2 + y2 , y = r sin θ, θ = arctan(y/x), z = z, z = z. (1.3) (1.4) and those for conversion from Cartesian to cylindrical coordinates are The formula θ = arctan(y/x) in (1.4) comes with a caveat: unless the point (x, y) lies in the ﬁrst quadrant of the xy-plane, this formula yields a reference angle which needs to be adjusted to obtain the actual value of θ. (This is similar to the conversion from Cartesian to polar coordinates in the plane: see §C.2 in Appendix C). Example 1.6. Find the Cartesian coordinates of the points P(2, π, e) and Q(4, arctan(−2), 2.7). Solution. The Cartesian coordinates of P are (x, y, e), where x = 2 cos π = −2, y = 2 sin π = 0. That is, P(−2, 0, e). The Cartesian coordinates of Q are (x, y, 2.7), where 4 8 x = 4 cos(arctan(−2)) = √ , y = 4 sin(arctan(−2)) = − √ . 5 5 4 8 That is, Q( √5 , − √5 , 2.7). Example 1.7. Find the cylindrical coordinates of the points P(3, 4, 5) and Q(−2, 2, −1). Solution. The cylindrical coordinates of P are (r, θ, 5), where √ 4 tan θ = , r = 32 + 42 = 5, 3 4 and θ lies in the ﬁrst quadrant. Hence, θ = arctan( 3 ) and the cylindrical coordinates of P are 4 (5, arctan( 3 ), 5). The cylindrical coordinates of Q are (r, θ, −1), where √ r = 22 + (−2)2 = 2 2, tan θ = −1, and θ lies in the second quadrant. Hence, θ = √ (2 2, 3π , −1). 4
3π 4

and the cylindrical coordinates of P are

4

z

y

y x

1 ( 2 , 0)

x

Figure 1.5. The cylinder x2 + y2 = x and its cross-section Example 1.8. Find the Cartesian equation of the surface r = cos θ. What kind of surface is this? Solution. We have r = cos θ x2 + y2 = x ⇐⇒ r2 = r cos θ
1 2 1 4

⇐⇒

x2 + y2 = x. (x − 1 )2 + y2 = 1 . 2 4

The last equation is the equation of a circle of radius ⇐⇒ x2 − x + 1 + y2 = 4

centered at the point ( 1 , 0) in the xy-plane: 2 ⇐⇒

Here, however, we use the equation x2 + y2 = x to describe a surface in space, so we view it as an equation in x, y, z. Since z does not appear in the equation, if the x- and y-coordinates of a point satisfy the equation x2 + y2 = x, then the point is on the surface. In other words, if a2 + b2 = a, then all points (a, b, z), whose x-coordinate is a and y-coordinate is b, are on the surface. We also know that all the points (x, y, 0), with x2 + y2 = x, are on the surface: these are the points on the circle in the xy-plane with the same equation. Putting these two observations together, we conclude that the given surface is the cylinder that consists of all the vertical lines passing through the points of the circle x2 + y2 = x in the xy-plane (see Figure 1.5). 1.4. Spherical coordinates We now introduce the so-called “spherical coordinates” in space. In order to deﬁne this set of coordinates, we choose a special point O, the origin, a plane that contains O, and a pair of mutually perpendicular axes passing through O—one that lies in the chosen plane and another that is orthogonal to the chosen plane. Since we want to relate the spherical coordinates of a point to its Cartesian coordinates, it is common to choose O to be the origin O(0, 0, 0) of the Cartesian coordinate system, the special plane to be the xy-plane, and the two axes to be the x-axis and the z-axis. With these choices, the spherical coordinates of a point P(x, y, z) in space are (ρ, θ, φ), where ρ = |OP|, θ is the angular polar coordinate of the point P0 (x, y) in the xy-plane, and φ is the angle between OP and the positive direction of the z-axis (see Figure 1.6); it is common to restrict φ to the range 0 ≤ φ ≤ π. The angles θ and φ are often called the longitude and the latitude of P, because with the right choice of the spherical coordinate system, the angles θ and 90◦ − φ for a point on the surface of the Earth are its geographic longitude and latitude.
5

z P(x, y, z) ρ z = ρ cos φ x θ r = ρ sin φ y P0 (x, y)

Figure 1.6. Spherical coordinates of a point in space Consider a point P(x, y, z) in space and let P0 (x, y, 0). Also, let (ρ, θ, φ) be the spherical coordinates of P. From the right triangle OPP0 , we ﬁnd that z = ±|P0 P| = |OP| cos φ = ρ cos φ, x = |OP0 | cos θ = ρ sin φ cos θ, x = ρ sin φ cos θ, ρ= x2 + y2 + z2 , |OP0 | = |OP| sin φ = ρ sin φ. y = |OP0 | sin θ = ρ sin φ sin θ. z = ρ cos φ. φ = arccos(z/ρ), (1.5) (1.6) On the other hand, using polar coordinates in the xy-plane, we ﬁnd that Thus, the formulas for conversion from spherical to Cartesian coordinates are y = ρ sin φ sin θ, θ = arctan(y/x), The formulas for conversion from Cartesian to spherical coordinates are where the formula for θ comes with some strings attached (the same as in the cases of polar and cylindrical coordinates). Example 1.9. Find the Cartesian coordinates of the point P(2, π/3, 3π/4). Solution. The Cartesian coordinates of P are √ 1 3 x = 2 sin(3π/4) cos(π/3) = √ , y = 2 sin(3π/4) sin(π/3) = √ , 2 2 √ z = 2 cos(3π/4) = − 2.

√ Example 1.10. Find the spherical coordinates of the points P(1, −1, − 6) and Q(−2, 2, −1). Solution. The spherical coordinates of P are √ √ √ √ √ 5π ρ = 1 + 1 + 6 = 8, φ = arccos − 6/ 8 = arccos − 3/2 = , 6 and the solution θ of tan θ = −1, 3π/2 < θ < 2π. √ Hence, θ = 7π/4 and the spherical coordinates of P are ( 8, 7π/4, 5π/6). The spherical coordinates of Q are √ ρ = 4 + 4 + 1 = 3, φ = arccos(−1/3), and the solution θ of tan θ = −1,
6

π/2 < θ < π.

Hence, θ = 3π/4 and the spherical coordinates of Q are (3, 3π/4, arccos(−1/3)). Example 1.11. Find the Cartesian equation of the surface ρ2 (sin2 φ − 2 cos2 φ) = 1. Solution. Since x2 + y2 = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ2 sin2 φ, we have ρ2 sin2 φ − 2ρ2 cos2 φ = 1 ⇐⇒ x2 + y2 − 2z2 = 1. The last equation represents a surface known as “one-sheet hyperboloid”; it is shown on Figure 1.7.

z

z

y x x

y

Figure 1.7. The hyperboloid x2 + y2 − 2z2 = 1 Figure 1.8. The region 2 ≤ ρ ≤ 3, π/2 ≤ φ ≤ π Example 1.12. Describe the solid deﬁned by the inequalities 2 ≤ ρ ≤ 3, π/2 ≤ φ ≤ π.

Solution. The second inequality restricts the points of the solid to the lower half-space. The inequality ρ ≤ 3 describes the points in space that are within distance of 3 from the origin, that is, the interior of the sphere x2 + y2 + z2 = 9. Similarly the inequality ρ ≥ 2 describes the exterior of the sphere x2 + y2 + z2 = 4. Thus, the given solid is obtained from the lower half of the interior of the sphere x2 + y2 + z2 = 9 by removing from it the interior of the smaller sphere x2 + y2 + z2 = 4 (see Figure 1.8). Example 1.13. Find the cylindrical and spherical equations of the sphere x2 + y2 + z2 + 4z = 0. Solution. By (1.4), x2 + y2 = r2 , so the cylindrical form of the given equation is x2 + y2 + z2 + 4z = 0 ρ2 + 4ρ cos φ = 0 ⇐⇒ r2 + z2 + 4z = 0. ρ = −4 cos φ. By (1.6), x2 + y2 + z2 = ρ2 , so the spherical form of the given equation is ⇐⇒
7

Exercises
1.1. Which of the points P(1, 3, 2), Q(0, −2, −1), and R(5, 3, −3): is closest to the xy-plane; lies in the xz-plane; lies in the yz-plane; is farthest from the origin? 1.2. Describe the set in R3 represented by the equation x + 2y = 2. 1.3. Find the side lengths of the triangle with vertices A(0, 6, 2), B(3, 4, 1) and C(1, 3, 4). Is this a special kind of a triangle: equilateral, isosceles, right? 1.4. Find an equation of the sphere that passes through P(4, 2, 1) and has center C(1, 1, 1). 1.5. Show that the equation x2 + y2 + z2 = x + 2y + 4z represents a sphere and ﬁnd its center and radius. Describe verbally the set in R3 represented by the given equation or inequality. 1.6. x = −3 1.11. xy = 0 1.7. z ≤ 2 1.8. −1 ≤ y ≤ 2 1.9. (x − 1)2 + y2 + z2 ≥ 4 1.10. 4 < x2 + y2 + z2 < 9 1.12. Write inequalities to describe the half-space consisting of all points to the right of the plane y = −2. 1.13. Write inequalities to describe the interior of the upper hemisphere of radius 2 centered at the origin. Find the Cartesian coordinates of the point with the given cylindrical coordinates. 1.17. (3, 3π/2, −1) 1.14. (2, 0, 3) 1.15. (3, π/4, −3) 1.16. (3, 7π/6, 2) Find the cylindrical coordinates of the point with the given Cartesian coordinates. 1.21. (3, 3, 2) 1.18. (1, −1, 4) 1.19. (−2, 2, 2) √ 1.20. ( 3, 1, −2) Find the Cartesian coordinates of the point with the given spherical coordinates. 1.22. (1, 0, 0) 1.25. (2, π/2, π/4) 1.23. (3, π/4, 3π/4) 1.24. (3, 7π/6, π/2) Find the spherical coordinates of the point with the given Cartesian coordinates. √ 1.26. (1, 1, 2) √ 1.29. ( 2, 1, −1) 1.27. (−1, 1, − 2) √ 1.28. ( 3, 1, 0) Find the cylindrical and spherical equations of the surface represented by the given Cartesian equation. 1.30. x2 + y2 = 2z 1.31. x2 + 2x + y2 + z2 = 1 1.32. z = x2 + y2 Find the Cartesian equation of the surface represented by the given cylindrical or spherical equation. If possible, describe the surface verbally. 1.33. ρ cos φ = −2 1.34. r = 2 sin θ 1.35. r2 + z2 = 4 1.36. ρ = 2 sin φ sin θ 1.37. ρ sin2 φ − 2 cos φ = 0 8 1.38. r cos θ − 2r sin θ = 3

Describe verbally the set in R3 represented by the given conditions. 1.39. θ = π/4 1.40. ρ = 2 1.41. r = 1, 0 ≤ z ≤ 1 1.42. 2 ≤ ρ ≤ 4, 0 ≤ φ ≤ π/2 1.43. ρ ≤ 1, −π/2 ≤ θ ≤ π/2 1.44. ρ sin φ ≤ 1, π/2 ≤ φ ≤ π

9

LECTURE 2

Vectors
2.1. Geometric vectors A vector is a quantity (such as velocity, acceleration, force, impulse, etc.) that has both magnitude and direction. In mathematics, we consider two types of vectors: geometric and algebraic. A geometric vector is a directed line segment, represented visually as an arrow: the length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. We shall denote vectors by small boldface letters: a, b, . . . , z. If v is a vector represented by an − − → arrow with initial point A and terminal point B, we may also write v = AB; this is consistent with an alternative notation for vectors: a, b, . . . , z. We say that two vectors u and v are equal, and write u = v, if u and v have equal magnitudes (lengths) and the same direction. The vector of zero length is called the zero vector and is denoted 0; it is the only vector with no speciﬁc direction.

2.2. Vector addition − − → − − → Given two vectors u = AB and v = BC (represented so that the terminal point of the ﬁrst vector is the initial point of the second), we deﬁne their sum u + v by (see Figure 2.1(a)) − − → − − → − − → u + v = AB + BC = AC. − − → −→ − When u = AB and v = CD are two vectors in general position, with C B, we ﬁrst represent v by −− −→ −− −→ an equal arrow BD and then apply the above deﬁnition: u + v = AD , as shown on Figure 2.1(b). C D v +

D v B C A u (b)

u

B A u (a) Figure 2.1. Vector addition

11

u

v

+

v

v

2.3. Multiplication by scalars Let v be a vector and let c be a real number (called also a scalar). The scalar multiple cv is the vector whose length is |c| times the length of v and whose direction is the same as the direction of v when c > 0, or opposite to the direction of v when c < 0. If c = 0 or v = 0, then cv = 0. The vector (−1)v has the same length as v but the opposite direction. We call this vector the negative (or opposite) of v and denote it by −v. For two vectors u and v, we deﬁne their diﬀerence u − v by u − v = u + (−v). We visualize the sum and the diﬀerence of two vectors u and v by drawing them so that they share the same initial point. We then consider the parallelogram they deﬁne. The sum u + v is represented by the diagonal of the parallelogram that passes through the common initial point of u and v and the diﬀerence u − v is represented by the other diagonal (see Figure 2.2).

u

+

u−

v

v

v u (a)

v u (b)

Figure 2.2. The sum and the diﬀerence of u and v

2.4. Coordinate representation of vectors. Algebraic vectors Often, it is convenient to introduce a coordinate system and to work with vectors algebraically − − → instead of geometrically. If we represent a (geometric) vector a by the arrow OA, where O is the origin of a Cartesian coordinate system, the terminal point A of a has coordinates (a1 , a2 ) or (a1 , a2 , a3 ), depending on the dimension. We call these coordinates the coordinates or the components of the algebraic vector a and write a = a1 , a2 or a = a1 , a2 , a3 . − − → We will use the same algebraic vector to represent all the geometric vectors equal to OA. For example, if A(1, −1), B(−1, 3), and C(0, 2), then − − → − − → BC = OA = 1, −1 . Thus, for any two points A(a1 , a2 , a3 ) and B(b1 , b2 , b3 ), the algebraic vector v representing the arrow − − → AB is v = b1 − a1 , b2 − a2 , b3 − a3 . From now on, we shall state all the formulas for three-dimensional vectors. One usually obtains the respective two-dimensional deﬁnition or result simply by dropping the last component of the vectors; any exceptions to this rule will be mentioned explicitly when they occur.
12

2.5. Operations with algebraic vectors Note that when we write a vector a algebraically, the distance formula provides a convenient tool for computing the length (also called the norm) a of that vector: if a = a1 , a2 , a3 , then a = a2 + a2 + a2 1 2 3
1/2

.

(2.1)

Furthermore, if a = a1 , a2 , a3 and b = b1 , b2 , b3 are vectors and c is a scalar, we have the following coordinate formulas for a + b, a − b, and ca: a + b = a1 + b1 , a2 + b2 , a3 + b3 , a − b = a1 − b1 , a2 − b2 , a3 − b3 , ca = ca1 , ca2 , ca3 . (2.2) (2.3) (2.4)

In other words, we add vectors, subtract vectors, and multiply vectors by scalars by simply performing the respective operations componentwise. Example 2.1. Consider the vectors a = 1, −2, 3 and b = 0, 1, −2 . Then 2a + b = 2, −4, 6 + 0, 1, −2 = 2, −3, 4 , √ 2a + b = 22 + (−3)2 + 42 = 29. The algebraic operations with vectors have several properties that resemble familiar properties of the algebraic operations with numbers. Theorem 2.1. Let a, b, and c denote vectors and let c, c1 , c2 denote scalars. Then: i) ii) iii) iv) v) vi) vii) viii) ix) x) xi) a+b=b+a a + (b + c) = (a + b) + c a+0=a a + (−a) = 0 c(a + b) = ca + cb (c1 + c2 )a = c1 a + c2 a (c1 c2 )a = c1 (c2 a) 1a = a a ≥ 0, with a = 0 if and only if a = 0 ca = |c| · a a + b ≤ a + b (the triangle inequality)

Proof. The proofs of parts i)–viii) are all similar: they use (2.2)–(2.4) to deduce i)–viii) from the respective properties of numbers. For example, to prove the distributive law vi), we write a = a1 , a2 , a3 and argue as follows: (c1 + c2 )a = (c1 + c2 )a1 , (c1 + c2 )a2 , (c1 + c2 )a3 = c1 a1 + c2 a1 , c1 a2 + c2 a2 , c1 a3 + c2 a3 = c1 a1 , c1 a2 , c1 a3 + c2 a1 , c2 a2 , c2 a3 = c1 a + c2 a
13

by (2.4) by the distributive law for numbers by (2.2) by (2.3).

The proofs of ix) and x) are also easy. For instance, ca = (ca1 )2 + (ca2 )2 + (ca3 )2 = c2 a2 + a2 + a2 1 2 3
1/2 1/2

=

√ c2 a2 + a2 + a2 1 2 3

1/2

= |c| · a .

The proof of xi) is slightly more involved, so we leave it to the exercises. 2.6. Unit vectors and the standard basis The simplest nonzero vectors are: e1 = 1, 0, 0 , e2 = 0, 1, 0 , e3 = 0, 0, 1 . These are called the standard basis vectors. Sometimes, especially in physics and engineering, the vectors e1 , e2 , and e3 are denoted by i, j, and k, respectively, but we shall stick with the above notation, because it is more common in mathematics (it is also more consistent with generalizations to higher dimensions). Because of (2.2)–(2.4), it is always possible to express a given vector as a combination of the three standard basis vectors. For example, 1, 2, 4 = 1, 0, 0 + 0, 2, 0 + 0, 0, 4 = 1, 0, 0 + 2 0, 1, 0 + 4 0, 0, 1 = e1 + 2e2 + 4e3 . More generally, if a = a1 , a2 , a3 , we have a = a1 , 0, 0 + 0, a2 , 0 + 0, 0, a3 = a1 e1 + a2 e2 + a3 e3 . Notice that e1 , e2 , and e3 have length 1. In general, vectors of length 1 are called unit vectors. We remark that if a is any nonzero vector (i.e., a 0), then there is a unique unit vector that points in the same direction as a—namely, the vector 1 a u= a= . a a Indeed, if c = a
−1

, then a and u = ca point in the same direction (since c > 0) and u = ca = |c| · a = a
−1

· a = 1.

This simple computation is known as normalization, and u is called the normalization of a. Example 2.2. Consider the vector a = 2, 1, 2 . Its length is √ a = 22 + 12 + 22 = 3, so the normalization of a is the vector
1 a 3

=

2 1 2 , , 3 3 3

.

2.7. The dot product We can now add two vectors and we can multiply a vector by a scalar. Is there a way to multiply two vectors? The ﬁrst way that comes to mind is to multiply the vectors componentwise: ab = a1 , a2 , a3 b1 , b2 , b3 = a1 b1 , a2 b2 , a3 b3 , but this way of multiplying two vectors turns out to be of little practical use. Instead, through the remainder of this lecture, we shall talk about two meaningful ways to multiply vectors: the dot product and the cross product.
14

Definition. Given vectors a = a1 , a2 , a3 and b = b1 , b2 , b3 , we deﬁne their dot product a · b to be a · b = a1 b1 + a2 b2 + a3 b3 . (2.5) Notice that the dot product of two vectors is a scalar, not a vector. Sometimes, the dot product is called also the inner product or the scalar product of a and b. Because of this, you should avoid the temptation to refer to the scalar multiple ca as the “scalar product of c and a”. Example 2.3. The dot product of a = 4, 2, −1 and b = −2, 2, 1 is a · b = (4)(−2) + (2)(2) + (−1)(1) = −5. The dot product of u = 2e1 + 3e2 + e3 and v = e1 + 4e2 is u · v = (2)(1) + (3)(4) + (1)(0) = 14. The next theorem summarizes some important algebraic properties of the dot product. Theorem 2.2. Let a, b, and c denote vectors and let c1 , c2 denote scalars. Then: i) a · a = a 2 ii) a · 0 = 0 iii) a · b = b · a iv) (c1 a) · (c2 b) = c1 c2 (a · b) v) (a + b) · c = a · c + b · c Proof. These follow from the deﬁnition of the dot product and from the properties of the real numbers. For example, if a = a1 , a2 , a3 and b = b1 , b2 , b3 , we have a · b = a1 b1 + a2 b2 + a3 b3 = b1 a1 + b2 a2 + b3 a3 =b·a That is, iii) holds. We said earlier that our deﬁnition of the dot product is motivated by applications. The next theorem relates the dot product of two vectors to their geometric representations. Theorem 2.3. If a and b are two vectors, then a · b = a b cos θ, where θ is the angle between a and b. Corollary 2.4. If a and b are two nonzero vectors and θ is the angle between them, then a·b . cos θ = a b Example 2.4. Find the angle between a = 1, 2, −1 and b = 0, 2, 1 . Solution. The angle θ between a and b satisﬁes (1)(0) + (2)(2) + (−1)(1) 3 3 √ cos θ = = √ √ = √ , 30 6 5 12 + 22 + (−1)2 02 + 22 + 12 so θ = arccos
√3 30

by (2.5) by the commutative law for real numbers by (2.5).

= 0.9911 . . . .
15

a − proju a

u

a proju a

Figure 2.3. The projection of a vector onto a unit vector We say that two vectors a and b are orthogonal (or perpendicular) if a · b = 0. By the corollary, a · b = 0 implies that the angle between a and b is 90◦ , so this deﬁnition makes perfect sense. Note that if a = a1 , a2 , a3 , then we can use the dot products of a and the standard basis vectors e j to extract the components of a: a1 = a · e1 , a2 = a · e2 , proju a = (a · u)u is a vector that is parallel to u. This vector is called the (orthogonal) projection of a onto u. It has the property that the diﬀerence between it and the original vector a is perpendicular to u: a − proju a · u = 0. The two-dimensional case of this is illustrated on Figure 2.3. Note that the size of the dot product a·u equals the length of the projection; the sign of that dot product indicates whether the projection and u have the same direction (positive sign) or opposite directions (negative sign). 2.8. The cross product The dot product of two vectors is a scalar. In this section, we deﬁne another product of two vectors: their “cross product”, which is a vector. Since our discussion requires basic familiarity with 2 × 2 and 3 × 3 determinants, a quick look at Appendix A may be helpful before proceeding further. Definition. Let a = a1 , a2 , a3 and b = b1 , b2 , b3 be two three-dimensional vectors. Then their cross product a × b is the vector a×b= a2 a3 a a a a e − 1 3 e + 1 2 e. b2 b3 1 b1 b3 2 b1 b2 3 a3 = a · e3 .

More generally, suppose that u is a unit vector. Then a · u is a scalar and

Remarks. There are a couple of things we should point out: 1. The cross product is deﬁned only for three-dimensional vectors. There is no analog of this operation for two-dimensional vectors. 2. The deﬁning formula may be confusing at ﬁrst. On the other hand, following the exact labeling is important because of the properties of the cross product (see Theorem 2.5
16

below). Another representation of the cross product that is much easier to remember is e1 e2 e3 a × b = a1 a2 a3 . b1 b2 b3 (2.6)

This expression is not exactly a determinant in the sense described in Appendix A (since its ﬁrst row consists of vectors instead of numbers), but if we expand this “determinant” as if we were dealing with the determinant of a regular 3 × 3 matrix, we get exactly the cross product of a and b. The following theorem summarizes the algebraic properties of the cross product. Theorem 2.5. Let a, b, and c denote three-dimensional vectors and let c, c1 , c2 denote scalars. Then i) b × a = −(a × b) ii) (c1 a) × (c2 b) = c1 c2 (a × b) iii) a × (b + c) = a × b + a × c iv) (a + b) × c = a × c + b × c v) a × a = 0 and a × 0 = 0 vi) a × (b × c) = (a × b) × c may fail. Example 2.5. We have e1 e2 e3 1 0 1 0 0 0 e = e3 , e2 + e1 − e1 × e2 = 1 0 0 = 0 1 3 0 0 1 0 0 1 0 and similarly, By property i), we also have e2 × e1 = −e3 , Note that and e1 × e3 = −e2 , e3 × e2 = −e1 . e2 × e3 = e1 , e3 × e1 = e2 .

e1 × (e2 × e1 ) = e1 × (−e3 ) = −(e1 × e3 ) = e2

(e1 × e2 ) × e1 = e3 × e1 = e2 , so e1 × (e2 × e1 ) = (e1 × e2 ) × e1 . On the other hand, e1 × (e1 × e2 ) = e1 × e3 = −e2 ,

whereas by property v), Thus, e1 × (e1 × e2 )

(e1 × e1 ) × e2 = 0 × e2 = 0. (e1 × e1 ) × e2 ; this illustrates property vi) above.

17

Example 2.6. Compute the cross product of a = 3, −2, 1 and b = 1, −1, 1 . Solution. By (2.6), e1 e2 e3 a × b = 3 −2 1 1 −1 1 = −2 1 3 1 3 −2 e − e + e −1 1 1 1 1 2 1 −1 3 = −e1 − 2e2 − e3 = −1, −2, −1 .

Second solution. We can also argue by using the algebraic properties in Theorem 2.5 and the values of the cross products of e1 , e2 , and e3 : (3e1 − 2e2 + e3 ) × (e1 − e2 + e3 ) = 3e1 × e1 − 2e2 × e1 + e3 × e1 − 3e1 × e2 + 2e2 × e2 − e3 × e2 + 3e1 × e3 − 2e2 × e3 + e3 × e3 = 0 + 2e3 + e2 − 3e3 + 0 + e1 − 3e2 − 2e1 + 0 = −e1 − 2e2 − e3 . Finally, we discuss the geometric meaning of the cross product. Theorem 2.6. Let a = a1 , a2 , a3 and b = b1 , b2 , b3 be three-dimensional vectors. Then: i) a × b is perpendicular to both a and b. The direction of a × b is determined by the right-hand rule. ii) a × b = a b | sin θ|, where θ is the angle between a and b. That is, the length of a × b is equal to the area of the parallelogram determined by a and b. iii) a × b = 0 if and only if a and b are parallel. iv) The volume V of the parallelepiped determined by three vectors a, b, and c which are not coplanar (that is, they do not lie in one plane) is given by the formula V = a · (b × c) . Furthermore, three vectors a, b, and c are coplanar if and only if a · (b × c) = 0. Example 2.7. Find two unit vectors perpendicular to a = 1, 2, 3 and b = 2, 1, 1 . Solution. We know from part i) of Theorem 2.6 that a × b is orthogonal to both a and b. We ﬁnd that e1 e2 e3 2 3 1 3 1 2 a×b= 1 2 3 = e − e + e = −e1 + 5e2 − 3e3 . 1 1 1 2 1 2 2 1 3 2 1 1 √ a × b = (−1)2 + 52 + (−3)2 = 35, and two unit vectors orthogonal to a and b are the normalization of a × b and its opposite: a×b −1 5 −3 1 −5 3 u= = √ , √ , √ and −u= √ , √ , √ . a×b 35 35 35 35 35 35 Hence,

18

Example 2.8. Find the area of the triangle with vertices P(1, 2, 3), Q(−1, 3, 2), and R(3, −1, 2). Solution. The area of PQR is one half of the area of the parallelogram determined by the − − → − − → vectors PQ = −2, 1, −1 and PR = 2, −3, −1 . We know from part ii) of Theorem 2.6 that the area of the parallelogram is the length of e1 e2 e3 − − → − − → 1 −1 −2 −1 −2 1 PQ × PR = −2 1 −1 = e1 − e2 + e = −4e1 − 4e2 + 4e3 . −3 −1 2 −1 2 −3 3 2 −3 −1 √ √ Therefore, the area of the triangle is 1 16 + 16 + 16 = 2 3. 2 2.9. Vectors in Mathematica 2.9.1. Representation. We represent a vector in Mathematica as a list of numbers (the components of the vector). For example, to deﬁne a vector a = a1 , a2 , a3 , we write a = {a1, a2, a3} To refer to a particular component of a vector (or a list), we use double square brackets. For example, the second component of the vector a deﬁned above can be accessed as a[[2]]. Although we are only interested in two- and three-dimensional vectors here, we should note that Mathematica can handle vectors of higher dimensions just as easily. On the geometric side, Mathematica has a built-in command Arrow, which produces a geometric vector in two or three dimensions. In three dimensions, its basic syntax is Arrow[{{x1, y1, z1}, {x2, y2, z2}}] This results in an arrow with initial point (x1 , y1 , z1 ) and terminal point (x2 , y2 , z2 ). (See Mathematica’s help for various formatting options and more advanced usage.) To get a two-dimensional vector, we simply use two-dimensional points as arguments: Arrow[{{x1, y1}, {x2, y2}}] For technical reasons, the Arrow command produces a “graphics primitive” instead of an actual plot of the geometric vector. For example, if we try to execute the command Arrow[{{1,2}, {4,-1}}] Mathematica will simply repeat it in response. To see the result of the above command, we need to “wrap” it in a Graphics command like this: Graphics[Arrow[{{1,2}, {4,-1}}]] Similarly, to display a three-dimensional geometric vector, we use a Graphics3D command. Here is an example: Graphics3D[Arrow[{{1,2,0}, {1,-1,2}}]] 2.9.2. Vector addition and scalar multiplication. Vector addition, vector subtraction and scalar multiplication are denoted in the usual way: if c is a scalar and a and b are two vectors (of equal dimensions), then a+b a-b c*a, c a denotes the vector sum a + b denotes the vector diﬀerence a − b denote the scalar multiple ca
19

For example, let us deﬁne the vectors a = 2, 7, −3 and b = −3, 0, 5 . a = {2,7,-3} b = {-3,0,5} We can calculate 2a, a − b, and a + 3b as follows: 2a a-b a+3*b {4,14,-6} {5,7,-8} {-7,7,12} 2.9.3. The dot and cross products. Mathematica has built-in commands for calculating the dot and cross products: Dot and Cross, respectively. These commands also have inﬁx operator forms. A period between vectors performs a dot product, and a × symbol between vectors performs a cross product. To enter the symbol ×, ﬁrst hit the Esc key, type “cross”, and then hit again the Esc key. As an example, let us use Mathematica to compute the dot and cross products of two generic three-dimensional vectors: a = {a1, a2, a3} b = {b1, b2, b3} Here are the two versions of the dot product: a.b Dot[a,b] a1 b1 + a2 b2 + a3 b3 a1 b1 + a2 b2 + a3 b3 And one of the versions of the cross product (see the Mathematica notebook ﬁle for the other version): Cross[a,b] {-a3 b2 + a2 b3, a3 b1 - a1 b3, -a2 b1 + a1 b2} 2.9.4. Graphical representation revisited. Using the Arrow command can be cumbersome when only basic drawing functionality is needed. In such situations, it is much simpler to use a user-deﬁned command that “wraps around” Arrow and makes it more user-friendly (but cuts some of its functionality). The command vector deﬁned below presents one such option. vector[a_, b_, rgb_:{0,0,0}] := Which[ Length[a] == Length[b] == 2, Graphics[{RGBColor[rgb[[1]],rgb[[2]],rgb[[3]]], Arrow[{a,b}]}], Length[a] == Length[b] == 3, Graphics3D[{RGBColor[rgb[[1]],rgb[[2]],rgb[[3]]], Arrow[{a,b}]}], True, Print["Bad dimensions!"] ]

20

The command vector can be called in two ways. In its more basic form, vector[{x1, y1}, {x2, y2}] vector[{x1, y1, z1}, {x2, y2, z2}] the vector command returns a graphical representation of a two- or three-dimensional arrow with the given initial and terminal points (similarly to Arrow). For example, try vector[{1,2}, {-1,0}] vector[{1,1,0}, {1,2,-1}] When the dimensions of the two inputs do not match, or if they match but are neither 2 nor 3, the vector command prints out the error message “Bad dimensions!”: vector[{1,2}, {1,2,-1}] Bad dimensions! The vector command allows the user to include a third, optional argument in the form of a predeﬁned color or a three-dimensional vector to be interpreted as an RGB color-speciﬁcation. For example, the following four commands will render their outputs in red, red, blue, and teal, respectively: vector[{1,2}, {-1,0}, Red] vector[{1,2}, {-1,0}, {1,0,0}] vector[{1,2}, {-1,0}, {0,0,1}] vector[{1,2}, {-1,0}, {0,0.5,0.5}] Exercises
− − → −→ − 2.1. Let O(0, 0), A(1, 2), B(3, 0), and M(−2, 1) be points in R2 . Draw the geometric vectors v = AB and w = MO and 1 use them to draw the following vectors v + 2w, w − v, and w + 2 v. Find b , a − 3b, a + 1 b , and 2a + b − 3e2 for the given vectors a and b. 2 2.2. a = 1, −2 , b = 2, 0 2.3. a = 3e1 + 2e2 , b = −e1 + e2 2.4. a = 2, 0, −1 , b = −1, 1, 1 2.5. a = e2 + 1 e3 , b = −2e1 + e2 + e3 2 2.6. Find a vector b that has the same direction as a = 1, −1 and length b = 3. Find a · b for the given vectors a and b. 2.7. a = 3, −2, 1 , b = 1, 0, 2 2.8. a = 2, 0, −1 , b = 2, 1, 4 1 2.9. a = 3e1 + 2e2 + 4e3 , b = −e1 + 2e2 + 2 e3 ◦ 2.10. a = 2, b = 3, (a, b) = 60 Find the angle between the given vectors. 2.11. a = 2, 0, −1 , b = √ 1, 4 2, 2.12. a = 2, 0 , b = 1, 3 2.13. a = 3e1 + 2e2 , b = −e1 + e3 2.14. a = e2 + e3 , b = e1 + 2e2 − 3e3 2.15. Find the three angles of the triangle with vertices A(1, 0, 0), B(2, 1, 0), C(0, 1, 2). Determine whether the given vectors are parallel, perpendicular, or neither.
21

2.16. a = 2, 1, −1 , b = 1 , 1, 2 2 2.17. a = 1, −2, 2 , b = −2, 4, √ √ −4 2.18. a = 1, 2, 3 , b = 1, 2, 3 2.19. Find the projection of a = 1, −1, 4 onto u =
2 1 2 3, 3, −3

.

2.20. Find the projection of a = 2, −1, 2 onto the direction of b = 2, −1, 0 .

22

Find a × b for the given vectors a and b. 2.21. a = 2, −3, 1 , b = 1, 1, 2 2.22. a = 3e1 + 2e2 + 2e3 , b = −e1 + 2e2 + e3 2.23. a = e1 + 2e2 , b = 3e2 + e3 2.24. a = 1, −1, 2 , b = 1, 0, −3 Let a = −1, 2, −4 , b = 7, 3, −4 , and c = −2, 1, 0 . Evaluate the given expression or explain why it is undeﬁned. 2.25. a · (2b + c) 2.26. b × (2a) 2.27. b × (c × a) 2.28. c · (−2a × b) 2.29. 2c (a × (a + 3b)) 2.30. a × (c · b) 2.31. Find a unit vector u orthogonal to both x = −2, 3, −6 and y = 2, 2, −1 . (a) by using the cross product; (b) without using the cross product. 2.32. Find the area of the triangle with vertices A(3, 2, 1), B(2, 4, 5), and C(3, 1, 4). 2.33. Find the volume of the parallelepiped with adjacent edges OA, OB, and OC, where A(2, 3, 0), B(3, −1, 2), and C(1, 0, −4). 2.34. Use the method from the second solution of Example 2.6 to ﬁnd r × s, where r = cos φ cos θe1 + cos φ sin θe2 − sin φe3 , u = e1 + a cos φe2 + a sin φe3 , s = − sin φ sin θe1 + sin φ cos θe2 . v = −b sin φe2 + b cos φe3 .

2.35. Use the method from the second solution of Example 2.6 to ﬁnd u × v, where 2.36. Let r = x, y, z and a = a, b, c . Describe the set of points P(x, y, z) such that r − a = R. 2.37. Prove the Cauchy–Schwarz inequality: If a, b are three-dimensional vectors, then |a · b| ≤ a b . 2.38. Prove the triangle inequality: If a, b are three-dimensional vectors, then a+b ≤ a + b . 2.39. Are the following statements true or false? (a) If a 0 and a · b = a · c, then it follows that b = c. (b) If a 0 and a × b = a × c, then it follows that b = c. (c) If a 0, a · b = a · c, and a × b = a × c, then it follows that b = c.

23

LECTURE 3

Lines and Planes
3.1. Equations of a straight line in space In our discussion of lines and planes, we shall use the notion of a position vector. A position vector is a vector whose initial point is at the origin. Thus, the components of a position vector are the same as the coordinates of its terminal point. We shall specify a line in space using a point on the line and the direction of the line. If r0 is the position vector of a point P0 (x0 , y0 , z0 ) on the line and if v is a vector parallel to the line, then the terminal point of a position vector r is on the line if and only if r = r0 + tv for some t ∈ R. (3.1)

We call (3.1) a parametric vector equation of the line through P0 and parallel to v. In this equation, we construct a generic point P on in two steps: ﬁrst we get on by going to the given point P0 ; then we slide along by a multiple of the given vector v until we reach P (see Figure 3.1). The value of the parameter t indicates the distance along (measured in units equal to v ) that we need to move to reach P: for example, t = 0 corresponds to P = P0 and t = −2 corresponds to a point that is at a distance 2 v from P0 in the direction opposite to v. P tv P0 r r0 O Figure 3.1. The parametric vector equation of a line Writing v and r in component form, v = a, b, c and r = x, y, z , we can express (3.1) in terms of the components of the vectors involved: x, y, z = x0 , y0 , z0 + t a, b, c = x0 + ta, y0 + tb, z0 + tc , or equivalently: x = x0 + ta, y = y0 + tb, z = z0 + tc (t ∈ R). (3.2) These are the parametric scalar equations of . In general, the vector v = a, b, c is determined up to multiplication by a nonzero scalar, that is, we can use any nonzero scalar multiple of v in place of v. This will change the coeﬃcients in equations (3.1) and (3.2), but not the underlying set
25

of points. We shall refer to the coordinates a, b, c of any vector v parallel to as direction numbers of . Another common way to specify a line in space are its symmetric equations. Given the scalar equations (3.2), with abc 0, we can eliminate t to obtain the equations x − x0 y − y0 z − z0 = = . (3.3) a b c These are the symmetric equations of . If one of a, b, or c is zero—say, b = 0, we obtain the symmetric equations of by keeping that scalar equation and eliminating t from the other two: x − x0 z − z0 = . y = y0 , a c If two among a, b, c are equal to zero, the corresponding scalar equations represent the symmetric equations of after discarding the third parametric equation: if a = b = 0, the symmetric equations of are x = x0 , y = y0 . Finally, we can describe a line by the coordinates of any two distinct points P1 (x1 , y1 , z1 ) and −− −→ P2 (x2 , y2 , z2 ) on the line. If P1 and P2 are such points, the vector P1 P2 = x2 − x1 , y2 − y1 , z2 − z1 is parallel to the line, so we can write the symmetric equations (3.3) in the form y − y1 z − z1 x − x1 = = , (3.4) x2 − x1 y2 − y1 z2 − z1 (or something like it, for other types of symmetric equations). Example 3.1. Find the equations of the line through the points P(1, 0, 3) and Q(2, −1, 4). − − → Solution. To obtain the symmetric equations, we note that PQ = 1, −1, 1 is a vector parallel to , so (3.3) can be written as x−1 y−0 z−3 = = ⇐⇒ x − 1 = −y = z − 3. 1 −1 1 − − → The parametric scalar equations (based again on PQ) are x = 1 + t, y = 0 + t(−1) = −t, z=3+t (t ∈ R), (t ∈ R). and the parametric vector equation is r(t) = 1, 0, 3 + t 1, −1, 1 = 1 + t, −t, 3 + t

Example 3.2. Find a parametric vector equation of the line passing through P(−1, 2, −3) and parallel to the line 2(x + 1) = 4(y − 3) = z. Solution. We can write the equations of the second line as x+1 y−3 = 1 = z, 1
2 1 1 , ,1 2 4 4

so v = is a vector parallel to the second line. Thus, v is also parallel to . That is, passes through P and is parallel to v. Its vector equation then is r(t) = −1, 2, −3 + t
1 1 , ,1 2 4 1 = −1 + 1 t, 2 + 4 t, −3 + t 2

(t ∈ R).

26

Example 3.3. Find the point of intersection and the angle between the lines with the vector parametrizations r1 (t) = 1 + t, −1 − t, −4 + 2t and r2 (u) = 1 − u, 1 + 3u, 2u .

Solution. If it exists, the point of intersection of the given lines will have coordinates (x, y, z) such that x = 1 + t = 1 − u, y = −1 − t = 1 + 3u, z = −4 + 2t = 2u, for some t, u. From the equations for x, we ﬁnd that t = −u. Hence, the other two equations between t and u become −1 + u = 1 + 3u, −4 − 2u = 2u =⇒ u = −1, t = −u = 1. Therefore, the point of intersection is (2, −2, −2). To ﬁnd the angle between the lines, we ﬁnd the angle θ between their parallel vectors: v = 1, −1, 2 for the ﬁrst line and u = −1, 3, 2 for the second. Thus, by Corollary 2.4, u·v −1 − 3 + 4 cos θ = = √ √ =0 =⇒ θ = 90◦ . u v 6 14 3.2. Equations of a plane We shall specify a plane in space using a point on the plane and a direction perpendicular to the plane. If r0 is the position vector of a point P0 (x0 , y0 , z0 ) on the plane and n = a, b, c is a vector perpendicular (also called normal) to the plane, then a generic point P(x, y, z) is on the plane if and −→ − only if P0 P is perpendicular to n (see Fig. 3.2). Introducing the position vector r of P, we can express this condition as −→ − P0 P · n = 0 ⇐⇒ (r − r0 ) · n = 0, (3.5) or equivalently, n · r = n · r0 . (3.6) These are the vector equations of the given plane. z P0 r − r0 P x O y n

Figure 3.2. The vector equation of a plane If we express the dot product in (3.5) in terms of the components of the vectors, we get a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0,
27

(3.7)

or the scalar equation of the given plane. Given the last equation, we can easily obtain the plane’s linear equation: ax + by + cz + d = 0, (3.8) simply by writing d = −ax0 − by0 − cz0 . Example 3.4. Find the equation of the plane that contains P(1, −2, 3) and is perpendicular to e2 + 2e3 . Solution. The scalar equation of this plane is 0(x − 1) + 1(y + 2) + 2(z − 3) = 0 ⇐⇒ y + 2z − 4 = 0. Example 3.5. Find the equation of the plane π that contains P(3, −1, 5) and is parallel to the plane with equation 4x + 2y − 7z + 5 = 0. Solution. A normal vector for the given plane is n = 4, 2, −7 . Hence, the scalar equation of π, which also has n as its normal vector, is 4(x − 3) + 2(y + 1) − 7(z − 5) = 0 ⇐⇒ 4x + 2y − 7z + 25 = 0.

Example 3.6. Find the equation of the plane that contains the points P(1, 2, 3), Q(−1, 3, 2), and R(3, −1, 2). − − → − − → Solution. The cross product PQ × PR is a normal vector for this plane. By Example 2.8, − − → − − → PQ × PR = −4e1 − 4e2 + 4e3 , so we can use the normal vector n = 1, 1, −1 . We obtain the scalar equation (x − 1) + (y − 2) − (z − 3) = 0 ⇐⇒ x + y − z = 0. Example 3.7. Find the angle between the planes 2x − y + 3z = 5 and 5x + 5y − z = 1. Solution. The angle between the two planes equals the angle between their normal vectors: n1 = 2, −1, 3 and n2 = 5, 5, −1 , respectively. If θ is the angle between n1 and n2 , we use the dot product to ﬁnd n1 · n2 10 − 5 − 3 2 cos θ = = √ √ = √ . n1 n2 714 14 51 Example 3.8. Find the line of intersection of the planes x + y + z + 1 = 0 and x − y + z + 2 = 0. Solution. We shall compute the symmetric equations of the given line. First, by adding the two equations, we obtain the equation z + 3/2 . 2x + 2z + 3 = 0 ⇐⇒ x = −z − 3/2 ⇐⇒ x = −1 On the other hand, subtracting the two given equations, we get 2y − 1 = 0 ⇐⇒ y = 1/2. Hence, the symmetric equations of the line of intersection are z + 3/2 x= , y = 1/2. −1
28

Note that these symmetric equations lead to the parametric scalar equations x = t, y = 1/2, z = −3/2 − t (t ∈ R).

Example 3.9. Find the distance from P(3, −5, 2) to the plane x − 2y + z = 5. First solution. The distance from P to the given plane is equal to the distance |PQ|, where Q is the intersection point of the plane and the line through P that is perpendicular to the plane. The normal vector to the plane n = 1, −2, 1 is parallel to the line PQ, so the symmetric equations of that line are x−3 y+5 z−2 = = . 1 −2 1 Combining these equations with the equation of the plane, we obtain a linear system x−3= y+5 , −2 z−2= y+5 , −2 x − 2y + z = 5,

1 whose solution ( 4 , − 5 , 3 ) is the point Q. Therefore, 3 3

|PQ| = (− 5 )2 + ( 10 )2 + (− 5 )2 3 3 3

1/2

=

5 3

√ 6.

We now give a second solution, which is somewhat trickier, but has the advantage that can be easily generalized (see the next example). Second solution. Let Q(u, v, w) be the same point as before. In this solution, we avoid the explicit computation of u, v, w. − − → − − → We have PQ = u − 3, v + 5, w − 2 . By the deﬁnition of Q, we know that PQ is normal to the − − → given plane, so PQ is parallel to the normal vector n = 1, −2, 1 . Hence, by the properties of the dot product, √ − − → − − → n · PQ = n |PQ| cos (n, PQ) = 6 |PQ|, since the angle between the two vectors is 0◦ or 180◦ . On the other hand, using the deﬁnition of the dot product, we have − − → n · PQ = |(u − 3) − 2(v + 5) + (w − 2)| = |u − 2v + w − 15| = |5 − 15| = 10. Here, we have used that u − 2v + w = 5, because Q lies on the given plane. Comparing the two √ − − → expressions for n · PQ , we conclude that |PQ| = 10/ 6. Example 3.10. Find the distance from P(x0 , y0 , z0 ) to the plane ax + by + cz + d = 0. Solution*. We essentially repeat the second solution of the previous example. Let Q(u, v, w) be the intersection point of the plane and the line through P that is perpendicular to the plane. Then − − → − − → PQ = u − x0 , v − y0 , w − z0 . Since PQ is normal to the given plane, it is parallel to its normal vector n = a, b, c . Hence, by the properties of the dot product, √ − − → − − → n · PQ = n |PQ| cos (n, PQ) = a2 + b2 + c2 |PQ|,
29

since the angle between the two vectors is 0◦ or 180◦ . On the other hand, using the deﬁnition of the dot product, we have − − → n · PQ = |a(u − x0 ) + b(v − y0 ) + c(w − z0 )| = |au + bv + cw − ax0 − by0 − cz0 | = | − d − ax0 − by0 − cz0 | = |ax0 + by0 + cz0 + d|. Here, we have used that au + bv + cw = −d, because Q lies on the given plane. Comparing the two − − → expressions for n · PQ , we conclude that the desired distance is |PQ| = |ax0 + by0 + cz0 + d| √ . a2 + b2 + c2

Exercises
3.1. Find the equations of the line that passes through the point (1, −2, 3) and is perpendicular to the plane with equation −2x + y − z + 4 = 0. 3.2. Find the equations of the line that passes through the points (0, −2, 3) and (1, 1, 5). 3.3. Let 1 be the line with equations x = −1 + 2t, y = 3t, z = 2 and let 2 be the line through the points (4, 3, 2) and (6, −1, 4). Determine whether these two lines are parallel, intersecting, or skew. If they intersect, ﬁnd their intersection point. x−3 y+1 = = z and the coordinate planes. 2 5 3.5. Find the angle between the lines whose respective equations are x+1 y+2 z−2 x+1 y+2 z−2 = = and = = . −3 2 5 2 1 −2 3.6. Explain why the two lines from Exercise 3.5 intersect. 3.4. Find the intersection points of the line with equations 3.7. Find an equation of the plane that passes through the point (1, −2, 3) and is perpendicular to the vector 3, 0, −2 . 3.8. Find an equation of the plane that contains the line with equations plane 3x − 2y = 5. 3.9. Find an equation of the plane that contains the line with equations line x = 1 − 2t, y = 4t, z = −3 + 2t. y+2 z−2 x−1 = = and is parallel to the 2 3 4

x−1 y+2 z = = and is perpendicular to the 2 −1 4 x−1 y+2 z = = and is perpendicular to 2 3 4

3.10. Find an equation of the plane that contains the line with equations the line x = 1 − 2t, y = 3t, z = −3 + 2t.

3.11. Find an equation of the plane through the points (1, −1, 0), (2, 0, 3), and (−2, 2, 1). 3.12. Find an equation of the plane through the points (1, 2, −1), (−1, 0, 1), and (2, 3, 0). 3.13. Find the equations of the intersection line of the planes x + y − z = 2 and 2x − 3y + z = 6. 3.14. Find an equation of the plane that contains both the line the line 2 with equations x = −2 − 3t, y = t, z = 4 + 4t. 3.15. Find an equation of the plane that contains both the line the line 2 with equations x = 1 + t, y = 2 − 3t, z = −1.
30
1

with equations x = 1 + 2t, y = −1 + 3t, z = −2t and with equations x = 1 − 2t, y = 2 + t, z = −1 − t and

1

Find the angle between the given planes. 3.16. 2x − y + 3z = 0, x + 2y + z = 4 3.17. x + y + z = 4, x + 2y − 3z = 6 3.18. The planes 2x − 2y − z = 1 and 6x − 6y − 3z = −5 are parallel. Find the distance between them.

31

LECTURE 4

Vector Functions
4.1. Deﬁnition A vector-valued function (or a vector function) is a function whose domain is a set of real numbers and whose range is a set of vectors. In general, a vector function has the form f(t) = f1 (t), f2 (t), f3 (t) = f1 (t)e1 + f2 (t)e2 + f3 (t)e3 . where f1 , f2 , and f3 are functions in the regular sense (functions for which both the inputs and outputs are real numbers). The functions f1 , f2 , and f3 are sometimes called the component functions of the vector function f. When we want to distinguish that the values of a function f are numbers (as opposed to vectors), we will refer to f as a scalar function. Example 4.1. As we know from Lecture #3, the parametric equation of the line in space passing through (1, 0, −5) and parallel to the vector 1, 2, 1 is, in fact, a vector function: r(t) = 1 + t, 2t, −5 + t (t ∈ R). The line is the set of values of this function, not the graph (the graph is four-dimensional). Here is another example of a vector function: f(t) = cos t, (t − 5)−1 , ln t ; its domain is (0, 5) ∪ (5, ∞). 4.2. Calculus of vector functions: Limits Vector functions can have limits, derivatives, antiderivatives, and deﬁnite integrals—just like scalar functions. As we will see, all of these can be computed componentwise. Because of that, once one is proﬁcient in the calculus of scalar-valued functions, one can easily master the calculus of vector functions. We begin with limits and continuity. The limit of a vector function is a vector. Definition. If f(t) = f1 (t), f2 (t), f3 (t) is deﬁned near a, then the limit of f as t → a is deﬁned by lim f(t) = lim f1 (t), lim f2 (t), lim f3 (t) , t→a t→a t→a t→a

provided that the limits of all three component functions exist. Definition. A vector function f(t) = f1 (t), f2 (t), f3 (t) , deﬁned near a, is continuous at a if lim f(t) = f(a). t→a Example 4.2. We have lim cos t, (t − 5)−1 , ln t = lim cos t, lim(t − 5)−1 , lim ln t = cos 7, 1 , ln 7 . 2 t→7 t→7 t→7 t→7

33

4.3. Calculus of vector functions: Derivatives Definition. If f(t) = f1 (t), f2 (t), f3 (t) is continuous at a, then its derivative f (a) at a is deﬁned f(a + h) − f(a) f(t) − f(a) = lim , h→0 t−a h provided that the limit on the right exists. If f (a) exists, we say that f is diﬀerentiable at a. f (a) = lim t→a by

Theorem 4.1. A vector function f(t) = f1 (t), f2 (t), f3 (t) is diﬀerentiable at a if and only if all three component functions f1 , f2 , f3 are. When f (a) exists, we have f (a) = f1 (a), f2 (a), f3 (a) . In other words, the derivative of a vector function can be computed using componentwise diﬀerentiation. Proof. We have f (a) = lim f(t) − f(a) t→a t−a f1 (t) − f1 (a), f2 (t) − f2 (a), f3 (t) − f3 (a) = lim t→a t−a f1 (t) − f1 (a) f2 (t) − f2 (a) f3 (t) − f3 (a) , , = lim t→a t−a t−a t−a = f1 (a), f2 (a), f3 (a) . f (t) = − sin t, −(t − 5)−2 , t−1 . Theorem 4.2. Let f and g be diﬀerentiable vector functions, let u be a diﬀerentiable scalar function, and let c be a scalar. Then the following formulas hold: i) f(t) + g(t) = f (t) + g (t); ii) cf(t) = cf (t); iii) u(t)f(t) = u(t)f (t) + u (t)f(t); iv) f(t) · g(t) = f(t) · g (t) + f (t) · g(t); v) f(t) × g(t) = f(t) × g (t) + f (t) × g(t); vi) f(u(t)) = f (u(t))u (t). Proof. We shall prove iv). If f(t) = f1 (t), f2 (t), f3 (t) and g(t) = g1 (t), g2 (t), g3 (t) , then f(t) · g(t) = f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t). Hence, the left side of iv) equals f(t) · g(t) = f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t) = f1 (t)g1 (t) + f1 (t)g1 (t) + f2 (t)g2 (t) + f2 (t)g2 (t) + f3 (t)g3 (t) + f3 (t)g3 (t). On the other hand, we have f (t) · g(t) = f1 (t), f2 (t), f3 (t) · g1 (t), g2 (t), g3 (t) = f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t), f(t) · g (t) = f1 (t), f2 (t), f3 (t) · g1 (t), g2 (t), g3 (t) = f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t),
34

Example 4.3. If f(t) = cos t, (t − 5)−1 , ln t , then

so the right side of iv) equals f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t) + f1 (t)g1 (t) + f2 (t)g2 (t) + f3 (t)g3 (t). This establishes iv). 4.4. Calculus of vector functions: Integrals The deﬁnition of the deﬁnite integral of a continuous function via Riemann sums (see Appendix B) can also be extended to vector functions. Here, we will adopt an equivalent deﬁnition, which is a little less transparent but considerably easier to use. Definition. If f(t) = f1 (t), f2 (t), f3 (t) is continuous in [a, b], then its deﬁnite integral is deﬁned by b b b b b a

f(t) dt

f(t) dt = a a

f1 (t) dt, a f2 (t) dt, a f3 (t) dt .

Remark. Note that

b a

f(t) dt is a vector (not a scalar and not a vector function).

Example 4.4. If f(t) = cos t, (t − 5)−1 , ln t , then
2 2 2 2

f(t) dt =
1 1

cos t dt,
1 2

(t − 5)−1 dt,
1 2 2 1

ln t dt

=

sin t 1 , ln |t − 5| 1 , t ln t − t

= sin 2 − sin 1, ln( 3 ), 2 ln 2 − 1 . 4 Theorem 4.3. Let f and g be continuous vector functions, let u be a vector, and let c be a scalar. Then the following formulas hold: i) ii) iii) iv) v) b a b a b a b a b a

f(t) + g(t) dt = a f(t) dt + b cf(t) dt = c a f(t) dt; b u · f(t) dt = u · a f(t) dt; b u × f(t) dt = u × a f(t) dt; f (t) dt = f(b) − f(a).

b

b a

g(t) dt;

Proof. We shall prove iv). Both sides of the identity are three-dimensional vectors, so it suﬃces to show that their respective components are equal. If u = u1 , u2 , u3 and f(t) = f1 (t), f2 (t), f3 (t) , then u × f(t) = u2 f3 (t) − u3 f2 (t), u3 f1 (t) − u1 f3 (t), u1 f2 (t) − u2 f1 (t) . Thus, the ﬁrst component of the left side of iv) is b b b

u2 f3 (t) − u3 f2 (t) dt = u2 a a

f3 (t) dt − u3 a f2 (t) dt.

Since the right side of this identity is also the ﬁrst component of the right side of iv), this shows that the ﬁrst components of the two sides of iv) match. The proof that the remaining two components of the two sides of iv) match is similar.
35

4.5. Vector functions in Mathematica Deﬁning and operating with vector functions in Mathematica is no diﬀerent from deﬁning and operating with scalar functions. For example, to deﬁne the function f(t) = cos t, (t − 5)−1 , ln t from Examples 4.2–4.4, we simply use the command f[t_] := {Cos[t], 1/(t-5), Log[t]} We can then use the commands Limit, D, Integrate and NIntegrate on the vector function f[t] just as we use them on scalar-valued functions: In[1]:= Limit[f[t], t->1] Out[1]:= {Cos[1], -(1/4), 0} In[2]:= f’[t] Out[2]:= {-Sin[t], -(1/(-5+t)ˆ2), 1/t} In[3]:= D[f[2t], t] Out[3]:= {-2Sin[2t], -(2/(-5+2t)ˆ2), 1/t} In[4]:= NIntegrate[f[t], {t,1,4}] Out[4]:= {-1.59827, -1.38629, 2.54518} In[5]:= Integrate[f[t], {t,0,1}] Out[5]:= {Sin[1], -Log[5/4], -1} Exercises
4.1. Evaluate lim t→a ln(t+1) sin 2t t t , t ,e

, where:

(a) a = 1;

(b) a = 0.

Find the derivative of the given vector function. 4.2. f(t) = ln t2 + 1 , arcsin 2t, et 2 2 4.3. f(t) = 5t3 + 2t, et sin t, et cos t 3 2t 4.4. f(t) = t√e1 − 3e e3 4.5. f(t) = t2 + 1e1 − cos(2t)e2 + sin(2t)e3 Evaluate the given integral. −1 −1 1 4.6. 0 t2 + 4 e1 + t t2 − 2 e3 dt 1 4.7. −1 5t3 + 2t, e2t , sin t cos4 t dt 4.8. Find f(t), if f (t) = 2te1 + et e2 + (ln t) e3 and f(1) = e1 + e2 .

36

LECTURE 5

Space Curves
Vector functions are a great tool for dealing with curves, both in the plane and in space. In this lecture, we use vector functions to give a uniﬁed treatment of plane curves and space curves. 5.1. Review of plane curves Recall that a curve γ in the plane is the set γ = (x, y) : x = x(t), y = y(t), t ∈ I , where I is some interval in R (ﬁnite or inﬁnite; open, closed, or semiopen) and x(t) and y(t) are continuous functions of t. We refer to the pair of equations x = x(t), y = y(t) (t ∈ I) (5.1) as parametric equations or a parametrization of γ. A curve γ has many parametrizations, and sometimes replacing one parametrization by another can change a particular property of the curve, even if the set of points stays the same. Thus, when we want to distinguish a particular parametrization, we talk about the parametric curve γ with equations (5.1). 5.2. Deﬁnition of space curve Next, we extend the above deﬁnitions to three dimensions. Definition. If x(t), y(t), z(t) are continuous functions deﬁned on an interval I, then the set γ = (x, y, z) : x = x(t), y = y(t), z = z(t), t ∈ I is called a space curve. The equations x = x(t), y = y(t), z = z(t) (t ∈ I) (5.2) are called parametric equations or a parametrization of γ; t is the parameter. Alternatively, we can think of the curve γ with equations (5.2) as the range of the vector function r(t) = x(t)e1 + y(t)e2 + z(t)e3 (t ∈ I). (5.3) We shall refer to (5.3) as the parametric (vector) equation of γ. Note that the plane curve (5.1) can be treated as a special case of a space curve given by the parametric equation r(t) = x(t)e1 + y(t)e2 + 0e3 (t ∈ I). A good way to visualize the connection between a space curve γ and its parametrization r(t) is to think of the vector r(t) as a position vector which moves as t varies. The terminal point of r(t) then traces the points on the curve (see Figure 5.1(a), or execute the Mathematica code on Figure 5.1(b) for an interactive demonstration).
37

z γ r(t) f[t_] := {t*Cos[t], t*Sin[t], t-1} Manipulate[ Show[{ ParametricPlot3D[f[t], {t, 0, 2Pi}], Graphics3D[{Red, Arrow[{{0,0,0}, f[t]}]}] }], {t, 0, 2Pi}] (b) Figure 5.1. Vector functions and curves in space Example 5.1. Consider the space curve γ with parametric equation r(t) = (cos t)e1 + (sin t)e2 + te3
2 2

x

O (a)

y

(0 ≤ t < ∞).

The point (cos t, sin t, 0) traces the unit circle x + y = 1 in the xy-plane in the counterclockwise direction, so the point (cos t, sin t, t) on γ moves in the positive direction of the z-axis along a spiral lying above that circle. This curve is known as the helix; it is shown on Figure 5.2. z z

x

O Figure 5.2. The helix

y x

O

y

Figure 5.3. r(t) = t cos t, t sin t, t

Example 5.2. Consider the space curve γ with parametric equation r(t) = (t cos t)e1 + (t sin t)e2 + te3 (t cos t)2 + (t sin t)2 = t2 , so the x, y, and z-coordinates of a point on γ always satisfy the equation x2 + y2 = z2 . Thus, γ is an inﬁnite spiral that starts at the origin and unwinds upward and outward; it is shown on Figure 5.3. Example 5.3. Given two points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ), the space curve r(t) = (1 − t) x1 , y1 , z1 + t x2 , y2 , z2
38

(0 ≤ t < ∞).

The initial point of the curve is the origin. For any given t, we have

(0 ≤ t ≤ 1)

is the line segment with endpoints P1 and P2 . Indeed, we can express r(t) in the form r(t) = x1 + t(x2 − x1 ), y1 + t(y2 − y1 ), z1 + t(z2 − z1 ) , −− −→ which is the parametric equation of the line through P1 that is parallel to P1 P2 . Thus, if the domain for t was the whole real line R, the curve would be the straight line through P1 and P2 . Restricting the domain for t to the closed interval [0, 1], we obtain the piece of the line between the endpoints of r(0) = x1 , y1 , z1 and r(1) = x2 , y2 , z2 ; that is precisely the line segment P1 P2 . 5.3. Smooth curves At ﬁrst glance, continuous curves appear to be a good model for the geometric objects that we think of as “curves”: that is, two- or three-dimensional objects that are essentially one-dimensional. However, it turns out that there are some rather peculiar continuous parametric curves. For example, in 1890 the Italian mathematician Giuseppe Peano stunned the mathematical community with his construction of a continuous plane curve that ﬁlls the entire unit square (see §5.7); there are also examples of continuous space curves that ﬁll an entire cube. In order to rule out such pathological curves, we shall restrict our attention to a special class of parametric curves called “piecewise smooth”. Definition. A curve γ given by a vector function r(t), t ∈ I, is called smooth if r (t) is continuous and r (t) 0 (except possibly at the endpoints of I). A curve γ consisting of a ﬁnite number of smooth pieces is called piecewise smooth. Example 5.4. For the helix r(t) = (cos t)e1 + (sin t)e2 + te3 , t ≥ 0, we have r (t) = (− sin t)e1 + (cos t)e2 + e3 Hence, the helix is a smooth curve. Example 5.5. Determine whether the curve γ deﬁned by the vector function r(t) = t2 e1 + t3 e2 + (cos t)e3 is smooth, piecewise smooth, or neither. Solution. The derivative r (t) = 2te1 + 3t2 e2 − sin te3 vanishes when t = 0. Thus, γ is not smooth. On the other hand, the curves γ1 : and γ2 : r(t) = t2 e1 + t3 e2 + (cos t)e3 (t ≤ 0) r(t) = t2 e1 + t3 e2 + (cos t)e3 (t ≥ 0) (t ∈ R) 0.

are smooth, since they fail the condition r (t) 0 only at an endpoint of the intervals on which they are deﬁned. Since γ is the union of γ1 and γ2 , it follows that γ is piecewise smooth.
39

z

P r (t) y

x

O

Figure 5.4. The tangent vector r (t) at P = r(t) 5.4. Tangent lines The requirement that r (t) 0 for a smooth curve is not arbitrary. To explain where it comes from, we now consider the following problem. Problem. Let γ be a smooth parametric curve with parametric equation r(t) = x(t)e1 + y(t)e2 + z(t)e3 (t ∈ I). Find an equation of the tangent line to γ at the point r(t0 ) = (x0 , y0 , z0 ), where t0 is an internal point of the interval I. Idea of solution. The tangent line will have the parametric equations x = x0 + at, y = y0 + bt, z = z0 + ct (t ∈ R), where u = a, b, c is any vector parallel to the tangent line. In other words, we need to determine the direction of the tangent line. Recall that the tangent line through P(x0 , y0 , z0 ) is the limit of the secant lines through P and a point Q(x, y, z) on the curve as Q → P. If Q = r(t), then a vector parallel to the secant line through P and Q is − − → PQ = x − x0 , y − y0 , z − z0 = r(t) − r(t0 ). Thus, direction of tangent line at P = lim direction of secant line through P, Q t→t0 = lim direction of r(t) − r(t0 ) t→t0 = lim direction of normalized r(t) − r(t0 ) t→t0 = lim t→t0 r(t) − r(t0 ) /(t − t0 ) r(t) − r(t0 ) = lim . t→t0 r(t) − r(t0 ) /(t − t0 ) r(t) − r(t0 )

By the deﬁnition of derivative, the limit of the numerator is r (a). By a slightly more complicated argument (which we omit), the limit of the denominator is r (a) . Thus, a vector parallel to the tangent line is r (t0 )/ r (t0 ) .
40

We now see the reason for the requirement of non-vanishing of the derivative in the deﬁnition of smoothness: with that requirement, a smooth curve has a tangent line at every point. Furthermore, the vector r (t) is parallel to the tangent line to the curve at the endpoint P of r(t). For this reason, r (t) is called also the tangent vector to the curve r(t) (see Figure 5.4). The unit tangent vector to r(t) is merely the normalization of r (t), and hence, is given by the formula T(t) = Example 5.6. For the helix, we have r (t) = (− sin t)e1 + (cos t)e2 + e3 , so the unit tangent vector is T(t) = r (t) = r (t) − sin t, cos t, 1 (− sin t)2 + (cos t)2 + 12 = − sin t cos t 1 √ , √ , √ 2 2 2 . r (t) . r (t)

Example 5.7. Find the equation of the tangent line to the helix at the point P corresponding to the value of the parameter t = π. Solution. We have P(−1, 0, π). The tangent line passes through P and is parallel to the tangent vector r (π) = 0, −1, 1 (see Example 5.6). Hence, its parametric equation is r(t) = −1, 0, π + t 0, −1, 1 = −1 + t, −t, π + t z−π y = . the symmetric equations of the tangent line are x = −1, −1 1 (t ∈ R);

Example 5.8. Find the angle between the helix and the curve γ deﬁned by the vector function r2 (t) = −te1 + (t − 1)2 e2 + πte3 at P(−1, 0, π). Solution. Let σ denote the helix. If P lies on both γ and σ, the angle between γ and σ at P is the angle between their tangent vectors. It is easy to see that both curves do pass through P: r1 (π) = r2 (1) = −1, 0, π . We know from Example 5.7 that the tangent vector to the helix at P is v1 = 0, −1, 1 . The tangent vector to γ at P is v2 = r2 (1) = −1, 2(t − 1), π Hence, if α is the angle between γ and σ at P, we have cos α = Thus, α = arccos √ π 2 + 2π2
41
t=1

(t ≥ 0)

= −1, 0, π .

v1 · v2 = v1 v2

0 · (−1) + (−1) · 0 + 1 · π 02 + (−1)2 + 12

π = √ . (−1)2 + 02 + π2 2 + 2π2

≈ 0.8315.

5.5. Normal vectors* We now want to ﬁnd vectors that are perpendicular to the curve r(t), that is, we want to ﬁnd vectors orthogonal to the tangent vector r (t), or equivalently, to the unit tangent vector T(t). One such vector is the unit normal vector T (t) . N(t) = T (t) In order to prove that N(t) and T(t) are orthogonal, we diﬀerentiate both sides of the identity T(t) · T(t) = T(t) We obtain T (t) · T(t) + T(t) · T (t) = 0 =⇒ 2T(t) · T (t) = 0 =⇒ T(t) · T (t) = 0. That is, T (t) is orthogonal to T(t). Another vector orthogonal to T(t) is the unit binormal vector B(t) = T(t) × N(t). By the properties of the cross product, B(t) is perpendicular to both T(t) and N(t) and has length equal to the area of the parallelogram with sides T(t) and N(t). Since this parallelogram is a square of side length 1, B(t) is a unit vector. Remark. Note that it is important to use T(t) and not r (t) when we compute the normal vector. In fact, r (t) need not be orthogonal to r (t), and in most cases it will not be. For example, if r(t) = 1, t, t2 , we have r (t) = 0, 1, 2t and r (t) = 0, 0, 2 , so r (t) · r (t) = 4t On the other hand, T(t) = 0, (1 + 4t2 )−1/2 , 2t(1 + 4t2 )−1/2 , and these are orthogonal. 5.6. Arc length You should be familiar with the formula for arc length of a parametric plane curve: if γ is a parametric curve in the xy-plane with parametric equations (5.1), where I = [a, b], then its length L(γ) is given by the formula b 2

= 1.

0

unless t = 0.

T (t) = 0, −4t(1 + 4t2 )−3/2 , 2(1 + 4t2 )−3/2 ,

L(γ) = a x (t)2 + y (t)2 dt.

(5.4)

In this section, we generalize this formula to space curves. We argue from scratch, so even if (5.4) appears new, this should not be an obstacle. First, we must say what we mean by “length” of a curve. We shall focus on piecewise smooth curves given by parametric equations of the form (5.3). The intuitive idea of length is that we can take the curve, “straighten” it into a line segment, and then measure the length of that line segment. Now, imagine that we live in a “rigid” world, where we cannot straighten our curve (or bend a thread so it would lay on top of the curve). We could try to cut a number of short straight line segments and piece them together so that the resulting polygon lies close to our original curve; we can then measure the total length of the polygon and argue that the length of the curve is about the same. This is the basic idea behind our deﬁnition of arc length.
42

z

x

O

y

Figure 5.5. Polygonal approximation to a curve by a 13-gon Let γ be a parametric curve given by (5.3) with I = [a, b]. For each n ≥ 1, set ∆n = (b − a)/n and deﬁne the numbers that partition [a, b] into n subintervals of equal lengths: t0 = a, t1 = a + ∆ n , t2 = a + 2∆n , ..., tn = a + n∆n = b. We write xi = x(ti ), yi = y(ti ), zi = z(ti ) for the coordinates of the point Pi (xi , yi , zi ) that corresponds to the value t = ti of the parameter. The points P0 , P1 , P2 , . . . , Pn partition γ into n arcs (see Figure 5.5). Let Ln be the length of the polygon with vertices at those points, that is, n n

Ln = i=1 |Pi−1 Pi | = i=1 (xi − xi−1 )2 + (yi − yi−1 )2 + (zi − zi−1 )2 .

If the sequence converges and lim Ln = L, we call L the arc length of the given curve. n→∞ Using the above deﬁnition and the properties of smooth functions and of deﬁnite integrals, we can prove the following formula for the length of a piecewise smooth curve. {Ln }∞ n=1 Theorem 5.1. Let γ be a piecewise smooth parametric curve with equation r(t) = x(t)e1 + y(t)e2 + z(t)e3 Then the arc length of γ is b b

(t ∈ [a, b]).

L(γ) = a x

(t)2

+y

(t)2

+z

(t)2

dt = a r (t) dt.

(5.5)

Remarks. 1. Note that (5.4) is a special case of (5.5): namely, the case when z(t) = 0. 2. Note that (5.5) gives the length of the parametric curve and not the length of the graph. For example, the parametric equation r(t) = cos te1 + sin te2 + 0e3 (0 ≤ t ≤ 4π) represents the unit circle in the xy-plane. If we apply (5.5) to this curve, we will ﬁnd that its length is 4π (and not 2π). The explanation for this is that the above curve traces the unit circle twice. Example 5.9. If f is continuous on [a, b], the length of the curve y = f (x) is b L= a 1 + f (x)2 dx.
43

This formula follows from the theorem by representing the given graph as a parametric curve with parametrization r(t) = te1 + f (t)e2 + 0e3 , a ≤ t ≤ b. Example 5.10. Find the length of the curve γ given by r(t) = 0 ≤ t ≤ 1. Solution. We have r (t) = 3t1/2 e1 − 2 sin 2te2 + 2 cos 2te3 , so (5.5) yields
1 1

2t3/2 , cos 2t, sin 2t , where

L(γ) =
0 1

r (t) dt =
0

3t1/2

2

+ (−2 sin 2t)2 + (2 cos 2t)2 dt

=

1 √ 9t + 4 sin2 2t + 4 cos2 2t dt = 9t + 4 dt 0 0 √ 2 13 13 − 8 1 13 1/2 u du = ≈ 2.879. = 9 4 27

5.7. Space-ﬁlling curves* In this section, we describe the general idea of the construction of Peano’s space-ﬁlling curve. We start with a piecewise linear curve contained in the unit square. This curve is then shrunk to a fraction of its original size and several copies of it are placed throughout the unit square in such a way that they do not intersect and can be easily connected one to another with straight lines. This yields a more complicated polygon that also lies within the unit square. The whole process (shrinking, copying, connecting) is then repeated iteratively. The ﬁrst two iterations of Peano’s original constructions are displayed on Figure 5.6. The result of the iterative process is a sequence of curves. It can be proved that the sequence of continuous functions underlying this sequence of curves has a limit, which is also a continuous function. It should be intuitively clear (and this can be proved rigorously) that the limit curve ﬁlls the entire square.

Figure 5.6. The ﬁrst two iterations in Peano’s construction
44

5.8. Space curves in Mathematica Mathematica can be quite useful when trying to visualize space curves. Mathematica’s main built-in command for plotting a space curve is ParametricPlot3D. Its basic syntax is ParametricPlot3D[f[t], {t,a,b}] where f is a three-dimensional vector function and a and b are two numerical expressions describing the endpoints of the interval for the parameter t. The function f can be given by its name as above, or explicitly, by a vector whose components are functions of t. The command ParametricPlot3D has many additional options which allow the user to format its output. The best source to learn about all of those is Mathematica’s help system. Here, we include only a few examples illustrating the basic syntax and some of the most common options: • To plot three loops of a helix, having already deﬁned the function r(t) = cos t, sin t, t/6 : r[t_] := {Cos[t], Sin[t], t/6} ParametricPlot3D[r[t], {t,0,6Pi}] • To plot two thick, black loops of a helix, using the explicit deﬁnition: ParametricPlot3D[{Cos[t], Sin[t], t/6}, {t,0,4Pi}, PlotStyle->{Black,Thick}] • To plot the spiral on Figure 5.3, without a coordinate frame, thick and in red: ParametricPlot3D[{t*Cos[t], t*Sin[t], t}, {t,0,8.5Pi}, PlotStyle->{Red,Thick}, Axes->False, Boxed->False] Exercises
5.1. Does the point P(0, π, 1) lie on the curve parametrized by the given vector function r(t)? If so, for what value(s) of t? (a) r(t) = sin t, 2 arccos t, t + 1 (b) r(t) = sin t, t, cos 2t (c) r(t) = te2 + tan te3 5.2. Do the curves r1 (t) = t, t2 , t3 and r2 (t) = t − 2, t2 − 3t + 2, t2 + t − 6 intersect? 5.3. Let two objects move in space along trajectories described by the vector functions r1 (t) = t, t2 , t3 and r2 (t) = t − 2, t2 − 3t + 2, t2 + t − 6 . Will the objects ever collide? That is, will they ever be at the same point in space at the same time? 5.4. Find the unit tangent vector T(t) to the curve r(t) = sin 3t, cos 3t, ln(t + 1) at the point with t = 0. 5.5. Find the equations of the tangent line to the curve r(t) = cos(πt), 3 sin(πt), t3 at the point (−1, 0, 1). 5.6. Find the equations of the tangent line to the curve r(t) = t3 − 2, t2 + 1, 3t + 1 at the point (−2, 1, 1). 5.7. Find the equations of the tangent line to the curve r(t) = 5 cos t, 5 sin t, 2 cos 2t at the point (−5, 0, 2). 5.8. Find the points on the curve r(t) = 5 cos t, 5 sin t, 2 cos 2t where the tangent line is parallel to the x-axis. Is the given curve smooth? If not, then is it piecewise smooth? 5.9. r(t) = t3 + t, sin 2t, 3t2 − 1 5.10. r(t) = 1, t2 , t3 , −1 ≤ t ≤ 1 5.11. r(t) = cos t, t2 , cos 2t 5.12. r(t) = cos t, cos 2t, cos 3t 5.13. Find the intersection point of the curves r1 (t) = 2t, t2 + t, t3 and r2 (t) = t, t2 − 3t + 4, t2 + t − 5 . Find the angle between the curves at that point, that is, ﬁnd the angle between the tangent vectors at the point. 5.14. The curves r1 (t) = t2 , 2t, ln(t2 − 3) and r2 (t) = t + 1, t2 − 5, t2 − 2t − 3 intersect at the point (4, 4, 0). Find the cosine of the angle between the two curves at that point.
45

Find the arc length of the curve r(t). 5.15. r(t) = 1, t2 , t3 , −1 ≤ t ≤ 1 √ 5.16. r(t) = t, ln t, 2 2t , 1 ≤ t ≤ 2 5.17. Let r(t), a ≤ t ≤ b, be a parametrization of a smooth curve γ. Let L = the new function t s(t) = a b a

r (t) dt be the length of γ and deﬁne

r (u) du

(a ≤ t ≤ b).

(a) Evaluate s(a) and s(b). (b) Use the Fundamental Theorem of Calculus to show that s (t) > 0 for all t with a < t < b. (c) Let t = t(s), 0 ≤ s ≤ L, be the inverse function of the function s(t), and deﬁne the vector function f(s) = r(t(s)), 0 ≤ s ≤ L. Show that f (s) = 1 for all s with 0 < s < L. (d) Convince yourself that the vector function f(s) from part (c) is another parametrization of the curve γ. This parametrization has a special property: f(s) is the position vector of the point on γ that is s units along the curve after the initial point. This parametrization is known as the parametrization with respect to arc length or the natural parametrization and plays an important role in the study of space curves. (e) Parametrize with respect to arc length the curve r(t) = e3t sin 4t, e3t cos 4t, 5 , 0 ≤ t ≤ 1 ln 4. 3

46

LECTURE 6

Multivariable Functions
6.1. Functions of two variables In this lecture we introduce multivariable functions. These are functions that depend on multiple inputs (usually 2, 3, or 4), but produce only a single output. We start our discussion with the simplest example: functions of two variables. For example, the temperature T at any point on the surface of the Earth at this very moment depends on two pieces of information: the longitude x and the latitude y of the point. Thus, we can think of T as a function T (x, y) of two variables x and y or as a function of the ordered pair (x, y). We now give a formal deﬁnition. Definition. A function f of two variables is a rule that assigns to each ordered pair of real numbers (x, y) in a set D a unique real number denoted f (x, y). The set D is the domain of f and the set of all values that f takes on is its range, that is, the range of f is { f (x, y) : (x, y) ∈ D}. 6.2. Functions of three and more variables The example that we used above to motivate the introduction of functions of two variables is somewhat artiﬁcial. A more natural (and also more useful) quantity is the temperature T at a point (x, y, z) in space at this very moment. Then T is a function of three variables—the coordinates (x, y, z) of the point. Note that with this function in hand we don’t have to assume that the temperature in a classroom and the temperature on the wing of an airplane ﬂying 39,000 feet above that classroom are the same. An even more realistic example is that of the temperature T at the point (x, y, z) in space at time t. This quantity is a function T (x, y, z, t) of four variables. Definition. A function f of n variables x1 , x2 , . . . , xn is a rule that assigns to each ordered ntuple of real numbers (x1 , x2 , . . . , xn ) in a set D a unique real number denoted f (x1 , x2 , . . . , xn ). The set D is the domain of f and the set of all values that f takes on is its range. In particular, a function f of three variables is a rule that assigns to each ordered triple of real numbers (x, y, z) in a set D (in R3 ) a unique real number denoted f (x, y, z). 6.3. Examples of domains and ranges of multivariable functions √ Example 6.1. The function f (x, y) = 1 − xy is deﬁned whenever 1 − xy ≥ 0 ⇐⇒ xy ≤ 1. If x > 0, the last inequality is satisﬁed for y ≤ x−1 ; if x < 0, the inequality is satisﬁed for y ≥ x−1 ; and if x = 0, it is satisﬁed for all y. That is, the domain of f contains the points in I quadrant on and below the graph of y = x−1 , the points in III quadrant on and above the graph of y = x−1 , and all the points in II and IV quadrants, including the x- and y-axes (see Figure 6.1). The range of f is [0, ∞). Clearly, f cannot attain negative values. On the other hand, if k ≥ 0, 1 − xy = k ⇐⇒ 1 − xy = k2
47

⇐⇒

xy = 1 − k2 .

Hence, if k ≥ 0, k 1, f (x, y) = k for all points (x, y) on the hyperbola xy = 1 − k2 ; also, f (x, y) = 1 for all points (x, y) on the coordinate axes. That is, every number k ∈ [0, ∞) does belong to the range of f . y z

x y x √ 1 − xy

Figure 6.1. The domain of z =

Figure 6.2. z = 3 − x2 − y2

Example 6.2. The domain of the function f (x, y, z) = ln(16 − x2 − y2 − 9z2 ) consists of the points (x, y, z) in space whose coordinates satisfy 16 − x2 − y2 − 9z2 > 0 ⇐⇒ x2 y2 z2 + + < 1. 16 16 16/9

Since ln x is an increasing function, the range of f consists of the logarithms of the positive numbers in the range of the function w = 16 − x2 − y2 − 9z2 . Because x2 + y2 + 9z2 can attain any non-negative value and cannot attain any negative value, it follows that w takes on the numbers in the interval (−∞, 16]. Thus, the range of f consists of the logarithms of the numbers in (0, 16], that is, the range of f is (−∞, ln 16]. 6.4. Graphs Definition. If f is a function of n variables x1 , x2 , . . . , xn with domain D, then the graph of f is the set of all points (x1 , x2 , . . . , xn , z) in Rn+1 with (x1 , x2 , . . . , xn ) ∈ D and z = f (x1 , x2 , . . . , xn ), that is, the graph of f is the set (x1 , x2 , . . . , xn , f (x1 , x2 , . . . , xn )) : (x1 , x2 , . . . , xn ) ∈ D . In particular, the graph of a function f (x, y) deﬁned on a set D in R2 is the surface given by (x, y, f (x, y)) : (x, y) ∈ D . Example 6.3. The graph of the function f (x, y) = 3 − x2 − y2 is the surface z = 3 − x2 − y2 . This surface is displayed on Figure 6.2.
48

6.5. Level curves and surfaces While the graph of a function of two variables provides an excellent visualization of the function, it might be quite diﬃcult to sketch without advanced technology. What can we do using only manual (and mental) labor? One answer is well-known to cartographers. When drawing topographic or termal maps, cartographers often produce two-dimensional drawings marked with curves of constant elevation or constant temperature. We can do the same for any function f of two variables: we can graph f in the xy-plane by marking its “level curves”. Definition. Let f be a function of two veriables, denoted x and y, and let k be a real number. The k-level curve of f is the set of all points (x, y) in the domain D of f which satisfy the equation f (x, y) = k. This idea becomes even more important in the case of functions of three variables. Note that the graph w = f (x, y, z) of such a function is a four-dimensional object, which we can study analytically but cannot represent graphically in our three-dimensional universe. However, we can draw the “level surfaces” of f . Definition. Let f be a function of three veriables, denoted x, y and z, and let k be a real number. The k-level surface of f is the set of all points (x, y, z) in the domain D of f which satisfy the equation f (x, y, z) = k. y

x −1 3 −1 2
−1 −2 2 1
1 2 1 3

Figure 6.3. The level curves of f (x, y) = x/(x2 + y2 ) Example 6.4. Describe the level curves of the function x f (x, y) = 2 . x + y2 Solution. We want to draw the curve x2 x =k + y2

for all k. If we exclude the origin from consideration (since the curve is undeﬁned there), we can represent this curve by the equation kx2 + ky2 − x = 0. When k = 0, this equation represents the y-axis x = 0, so the level curve f (x, y) = 0 is the y-axis with the origin removed from it. When k 0, we can transform the equation of the level curve
49

further to x x − + y2 = 0 k
2

⇐⇒

1 x− 2k

2

+ y2 =

1 , 4k2

1 1 which is the equation of a circle centered at ( 2k , 0) of radius 2|k| (k can be negative). Since the origin does lie on this circle, the level curve f (x, y) = k is the circle with the origin removed from it. A contour map of f showing the level curves with k = 0, ± 1 , ± 1 , ±1, ±2 is shown on Figure 6.3. 3 2

6.6. Limits of multivariable functions Since all the important concepts in single-variable calculus were deﬁned in terms of limits, if we are to generalize those concepts to functions of two and more variables, we must ﬁrst deﬁne the notion of a limit of a multivariable function. The following deﬁnition generalizes the formal deﬁnition of limit to functions of two variables. Definition. Let f be a function of two variables whose domain D includes points arbitrarily close to (a, b). We say that the limit of f (x, y) as (x, y) approaches (a, b) is L and write
(x,y)→(a,b)

lim

f (x, y) = L

or

f (x, y) → L as (x, y) → (a, b),

if for any given number ε > 0, there is a corresponding number δ = δ(ε) > 0 such that 0< (x − a)2 + (y − b)2 < δ =⇒ | f (x, y) − L| < ε.

This deﬁnition is very formal. It is also very useful in proofs (something we will not be concerned with) and very imposing at ﬁrst sight. A more intuitive deﬁnition reads as follows. Definition. Let f be a function of two variables whose domain D includes points arbitrarily close to (a, b). We say that the limit of f (x, y) as (x, y) approaches (a, b) is L and write
(x,y)→(a,b)

lim

f (x, y) = L

or

f (x, y) → L as (x, y) → (a, b),

if f (x, y) approaches L as (x, y) approaches (a, b), independent of how (x, y) approaches (a, b). We can deﬁne limits of functions of three or more variables in a similar fashion. Definition. Let f be a function of three variables whose domain D includes points arbitrarily close to (a, b, c). We say that the limit of f (x, y, z) as (x, y, z) approaches (a, b, c) is L and write
(x,y,z)→(a,b,c)

lim

f (x, y, z) = L

or

f (x, y, z) → L as (x, y, z) → (a, b, c),

if f (x, y, z) approaches L as (x, y, z) approaches (a, b, c), independent of how (x, y, z) approaches (a, b, c). More precisely, the limit of f (x, y, z) as (x, y, z) approaches (a, b, c) is L, if for any given number ε > 0, there is a corresponding number δ = δ(ε) > 0 such that 0< (x − a)2 + (y − b)2 + (z − c)2 < δ =⇒ | f (x, y, z) − L| < ε.

Limits of multivariable functions can be much trickier than those of single-variable functions. The primary reason for that is the inﬁnitude of ways in which the arguments (x, y) (or (x, y, z)) can approach the point (a, b) (or (a, b, c)). This point is best illustrated by examples of limits that do not exist.
50

x + y2 does not exist. Indeed, we know from Example 6.4 (see also Figure 6.3) that the level curves of this function all want to pass through (0, 0) (but since the point is not in the domain of f , they can’t). Thus, if we approach (0, 0) along two distinct level curves (say, along f (x, y) = 1 and f (x, y) = −1), the function values will approach distinct values (1 and −1, respectively). However, if the limit existed, f (x, y) would have to approach the same value independent of the way (x, y) approaches (0, 0). Therefore, the limit does not exist.
(x,y)→(0,0)

Example 6.5. The limit

lim

x2

Example 6.6. Show that the limit

(x,y)→(0,0)

lim

xy4 does not exist. x2 + y8

Solution. First, consider the behavior of the function as (x, y) → (0, 0) along the straight line y = x. We have x5 x3 lim f (x, x) = lim 2 = lim = 0. x→0 x→0 x + x8 x→0 1 + x6 On the other hand, if we let (x, y) → (0, 0) along the curve x = y4 , we get y8 1 = . 8 + y8 y→0 y→0 y 2 Since diﬀerent approach paths lead to diﬀerent limiting values, the limit does not exist. lim f (y4 , y) = lim Example 6.7. For any numbers a, b, c, we have
(x,y)→(a,b)

lim

x = a,

(x,y)→(a,b)

lim

y = b,

(x,y)→(a,b)

lim

c = c.

Using these facts and the two-variable versions of the limit laws:
(x,y)→(a,b)

lim [ f (x, y) ± g(x, y)] = xy =

(x,y)→(a,b)

lim

f (x, y) ±

(x,y)→(a,b)

lim

g(x, y),

etc.,

we can compute the limit of any polynomial in x and y at any point (a, b) in the plane. For example,
(x,y)→(2,3)

lim

(x,y)→(2,3)

lim

x

(x,y)→(2,3)

lim

y = 2 · 3 = 6.

Next, we look at some examples of evaluation of limits of functions of two and three variables. Example 6.8. Evaluate sin(x2 + y2 ) . (x,y)→(0,0) x2 + y2 lim

Solution. If we change the coordinates of the point (x, y) from Cartesian to polar, the condition (x, y) → (0, 0) is equivalent to r → 0 and the limit takes the form sin(x2 + y2 ) sin(r2 ) 2r cos(r2 ) = lim = lim = 1, (x,y)→(0,0) r→0 r→0 x 2 + y2 r2 2r where we used l’Hospital’s rule to pass from the second limit to the third. lim

51

Example 6.9. Evaluate

(x,y,z)→(0,0,0)

lim

(x2 + y2 )z . x2 + y2 + z2

Solution. This time, we change the coordinates of the point (x, y, z) from Cartesian to spherical, and the condition (x, y, z) → (0, 0, 0) becomes ρ → 0. Hence,
(x,y,z)→(0,0,0)

lim

(ρ sin φ)2 (ρ cos φ) (x2 + y2 )z = lim x2 + y2 + z2 ρ→0 ρ2 = lim ρ sin2 φ cos φ = 0. ρ→0 Example 6.10. Evaluate

(x,y)→(1,2)

lim

x2 y − 2x2 − 2xy + 6x + 5y − 12 . x2 + y2 − 2x − 4y + 5

Solution. Both the numerator and the denominator of the function vanish at the point (1, 2), so the limit is an indeterminate form of type 0/0. We will try to argue similarly to the previous example, but before we can do so, we need to change the limit from one at the point (1, 2) to one at the origin. Let us deﬁne new variables u and v by x = 1 + u, y = 2 + v. The condition (x, y) → (1, 2) is equivalent to (u, v) → (0, 0), so we can rewrite the original limit as x2 y − 2x2 − 2xy + 6x + 5y − 12 (x,y)→(1,2) x2 + y2 − 2x − 4y + 5 (1 + u)2 (2 + v) − 2(1 + u)2 − 2(1 + u)(2 + v) + 6(1 + u) + 5(2 + v) − 12 = lim (u,v)→(0,0) (1 + u)2 + (2 + v)2 − 2(1 + u) − 4(2 + v) + 5 u2 v + 2u + 4v = lim . (u,v)→(0,0) u2 + v2 We can now change the coordinates of the point (u, v) from Cartesian to polar. We ﬁnd that lim u2 v + 2u + 4v r3 cos θ sin θ + 2r cos θ + 4r sin θ = lim (u,v)→(0,0) r→0 u2 + v2 r2 = lim r cos θ sin θ + (2 cos θ + 4 sin θ)r−1 lim r→0 = lim(2 cos θ + 4 sin θ)r−1 r→0  +∞ when 2 cos θ + 4 sin θ > 0,  = 0 when 2 cos θ + 4 sin θ = 0,  −∞ when 2 cos θ + 4 sin θ < 0. Hence, the given limit does not exist. 6.7. Continuity of multivariable functions We can use the notion of limit of a multivariable function to deﬁne the notion of continuity. Definition. Let f be a function of two variables whose domain D includes (a, b) and points arbitrarily close to (a, b). We say that f (x, y) is continuous at (a, b), if
(x,y)→(a,b)

lim

f (x, y) = f (a, b).
52

We say that f is continuous in D, if it is continuous at every point (a, b) ∈ D. Definition. Let f be a function of three variables whose domain D includes (a, b, c) and points arbitrarily close to (a, b, c). We say that f is continuous at (a, b, c), if
(x,y,z)→(a,b,c)

lim

f (x, y, z) = f (a, b, c).

We say that f is continuous in D, if it is continuous at every point (a, b, c) ∈ D. We want also to extend the notion of piecewise continuity to multivariable functions, and in particular, to functions of two variables. Recall that a single-variable function f is called piecewise continuous on [a, b] if it is continuous everywhere in [a, b] except for a ﬁnite number of points, where it can have only jump or removable discontinuities. The two-dimensional version of this deﬁnition reads as follows. Definition. Let f be a function of two variables with domain D. We say that f is piecewise continuous in D, if it is continuous at every point (a, b) ∈ D, except possibly for the points on a ﬁnite number of smooth curves in D, say γ1 , γ2 , . . . , γn , where f may have “jump discontinuities.” Remark. As in the case of single-variable functions, polynomials in several variables are continuous. Likewise, rational functions in several variables are continuous except for the points where the denominator is 0. Finally, if f is a continuous function (of a single variable) and g is a continuous function of the variables x, y, . . . , the composition f (g(x, y, . . . )) is continuous at all points (x, y, . . . ) where it is deﬁned. Example 6.11. The function f (x, y) = cos x2 − y2 x2 + y2

is continuous everywhere in its domain, which is all the points in R2 except (0, 0). Indeed, f is the composition of the continuous function w = cos t and the rational function t= x2 − y2 . x2 + y2

Example 6.12. Show that the function  2 2  sin(x + y ) f (x, y) = x2 + y2  1 is continuous everywhere. Solution. Recall that Hence, the function

if (x, y)

(0, 0),

if (x, y) = (0, 0),

sin x = 1. x→0 x   sin x if x 0, g(x) = x 1 if x = 0, lim

is continuous everywhere. Furthermore, the polynomial x2 + y2 is also continuous everywhere. Therefore, the function f (x, y) = g(x2 + y2 ) is also continuous everywhere.
53

Exercises
Find the domain and the range of the given function. 2 6.2. f (x, y) = ln x + y2 6.1. f (x, y) = y2 e−x /2 z 6.3. f (x, y, z) = 2 x − y2 − 1 Sketch the domain of the given function. 6.4. f (x, y) = ln x + y2 √ x + 2y 6.5. f (x, y) = 2 x + y2 − 1 For each function, draw the level curves for levels −1, 0, 1, 2, and 3; then sketch the graph z = f (x, y). 6.6. f (x, y) = 1 + y2 6.7. f (x, y) = 2x + 3y − 5 6.8. f (x, y) = x2 + y2 Find the given limit or show that it does not exist. 6.9. lim x4 − 4xy2 + 2x3 y (x,y)→(1,3) xy 6.10. lim 2 + y2 (x,y)→(0,0) x sin(2xy) 6.11. lim (x,y,z)→(1,0,2) 1 − zy2 3x2 y 6.12. lim (x,y)→(0,0) x4 + y2 xy 6.13. lim (x,y)→(0,0) x2 + y2 xy 6.14. lim √ (x,y)→(0,0) 1 + xy − 1 Find the set of points in R2 or R3 where the given function is continuous. 6.15. f (x, y) = sin x4 + x2 y2   xy if (x, y) (0, 0), 6.16. f (x, y) = x2 + y2 1 if (x, y) = (0, 0). 2 6.17. f (x, y, z) = ln 4 − x − y2 − 4z2   √ xy if (x, y) (0, 0), 1 + xy − 1 6.18. f (x, y) = 1 if (x, y) = (0, 0).

54

LECTURE 7

Surfaces
7.1. Introduction In single-variable calculus, the interplay between curves and functions is essential: we use geometrical questions about curves to motivate the study of calculus and then apply our knowledge of calculus to answer questions about plane curves. A similar symbiosis occurs in multivariable calculus between functions and surfaces. However, while we are familiar with a fair number of plane curves, our supply of known surfaces is limited essentially to planes, spheres, and graphs z = f (x, y). We introduced also the level surfaces of functions of three variables: these are the sets Σ = (x, y, z) ∈ R3 : F(x, y, z) = k . However, there are hardly any level surfaces that we recognize. The main goal of this lecture is to introduce some other surfaces that are commonly used in multivariable calculus. 7.2. Cylinders Definition. A cylinder is a surface that consists of all lines—called rullings—that are parallel to a given line and pass through a given plane curve. Example 7.1. We have seen already one example of a cylinder: in Example 1.8, we explained that the equation x2 +y2 = x describes the surface Σ in space that consists of all the points on vertical lines that pass through a circle in the xy-plane. Note that by a similar argument, any equation of one of the forms f (x, y) = 0, f (x, z) = 0, or f (y, z) = 0 represents a cylinder in space. For example, the equation 2y2 + z2 = 1 represents the cylinder that passes through the ellipse 2y2 + z2 = 1 in the yz-plane and has rullings parallel to the x-axis. Note that the three-dimensional equations of the ellipse are 2y2 + z2 = 1, x = 0. Similarly, in space, the equation x = y2 represents the cylinder that passes through the parabola x = y2 in the xy-plane and has rullings parallel to the z-axis. 7.3. Quadric surfaces Quadric surfaces are generalizations of the conic sections to three dimensions. Definition. A quadric surface is the set of all points (x, y, z) ∈ R3 that satisfy a second-degree equation in the variables x, y, and z: A11 x2 + A12 xy + A13 xz + A22 y2 + A23 yz + A33 z2 + B1 x + B2 y + B3 z + B0 = 0, where A11 , . . . , A33 , B0 , . . . , B3 are constants.
55

In other words, a quadric surface is a level surface of a quadratic polynomial in three variables. In the next couple of examples, we visualize some quadric surfaces by sketching their traces in planes parallel to the coordinate planes. Example 7.2. Consider the quadric surface x2 + y2 + z2 = 1. 4 The trace of this surface in the horizontal plane z = k has equations x2 + y2 = 1 − k2 , 4 which represent an ellipse, provided that 1 − k2 > 0 ⇐⇒ z = k,

−1 < k < 1.

Similarly, the traces of the surface in the planes y = k, −1 < k < 1, and x = k, −2 < k < 2, are also ellipses. Thus, the given surface is called an ellipsoid. Example 7.3. Consider the quadric surface z = x2 − 2y2 . The trace of this surface in a plane x = k is z = k2 − 2y2 , x = k, which is a parabola open downward. Likewise, the trace in a plane y = k is a parabola open upward, z = x2 − 2k2 , y = k. Lastly, the trace in a vertical plane z = k is the hyperbola x2 − 2y2 = k, This surface is called a hyperbolic paraboloid. Remark. At ﬁrst, one might think that there is no neat classiﬁcation of all quadric surfaces, but in fact there is. First of all, by a change of the coordinates, every quadric surface can be brought into the form Ax2 + By2 + Cz2 + Dz + E = 0, where either D = 0 or C = E = 0. If one of the variables is missing—say, if A = 0 or B = 0, then the quadric surface is a cylinder parallel to the respective coordinate axis. This leaves only six possibilities, which lead to the following six surfaces: cone, ellipsoid, elliptic paraboloid, hyperbolic paraboloid, one-sheet hyperboloid, and two-sheet hyperboloid. These surfaces in standard positions are displayed on Figure 7.1 and some basic facts about them are summarized in Table 7.1. Example 7.4. Classify the quadric surface Σ with equation 4y2 + z2 − x + 16y − 4z + 19 = 0.
56

z = k.

Surface

Equation xy

Traces yz Parabolas Parabolas Ellipses xz Parabolas Parabolas Ellipses

Axis

z x2 y2 = + c a2 b2 2 y2 Hyperbolic z x = 2− 2 Paraboloid c a b 2 2 x y z2 Ellipsoid + 2+ 2 a2 b c 2 2 x y z2 One-Sheet + 2− 2 Hyperboloid a2 b c 2 2 y z2 x + 2− 2 Cone a2 b c 2 2 x y z2 Two-Sheet + 2− 2 Hyperboloid a2 b c Elliptic Paraboloid

Ellipses Hyperbolas =1 =1 =0 = −1 Ellipses Ellipses Ellipses Ellipses

z-axis z-axis

Hyperbolas Hyperbolas z-axis Hyperbolas Hyperbolas z-axis Hyperbolas Hyperbolas z-axis

Table 7.1. The non-degenerate quadric surfaces z z z

y x (a) Elliptic Paraboloid z y x (b) Hyperbolic Paraboloid z x (c) Ellipsoid z

y

y x x

y x

y

(d) One-Sheet Hyperboloid

(e) Cone

(f) Two-Sheet Hyperboloid

Figure 7.1. The non-degenerate quadric surfaces

57

Solution. We can rewrite this equation in the form 4(y + 2)2 + (z − 2)2 − (x + 1) = 0. If we now change the coordinates to X = y + 2, Y = z − 2, Z = x + 1, we ﬁnd that a point in P(x, y, z) in the xyz-space is on Σ if and only if P(X, Y, Z) is on the surface Z = 4X 2 + Y 2 in the XYZ-space. The latter surface is an elliptic paraboloid with a = 1 , b = 1, 4 and c = 1. Note that in the xyz-space the vertex of the paraboloid is at (−1, −2, 2) (these are the xyz-coordinates of the origin of the XYZ-coordinate system) and its axis is the line y = −2, z = 2 (these are equations of the Z-axis in xyz-coordinates). 7.4. Parametric surfaces By now, you have probably come to the realization that a “surface” is a three-dimensional geometric object that is essentially two-dimensional. In the case of graphs z = f (x, y) this is quite obvious: the points (x, y, z) on the graph are in one-to-one correspondence with the points (x, y) in the domain of f (x, y). For example, the points on the elliptic paraboloid z = x2 + 2y2 (see Figure 7.1(a)) are in a one-to-one correspondence with the points in the xy-plane. On the other hand, the two-dimensionality of a level surface, such as the ellipsoid x2 + 2y2 + z2 = 4, is less obvious from its equation (though it is quite intuitive from Figure 7.1(c)). Level surfaces are also hard to plot, because there is no easy way to generate points on a level surface. On the other hand, level surfaces are the natural analytic representation for some basic examples: the cone x2 + y2 − z2 = 0, the cylinder x2 + y2 = 1, and the sphere x2 + y2 + z2 = 1 are all level surfaces that cannot be represented as graphs of functions of two variables. In this section, we introduce the notion of a “parametric surface”. Definition. If x = x(u, v), y = y(u, v) and z = z(u, v) are three continuous functions deﬁned on a set D in R2 , then the set Σ of the points (x, y, z) satisfying the parametric equations x = x(u, v), y = y(u, v), z = z(u, v) ((u, v) ∈ D), (7.1)

is called a parametric surface; the variables u, v are called the parameters. Equations (7.1) and the vector function r : R2 → R3 , given by r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 are referred to as parametrizations of Σ. Remark. Like level surfaces, parametric surfaces generalize of the graphs z = f (x, y) of functions of two variables. Indeed, suppose that Σ is the surface z = f (x, y), (x, y) ∈ D. Then we can parametrize Σ by the equations x = u, y = v, z = f (u, v) ((u, v) ∈ D). ((u, v) ∈ D), (7.2)

On the other hand, a parametric surface is almost as easy to plot as the graph of a function: to plot z = f (x, y), we let the point (x, y) roam through the domain D and plot the points (x, y, f (x, y)); to plot the surface Σ parametrized by (7.2), we let (u, v) roam through D and plot the endpoints of the position vectors r(u, v).
58

Example 7.5. Identify the surface Σ, parametrized by the vector function r(u, v) = ue1 + u cos ve2 + u sin ve3 , where −∞ < u < ∞ and 0 ≤ v ≤ 2π. Solution. For a point (x, y, z) on Σ, we have y2 + z2 = (u cos v)2 + (u sin v)2 = u2 = x2 . This is the equation of a cone whose axis of symmetry is the x-axis. Example 7.6. Find a parametrization of the plane Π through the points A(1, 2, 3), B(1, −1, −2), and C(−2, 0, 1). − − → Solution. Let us ﬁrst ﬁnd an equation of the plane. Two vectors parallel to Π are AB = − − → − − → − − → 0, −3, −5 and BC = −3, 1, 3 . Hence, the vector n = AB × BC is normal to Π. We have n = (−3e2 − 5e3 ) × (−3e1 + e2 + 3e3 ) = 9(e2 × e1 ) − 9(e2 × e3 ) + 15(e3 × e1 ) − 5(e3 × e2 ) = −4e1 + 15e2 − 9e3 , −4(x − 1) + 15(y − 2) − 9(z − 3) = 0. Let us solve this equation for one of the variables, say for x: −4(x − 1) + 15(y − 2) − 9(z − 3) = 0 We see that Π is the graph of the function x= so it is parametrized by the vector function r(y, z) = where −∞ < y, z < ∞. Example 7.7. Let a = a1 , a2 , a3 and b = b1 , b2 , b3 be two vectors that are not parallel to each other. Find a parametrization of the plane Π through the origin that is parallel to a and b. Solution. We can try to argue similarly to the previous problem. However, the equation of Π a2 b3 − a3 b2 x + a3 b1 − a1 b3 y + a1 b2 − a2 b1 z = 0, and it is not clear which variable to solve for: any of the coeﬃcients can be zero. Thus, we will use a diﬀerent approach. Let P(x, y, z) be an arbitrary point in Π. We consider a parallelogram OAPB in Π, whose sides − − → − − → OA and OB are parallel to a and b, respectively (see Figure 7.2). Then OA = ua and OB = vb for some numbers u, v. If we denote the position vector of P by r, we conclude that r = ua + vb. By varying u, v we get the entire plane Π, so one possible parametrization is r(u, v) = (a1 u + b1 v)e1 + (a2 u + b2 v)e2 + (a3 u + b3 v)e3 (−∞ < u, v < ∞). is
15 y 4 15 y 4 9 − 4z + 1, 4

so an equation for Π is

⇐⇒ ⇐⇒

x−1= x=
15 y 4

15 (y − 2) − 9 (z 4 4 9 1 − 4z + 4.

− 3)

− 9z + 4

1 4

e1 + ye2 + ze3 ,

59

P B b r a A

O

Figure 7.2. The parallelogram OAPB Example 7.8. Find a parametrization of the elliptic cylinder x2 + 4z2 = 1, x = cos u, z = 1 sin u 2 y ≥ 5. Solution. A parametrization of the ellipse x2 + 4z2 = 1 in the xz-plane is given by the functions (0 ≤ u ≤ 2π). The points on the given cylinder must have x- and z-coordinates that satisfy these two equations and a y-coordinate with y ≥ 5. Thus, we can parametrize the cylinder by r(u, v) = cos ue1 + ve2 + 1 sin ue3 , 2 where 0 ≤ u ≤ 2π and v ≥ 5. Example 7.9. Find a parametrization of the part of the plane 3x + 2y + z = 4 that lies within the paraboloid z = x2 + 2y2 + 1. Solution. By solving the equation of the plane for z, z = 4 − 3x − 2y, we can easily obtain a parametrization of the whole plane: r(u, v) = ue1 + ve2 + (4 − 3u − 2v)e3 (−∞ < u, v < ∞).

To parametrize just the given part of the plane, we restrict u, v to its projection onto the xy-plane. The plane and the paraboloid intersect along the curve that contains the simultaneous solutions of the equations 3x + 2y + z = 4, z = x2 + 2y2 + 1. Eliminating z, we ﬁnd that the x- and y-coordinates of the points on this curve satisfy 4 − 3x − 2y = x2 + 2y2 + 1 =⇒ x+
3 2 2

+2 y+

1 2 2

=

23 . 4

Thus, the intersection curve of the plane and the ellipsoid consists of all points on the plane whose x- and y-coordinates satisfy the last equation. Consequently, the intersection curve projects onto an ellipse in the xy-plane, and the part of the plane 3x + 2y + z = 4 that lies within the given paraboloid projects onto the interior of that ellipse. We obtain the parametrization r(u, v) = ue1 + ve2 + (4 − 3u − 2v)e3 , u+
3 2 2

+2 v+

1 2 2

≤

23 . 4

7.5. Plotting surfaces using Mathematica Mathematica provides several commands for plotting surfaces, which we survey in this section.
60

7.5.1. Graphs of functions of two variables: Plot3D. To plot the surface z = f (x, y), where f is a function of two variables, we use the command Plot3D. The basic form of this command is Plot3D[f[x,y], {x,a,b}, {y,c,d}] where a, b, c and d are numbers or numerical expressions. This will generate the part of the graph of f where a ≤ x ≤ b and c ≤ y ≤ d. For example, the following command will plot the surface z = x2 − y2 over the rectangular region given by −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1: Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}] Similarly to the Plot command, Plot3D has many options. You can get a list of these options and their default settings by executing Options[Plot3D] or by reading the help on Plot3D. Here we demonstrate a few of the more commonly used options. • To label the axes, we use the AxesLabel option: Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, AxesLabel->{"x","y","z"}] • To change the aspect ratio of the plot, we specify values to the parameter BoxRatios. The option BoxRatios speciﬁes the ratio of the side lengths for the box that encloses the graph. The setting BoxRatios->Automatic gives identical scales on all three axes. Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, BoxRatios->{1,2,3}] Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, BoxRatios->Automatic] • To suppress the box surrounding the surface we set the option Boxed to False. Similarly, the axes can be suppressed by setting Axes to False. Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, Boxed->False, Axes->False] • The colors of the surface may be changed using the options PlotStyle and ColorFunction and the color-specifying commands RGBColor or Hue as shown in the examples below. Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, PlotStyle->Red] Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, PlotStyle->RGBColor[1,0,0]] Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, PlotStyle->Hue[0.5]] Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, ColorFunction->Function[{x,y,z}, RGBColor[xˆ2,0,yˆ2]]] Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, ColorFunction->Function[{x,y,z}, RGBColor[xˆ2,0,yˆ2]], ColorFunctionScaling->False] • To eliminate the colors altogether, we set the option ColorFunction to (White&). We can also suppress the mesh that Mathematica lays on the plot by setting the option Mesh to False. Plot3D[xˆ2-yˆ2, {x,-1,1}, {y,-1,1}, ColorFunction->(White&), Mesh->False] Rotating the surface around can, sometimes, reveal its shape and characteristics more clearly. To rotate a surface plot in Mathematica, simply left-click on it and start moving the mouse without releasing the mouse button. Sliding the mouse up and down while holding the Ctrl key allows you to zoom in and out the surface.
61

7.5.2. Level surfaces: ContourPlot3D. The basic syntax for plotting a level surface using Mathematica is ContourPlot3D[f[x,y,z]==k, {x,x1,x2}, {y,y1,y2}, {z,z1,z2}] This command plots the part of the surface f (x, y, z) = k that lies within the box given by the inequalities x1 ≤ x ≤ x2 , y1 ≤ y ≤ y2 , z1 ≤ z ≤ z2 . For convenience, the command ContourPlot3D also accepts surfaces given by the “more general” type of equation f (x, y, z) = g(x, y, z): ContourPlot3D[f[x,y,z]==g[x,y,z], {x,x1,x2}, {y,y1,y2}, {z,z1,z2}] Finally, this command can plot simultaneously several level surfaces of the same function f . For example, to plot the −1-, 3- and 6-level surfaces of the function x2 − y2 − 3z (in blue, red and green, respectively), we use ContourPlot3D[xˆ2-yˆ2-3z, {x,-2,2}, {y,-2,2}, {z,-2,2}, Contours->{{-1,Blue}, {3,Red}, {6,Green}} 7.5.3. Parametric surfaces: ParametricPlot3D. To plot the parametric surface with parametrization (7.1), we use a version of the command ParametricPlot3D, which we used in Lecture #5 to plot a parametric space curve. When used to plot a parametric surface, the basic syntax of this command is ParametricPlot3D[{x[u,v],y[u,v],z[u,v]}, {u,u1,u2}, {v,v1,v2}] Alternatively, if the surface is described in the vector form (7.2) and we have deﬁned the vector function r(u, v) in Mathematica as r[u,v], we can use the form ParametricPlot3D[r[u,v], {u,u1,u2}, {v,v1,v2}] Below we give examples of both forms of ParametricPlot3D. Both commands plot the surface with parametrization r(u, v) = u cos ve1 + u sin ve2 + (v/2)e3 , where 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π. ParametricPlot3D[{u*Cos[v], u*Sin[v], v/2}, {u,0,1}, {v,0,2Pi}, PlotStyle->Red] r[u_,v_] := {u*Cos[v], u*Sin[v], v/2} ParametricPlot3D[r[u,v], {u,0,1}, {v,0,2Pi}, Mesh->None] The above examples illustrate that ContourPlot3D and ParametricPlot3D accept some options that resemble those for Plot and Plot3D. This is true for most plotting options and the best way to learn about such matters is to take a look at Mathematica’s help and the examples there. 7.5.4. Non-rectangular domains: RegionFunction. Sometimes, one wants to plot a function or a parametric surface with a domain other than a rectangular box. Or, perhaps, one wants to plot a part of a level surface, but that part is restricted by a shape other than a box (for example, an ellipsoid). In such situation, one can use the option RegionFunction of the above plotting commands to restrict the variables to regions other than rectangular boxes. For example, the following command restricts the plot of the function z = x2 − y2 to the interior of the ellipse x2 + y2 /4 ≤ 1: Plot3D[xˆ2-yˆ2, {x,-3,3}, {y,-3,3}, RegionFunction->Function[{x,y,z}, xˆ2+yˆ2/4Red, Mesh->None]
62

Note that the value of RegionFunction is set to Function[{x,y,z}, ...] where in place of the dots we have included the restriction on the variables. That restriction can involve double inequalities or even several simultaneous inequalities (see the next example). Here is another example of use of RegionFunction, this time inside a ContourPlot3D command: ContourPlot3D[xˆ2+2yˆ2-zˆ2==1, {x,-3,3}, {y,-3,3}, {z,-3,3}, RegionFunction->Function[{x,y,z}, 1None] This command plots the part of the hyporboloid x2 + 2y2 − z2 = 1 that lies between the cylinders x2 + y2 = 1 and x2 + y2 = 4 and above the plane x + z = −2. The usage of RegionFunction to restrict the parameters of a parametric surface is similar, but somewhat more subtle: the region function generally takes the form Function[{x,y,z,u,v}, ...] with a restriction on the parameters u and v only. (Though it is possible to include also restrictions on x, y and z, which would have the same eﬀect as described above.) The dynamic command below plots the parametric surface r(u, v) = u cos ve1 + u sin ve2 + (v/2)e3 , where 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π and u2 + v2 /4 ≤ a, with a allowed to change between 0.1 and 10. Manipulate[ ParametricPlot3D[{u*Cos[v], u*Sin[v], v/2}, {u,0,1}, {v,0,2Pi}, RegionFunction->Function[{x,y,z,u,v}, uˆ2+vˆ2/4{{-1,1}, {-1,1}, {0,Pi}}, Mesh->None], {a,1/10,10}] It is instructive to execute it and experiment with various values of a in order to see the eﬀect on the plot from changing the restrictions on the parameters. Exercises
Find the traces of the given quadric surface in the planes x = k, y = k, and z = k. Use those traces and Table 7.1 to classify the surface. 7.1. z = x2 − y2 7.2. 4y2 − x2 − z2 = 1 7.3. y2 + 4z2 = 5 − x2 7.4. y2 + 4z2 = 5 + x2 Rewrite the equation of the given quadric surface in a standard form and use the result and Table 7.1 to classify the surface. 7.5. 3x + 6y2 − 3z2 = 9 7.6. x2 + 4y2 + z2 − 4x + 2z = 4 7.7. Parametrize the curve of intersection of the cylinder x2 + y2 = 9 and the plane z = x + y + 2. Identify the given parametric surface. 7.8. r(u, v) = u cos ve1 + u sin ve2 + u2 e3 7.9. r(θ, z) = 2 cos θe1 + sin θe2 + ze3 7.10. Parametrize the plane that contains the lines x−1 y+1 z x+2 z−4 = = and =y= . 2 3 −2 −3 4
63

7.11. Parametrize the half of the hyperboloid x2 − 4y2 − 4z2 = 16 that lies in the half-space x ≤ 0.

7.12. Parametrize the part of the hemisphere x2 + y2 + z2 = 5, z ≥ 0 that lies inside the cylinder x2 + y2 = 1. 7.13. Parametrize the part of the paraboloid z = 4 − x2 − y2 that lies above the xy-plane by using: (a) x and y as parameters; (b) the polar coordinates in the plane as parameters. 7.14. Let 0 < a < b. A torus of radii a and b is the surface of revolution obtained by rotating about z-axis the circle (x − b)2 + z2 = a2 in the xz-plane. Parametrize the torus of radii a and b.

64

LECTURE 8

Partial Derivatives
In this lecture, we introduce the derivatives of multivariable functions—the so-called “partial derivatives.” 8.1. First-order partial derivatives Definition. If f is a function of two variables, its partial derivatives f x and fy at (a, b) are deﬁned by f (a + h, b) − f (a, b) ∂f (a, b) = ∂ x f (a, b) = lim f x (a, b) = h→0 ∂x h and f (a, b + h) − f (a, b) ∂f (a, b) = ∂y f (a, b) = lim . fy (a, b) = h→0 ∂y h If f is a function of n variables x1 , x2 , . . . , xn , its partial derivatives f x1 , f x2 , . . . , f xn at (a1 , a2 , . . . , an ) are deﬁned by ∂f (a1 , a2 , . . . , an ) = ∂ x j f (a1 , a2 , . . . , an ) f x j (a1 , a2 , . . . , an ) = ∂x j f (a1 , . . . , a j + h, . . . , an ) − f (a1 , . . . , a j , . . . , an ) ( j = 1, 2, . . . , n). = lim h→0 h To illustrate the meaning of the partial derivatives, we take a closer look at the two-dimensional case. Given a function f (x, y) and a point (a, b) in its domain, we obtain two (single-variable) functions F1 (x) = f (x, b) and F2 (y) = f (a, y), by ﬁxing, respectively, the variables y and x in f (x, y). The points on the surface z = f (x, y) with y = b lie on the curve of intersection of that surface and the vertical plane y = b (see the blue curve

Figure 8.1. The partial derivatives as slopes
65

on Figure 8.1). That curve has exactly the same shape as the graph of the function F1 (x) in the plane. Likewise, the graph of F2 (y) can be visualized as the curve of intersection of the surface z = f (x, y) and the vertical plane x = a (see the red curve on Figure 8.1). Using the functions F1 (x) and F2 (y), we can express the partial derivatives of f (x, y) at (a, b) as f x (a, b) = lim h→0 F1 (a + h) − F1 (a) f (a + h, b) − f (a, b) = lim = F1 (a) h→0 h h

(8.1)

and f (a, b + h) − f (a, b) F2 (b + h) − F2 (b) = lim = F2 (b). (8.2) h→0 h→0 h h That is, the partial derivatives f x and fy of f are the derivatives of the functions F1 (x) and F2 (y). Recall that F1 (a) and F2 (b) are the slopes of the graphs of those functions at x = a and y = b, respectively. Looking at the copies of those graphs on the surface z = f (x, y) (the blue and red curves on Figure 8.1), we see that: fy (a, b) = lim The partial derivatives f x (a, b) and fy (a, b) represent the slopes at (a, b) of the sections of the graph z = f (x, y) with the vertical planes y = b and x = a, respectively. That is, f x (a, b) and fy (a, b) represent the rates of change of f at (a, b) in the directions of the x- and the y-axes, respectively. Furthermore, equations (8.1) and (8.2) can be extended to functions of more than two variables. In that more general form, they lead to the following practical observation. To compute the partial derivative ∂ x j f (x1 , . . . , xn ) with respect to one of the variables, one can treat all other variables as constants and diﬀerentiate the resulting single-variable function F(x j ) using the familiar rules from single-variable calculus. The following example demonstrates how we use this simple observation to calculation of partial derivatives. Example 8.1. Find the ﬁrst-order partials of the functions f (x, y) = x2 x , + y2 g(x, y, z) = e xy sin xz.

Solution. The partial derivatives of f are f x (x, y) = and fy (x, y) = x · The partials of g are g x (x, y, z) = ye xy sin xz + e xy (z cos xz), gy (x, y, z) = xe xy sin xz, gz (x, y, z) = e xy (x cos xz).
66

y2 − x2 1 · (x2 + y2 ) − x · (2x) = 2 (x2 + y2 )2 (x + y2 )2 −2xy −(2y) = 2 . (x2 + y2 )2 (x + y2 )2

8.2. Higher-order partial derivatives Just as we can diﬀerentiate the derivative f (x) of a single-variable function to obtain its second derivative, we can diﬀerentiate the partial derivatives of a multivariable function to obtain its higher-order partials. For example, a function f of two variables has four second-order partial derivatives f xx , f xy , fyx , and fyy : f xx = ∂ ∂f ∂2 f = , 2 ∂x ∂x ∂x ∂2 f ∂ ∂f fyx = = , ∂x∂y ∂x ∂y f xy = ∂2 f ∂ ∂f = , ∂y∂x ∂y ∂x ∂ ∂f ∂2 f fyy = 2 = . ∂y ∂y ∂y

Example 8.2. Find the second-order partials of the function f (x, y) = x2 cos y + y2 sin x. Solution. We have f x (x, y) = 2x cos y + y2 cos x, The second-order partials of f then are: f xx = 2 cos y − y2 sin x, fyx = −2x sin y + 2y cos x, f xy = −2x sin y + 2y cos x, fyy = −x2 cos y + 2 sin x. fy (x, y) = −x2 sin y + 2y sin x.

Remark. Notice that the second-order partials of the function in the example satisfy f xy = fyx . In other words, for this function it is not important whether we diﬀerentiate ﬁrst with respect to x and then with respect to y or vice versa. One might ask whether this is always true. The answer is: “only for ‘nice’ functions”. The following theorem gives a necessary condition for this to hold. Theorem 8.1. Suppose that f is a function of two variables deﬁned near (a, b) and having second-order partials f xy and fyx that are continuous at (a, b). Then f xy (a, b) = fyx (a, b). The following example illustrates the usefulness of the above theorem. Example 8.3. Find the second-order partials of f (x, y, z) = sin(x2 + 2y2 − z). Solution. The ﬁrst-order partials of f are: f x (x, y, z) = 2x cos(x2 + 2y2 − z), fy (x, y, z) = 4y cos(x2 + 2y2 − z), fz (x, y, z) = − cos(x2 + 2y2 − z). There are nine second-order partials, but we really have to compute only six of them. We have f xx (x, y, z) = 2 cos(x2 + 2y2 − z) + 2x − 2x sin(x2 + 2y2 − z) , fyy (x, y, z) = 4 cos(x2 + 2y2 − z) + 4y − 4y sin(x2 + 2y2 − z) , fzz (x, y, z) = − sin(x2 + 2y2 − z).
67

For the mixed partials, we have f xy (x, y, z) = 2x − 4y sin(x2 + 2y2 − z) = −8xy sin(x2 + 2y2 − z), fzy (x, y, z) = − − 4y sin(x2 + 2y2 − z) = 4y sin(x2 + 2y2 − z), fzx (x, y, z) = − − 2x sin(x2 + 2y2 − z) = 2x sin(x2 + 2y2 − z). Note that the last three functions are continuous (they are products of polynomials in x, y, z and compositions of sine functions and polynomials in x, y, z). Furthermore, it is clear that the partials fyx , f xz , and fyz will also be continuous. But then the theorem implies that, in fact, we have fyx = f xy , f xz = fzx , fyz = fzy . Example 8.4. Find f xy (0, 0) and fyx (0, 0) for  2 2  xy(y − x ) f (x, y) = x2 + y2  0

if (x, y)

(0, 0),

if (x, y) = (0, 0).

Solution. The ﬁrst order partials of f are:  5 4 2 3  y − x y − 4x y f x (x, y) = (x2 + y2 )2  0 and

if (x, y)

(0, 0),

if (x, y) = (0, 0),

 4 3 2 5  xy + 4x y − x if (x, y) (0, 0), fy (x, y) = (x2 + y2 )2  0 if (x, y) = (0, 0). These formulas are straightforward everywhere except at (x, y) = (0, 0), where we use the deﬁnition of partial derivatives: f (h, 0) − f (0, 0) 0−0 f x (0, 0) = lim = lim =0 h→0 h→0 h h and f (0, h) − f (0, 0) 0−0 fy (0, 0) = lim = lim = 0. h→0 h→0 h h Then f x (0, h) − f x (0, 0) h−0 f xy (0, 0) = lim = lim =1 h→0 h→0 h h and fy (h, 0) − fy (0, 0) −h − 0 fyx (0, 0) = lim = lim = −1, h→0 h→0 h h that is, f fails to satisfy f xy (0, 0) = fyx (0, 0). On the other hand, when (x, y) (0, 0), we have y6 + 9x2 y4 − 9x4 y2 − x6 . (x2 + y2 )3 This is not surprising, considering that the function f xy (x, y) = fyx (x, y) = y6 + 9x2 y4 − 9x4 y2 − x6 (x2 + y2 )3
68

is continuous everywhere except at (0, 0) (since it is a rational function in x and y, whose denominator vanishes only at (0, 0)). 8.3. Tangent planes Recall that the equation of the tangent line to the curve y = f (x) at a point P(x0 , y0 ) on the curve is given by y − y0 = f (x0 )(x − x0 ). The following theorem is the analog of this formula for functions of two variables. It describes the equation of the tangent plane to the surface z = f (x, y) at a point P(x0 , y0 , z0 ) on that surface in terms of the partials of f . Strictly speaking, we should ﬁrst deﬁne what a tangent plane is, but we will leave that for Lecture #10. The same applies to the proof of the theorem. Theorem 8.2. Suppose that f has continuous partial derivatives at (x0 , y0 ). Then the equation of the tangent plane to the surface z = f (x, y) at the point P(x0 , y0 , z0 ) is z − z0 = f x (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ). Example 8.5. Find the tangent plane to the surface z = e x −y at (−1, 0, e).
2 2

Solution. First, observe that the point is indeed on the surface: e = e(−1) −0 . The partials of the given function are ∂z ∂z 2 2 2 2 = 2xe x −y , = −2ye x −y , ∂x ∂y whence ∂z ∂z (−1, 0) = −2e, (−1, 0) = 0. ∂x ∂y Applying the theorem, we ﬁnd that the equation of the tangent plane is z − e = −2e(x − (−1)) + 0(y − 0) ⇐⇒ 2ex + z + e = 0.
2 2

8.4. Linear approximations and diﬀerentiable functions* Recall that for a single-variable function f which has a derivative f (a) at x = a (we called such functions “diﬀerentiable”), the tangent line at (a, f (a)) always exists and has equation y = f (a) + f (a)(x − a). That is, the linear function f (a) + f (a)(x − a) provides a very good approximation to f (x) when x is near a: f (x) ≈ f (a) + f (a)(x − a). When we deal with functions of two variables, we may encounter functions f (x, y) for which both partial derivatives f x (a, b) and fy (a, b) exist, but the plane z = f (a, b) + f x (a, b)(x − a) + fy (a, b)(y − b) is not tangent to the surface z = f (x, y) and does not provide a good approximation to f (x, y) for (x, y) near (a, b). Thus, in the case of functions of two or more variables, we reserve the term “diﬀerentiable function” for functions for which this type of approximation (known as the linear approximation to f at (a, b)) does hold.
69

Definition. A function f of two variables is called diﬀerentiable at (a, b), if near (a, b) we have f (x, y) = f (a, b) + f x (a, b)(x − a) + fy (a, b)(y − b) + ε1 (x, y)(x − a) + ε2 (x, y)(y − b), where
(x,y)→(a,b)

lim

ε1 (x, y) =

(x,y)→(a,b)

lim

ε2 (x, y) = 0.

Similarly, a function f of three variables is called diﬀerentiable at (a, b, c), if near (a, b, c) we have f (x, y, z) = f (a, b, c) + f x (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − c) + ε1 (x, y, z)(x − a) + ε2 (x, y, z)(y − b) + ε3 (x, y, z)(z − c), where
(x,y,z)→(a,b,c)

lim

ε1 (x, y, z) =

(x,y,z)→(a,b,c)

lim

ε2 (x, y, z) =

(x,y,z)→(a,b,c)

lim

ε3 (x, y, z) = 0.

Example 8.6. Show that the function f (x, y) = x3 + xy − 2y3 is diﬀerentiable at (1, −1). Solution. We have f (1, −1) = 2, Hence, we are looking for functions f x (1, −1) = 2, and
2 (x, y)

fy (1, −1) = −5. such that

1 (x, y)

f (x, y) = 2 + 2(x − 1) − 5(y + 1) + 1 (x, y)(x − 1) + 2 (x, y)(y + 1), lim lim 1 (x, y) = 2 (x, y) = 0.
(x,y)→(1,−1) (x,y)→(1,−1)

To ﬁnd such functions

1 (x, y)

and

2 (x, y),

we will transform the expression

f (x, y) − 2 − 2(x − 1) + 5(y + 1). Similarly to Example 6.10, it is convenient to rewrite this expression in terms of the variables u = x − 1 and v = y + 1. We have f (u + 1, v − 1) − 2 − 2u + 5v = u3 − 2v3 + 3u2 + uv + 6v2 = (u2 + 3u + v)u + (−2v2 + 6v)v. Returning back to the original variables x and y, we now obtain f (x, y) − 2 − 2(x − 1) + 5(y + 1) = where
1 (x, y) 1 (x, y)(x

− 1) + 2 (x, y)(y + 1), = −2(y + 1)2 + 6(y + 1).

= (x − 1)2 + 3(x − 1) + (y + 1),

2 (x, y)

Thus, we have now found functions 1 (x, y) and 2 (x, y) satisfying (8.4). Moreover, it is easy to see that these functions satisfy also (8.4). Therefore, the function f (x, y) is diﬀerentiable at (1, −1). Next, we state an alternative deﬁnition of diﬀerentiability, which is sometimes more convenient to use. For example, this deﬁnition is easier to apply when one aims to show that a given function is not diﬀerentiable (see Exercises 8.30 and 8.31). Definition. A function f of two variables is called diﬀerentiable at (a, b), if
(x,y)→(a,b)

lim

f (x, y) − f (a, b) − f x (a, b)(x − a) − fy (a, b)(y − b) (x − a)2 + (y − b)2
70

= 0.

Similarly, a function f of three variables is called diﬀerentiable at (a, b, c), if f (x, y, z) − f (a, b, c) − f x (a, b, c)(x − a) − fy (a, b, c)(y − b) − fz (a, b, c)(z − c) lim = 0. (x,y,z)→(a,b,c) (x − a)2 + (y − b)2 + (z − c)2 As we mentioned above (and as Exercises 8.30 and 8.31 demonstrate), the existence of partial derivatives does not guarantee the diﬀerentiability of f . It is therefore natural to ask what condition does? It turns out that the existence of continuous partials suﬃces. For the sake of simplicity, we state the relevant theorem only for functions of two variables. Theorem 8.3. If the partial derivatives f x and fy exist near (a, b) and are continuous at (a, b), then f is diﬀerentiable at (a, b). 8.5. The chain rule Recall that the chain rule in single-variable calculus gives a formula for the derivative of a composite function: if y = f (x(t)), then dy d f dx = . dt dx dt In this section, we discuss the generalizations of the chain rule to functions of two and more variables. Our ﬁrst result concerns composite functions of the form F(t) = f (x1 (t), . . . , xn (t)), where f is a diﬀerentiable function of n variables and x j = x j (t), 1 ≤ j ≤ n, are diﬀerentiable single-variable functions. Theorem 8.4. Let f be a diﬀerentiable function of n variables and let x1 (t), . . . , xn (t) be diﬀerentiable single-variable functions. Then z = f (x1 (t), . . . , xn (t)) is a diﬀerentiable function of t, and its derivative is dz ∂ f dx1 ∂ f dxn = + ··· + . dt ∂x1 dt ∂xn dt
∂ Remark. Take a note that we use “partial d’s,” ∂x j , to denote derivatives of functions of two d variables, but we use “straight d’s,” dt , to denote derivatives of functions of a single variable. It is important to follow this convention when dealing with a mixture of multivariable and singlevariable functions.

Example 8.7. Let z = f (x, y) = x ln(x + 2y), x = cos t, and y = sin t. Find z (0). Solution. The partials of f are f x (x, y) = ln(x + 2y) + and the derivatives of x(t) and y(t) are x (t) = − sin t, y (t) = cos t. Hence, by the above theorem, the derivative z (0) of the composite function z = f (x(t), y(t)) at t = 0 is dz ∂z dx ∂z dy (0) = (1, 0) (0) + (1, 0) (0) dt ∂x dt ∂y dt 1 2 = ln(1 + 0) + (−0) + (1) = 2. 1+0 1+0
71

x , x + 2y

fy (x, y) =

2x , x + 2y

In general, we have dz = dt ln(cos(t) + 2 sin t) + cos t cos t + 2 sin t (− sin t) + 2 cos t cos t + 2 sin t (cos t).

We now state the general version of the chain rule for compositions of multivariable functions. Theorem 8.5. Let f be a diﬀerentiable function of n variables and let each x j = x j (t1 , . . . , tm ), 1 ≤ j ≤ n, be a diﬀerentiable function of (the same) m variables. Then u = f (x1 (t1 , . . . , tm ), . . . , xn (t1 , . . . , tm )) is a diﬀerentiable function of t1 , . . . , tm , with partial derivatives ∂u ∂ f ∂x1 ∂ f ∂xn = + ··· + ∂ti ∂x1 ∂ti ∂xn ∂ti (i = 1, . . . , m).

Example 8.8. Let z = x2 − xy, x = s sin t, and y = t sin s. Find ∂z/∂s and ∂z/∂t. Solution. We have z x (x, y) = 2x − y, and x s (s, t) = sin t, xt (s, t) = s cos t, y s (s, t) = t cos s, Hence, the partials of the composite function z(x(s, t), y(s, t)) are yt (s, t) = sin s. zy (x, y) = −x,

z s (s, t) = z x (x(s, t), y(s, t))x s (s, t) + zy (x(s, t), y(s, t))y s (s, t) = (2s sin t − t sin s) sin s + (−s sin t)(t cos s) and zt (s, t) = z x (x(s, t), y(s, t))xt (s, t) + zy (x(s, t), y(s, t))t s (s, t) = (2s sin t − t sin s)(s cos t) + (−s sin t) sin s. Remark. Often, we do not substitute x(s, t) and y(s, t) in the formulas for the partials of z(s, t). In such cases, the answer in the above example would take the form z s (s, t) = (2x − y) sin s − xt cos s, zt (s, t) = (2x − y)s cos t − x sin s.

Example 8.9. Find the partial derivatives of w(x, y, t), where √ w = u2 + v2 , u = y + x cos t, v = x + y sin t. Solution. The partial derivatives of these functions are: ∂w u = √ , ∂u u2 + v2 ∂u = cos t, ∂x ∂v = 1, ∂x ∂w v = √ , ∂v u2 + v2 ∂u = −x sin t, ∂t ∂v = x cos t. ∂t

∂u = 1, ∂y ∂v = sin t, ∂y
72

Hence, the partial derivatives of the composite function w(x, y, t) are: ∂w ∂w ∂u ∂w ∂v = + = ∂x ∂u ∂x ∂v ∂x ∂w ∂w ∂u ∂w ∂v = + = ∂y ∂u ∂y ∂v ∂y ∂w ∂w ∂u ∂w ∂v = + = ∂t ∂u ∂t ∂v ∂t v u cos t √ + √ = u2 + v2 u2 + v2 u v sin t √ + √ = u2 + v2 u2 + v2 −ux sin t vx cos t √ + √ = u2 + v2 u2 + v2 u cos t + v √ , u2 + v2 u + v sin t √ , u2 + v2 −ux sin t + vx cos t √ . u2 + v2

Example 8.10. Find the partial derivative wyx of w(x, y, t), where √ w = u2 + v2 , u = y + x cos t, v = x + y sin t. Solution. To ﬁnd a higher-order partial derivative of a composite function, we “mix” the chain rule with the other rules for partial diﬀerentiation. We have ∂2 w ∂ ∂w ∂ u + v sin t √ = = ∂x∂y ∂x ∂y ∂x u2 + v2 ∂ u + v sin t ∂v ∂ u + v sin t ∂u √ √ + . = 2 + v2 ∂u ∂x ∂v u u2 + v2 ∂x We now compute the partials with respect to u and v: ∂ u + v sin t √ ∂u u2 + v2 = ∂ (u + v sin t)(u2 + v2 )−1/2 ∂u = (u2 + v2 )−1/2 + (u + v sin t)(−u)(u2 + v2 )−3/2 v2 − uv sin t , (u2 + v2 )3/2

= (u2 + v2 )−3/2 (u2 + v2 ) − u(u + v sin t) = ∂ u + v sin t √ ∂v u2 + v2 =

∂ (u + v sin t)(u2 + v2 )−1/2 ∂v = (sin t)(u2 + v2 )−1/2 + (u + v sin t)(−v)(u2 + v2 )−3/2 = (u2 + v2 )−3/2 (u2 + v2 ) sin t − v(u + v sin t) = u2 sin t − uv . (u2 + v2 )3/2

Substituting these partials into the expression for wyx , we get ∂2 w ∂ u + v sin t ∂u ∂ u + v sin t ∂v √ √ = + ∂x∂y ∂u u2 + v2 ∂x ∂v u2 + v2 ∂x 2 2 v − uv sin t u sin t − uv = 2 · cos t + 2 ·1 2 )3/2 (u + v (u + v2 )3/2 u2 sin t − uv(1 + sin t cos t) + v2 cos t = . (u2 + v2 )3/2
73

8.6. Partial derivatives in Mathematica Recall that if f (x) is a single-variable function, we can calculate its derivative in Mathematica using either of the following two commands: f’[x] D[f[x],x] The D command that appears in the latter option is very versatile and is used also to calculate partial derivatives. Its basic form for multivariable functions is D[f[x1 , x2 , . . . , xn ], xi , x j , . . . ] For a function f (x1 , . . . , xn ), this command will calculate the partial derivative f xi ,x j ,... (x1 , . . . , xn ). Here are some examples: D[f[x,y,z],x] f x (x, y, z) D[f[x,y,z],x,z] f xz (x, y, z) D[f[x,y],x,x,y] f xxy (x, y) D[f[x,u,v],v,v,x,u] fvvxu (x, u, v) The D command has also two other forms that provides useful shortcuts for special kinds of partial derivatives. If f is a function of several variables, one of which is x, then D[f[x, . . . ,], x, n] is a quick way to obtain the n-th order partial ∂n f /∂xn . Furthermore, if f is a multivariable function, the command D[f[x1 , x2 , . . . , xn ], x1 , x2 , . . . , xn ] will calculate its gradient f (x1 , . . . , xn ) (see Lecture #9). There is one use of the D command that requires extra care. Suppose that we have a function f (x, y) = x2 − y2 and we want to deﬁne a new function g by g(x, y) = f x (x, y) + fy (x, y). It may feel natural to use the following Mathematica commands: f[x_,y_] := xˆ2 - yˆ2 g[x_,y_] := D[f[x,y],x] + D[f[x,y],y] However, the function g deﬁned this way behaves rather bizarrely. For instance, asking Mathematica to execute simple commands like g[0,0] Plot3D[g[x,y], {x,-1,1}, {y,-1,1}] results in cryptic error messages like “General::ivar: 0 is not a valid variable.” The proper way to deﬁne g in this case is to use “=” instead of “:=” in the above code g[x_,y_] = D[f[x,y],x] + D[f[x,y],y] Try the above commands with g deﬁned this way. Exercises
Find the ﬁrst-order partial derivatives of the given function. 8.1. f (x, y) = x4 − 4x2 y3/2 8.2. f (x, y, z) = e2xy sin xz 8.3. f (x, y, z) = ln(3x + 7yz) x 8.4. f (x, y) = x2 + xy + y2 √ 8.5. f (x, z) = x2 + z2 + 1 8.6. f (r, θ) = r tan2 θ
74

Find the second-order partial derivatives of the given function. 8.7. f (x, y) = x4 − 4x2 y3/2 8.8. f (x, y) = cos x2 + xy + y2 √ 8.9. f (x, z) = x2 + z2 + 1 Find the indicated partial derivative of the given function. 8.10. f (x, y) = x4 − 4x2 y3/2 ; f xyx √ 8.11. f (x, z) = x2 + z2 + 1; ∂3 f /∂x∂2 z 8.12. f (x, y) = cos x2 + xy + y2 ; f xxy 8.13. f (x, y, z) = xz2 e2yz ; f xyz , fzzy Use the chain rule to calculate dF/dt for the given function F. 8.14. F(t) = f (x(t), y(t)), where f (x, y) = 2xy + y3 , x(t) = t2 + 3t + 4, y(t) = t2 2 2 8.15. F(t) = f (x(t), y(t)), where f (x, y) = xye x +4y , x(t) = 2 cos t, y(t) = sin t 8.16. F(t) = f (x(t), y(t), z(t)), where f (x, y, z) = x2 + y2 − 3xyz, x(t) = t sin t, y(t) = t cos t, z(t) = 1 t2 2 8.17. F(t) = f (x(t), y(t), z(t)), where f (x, y, z) = ln(xy + 2yz − 3xz), x(t) = 2et , y(t) = 3e−t , z(t) = e2t Use the chain rule to calculate ∂F/∂s and ∂F/∂t for the given function F. 8.18. F(s, t) = f (x(s, t), y(s, t)), where f (x, y) = 2x2 y − xy2 , x(s, t) = t2 + 3s2 , y(s, t) = 3st 8.19. F(s, t) = f (x(s, t), y(s, t)), where f (x, y) = arctan x2 − 4y2 , x(s, t) = 2set + 2se−t , y(s, t) = te s − te−s 8.20. F(s, t) = f (x(s, t), y(s, t), z(s, t)), where f (x, y, z) = ln x2 + 2y2 + 3z2 , x(s, t) = s cos t, y(s, t) = s sin t, z(s, t) = t2 8.21. F(s, t) = f (x(s, t), y(s, t), z(s, t)), where f (x, y, z) = cos xy + y3 x + z2 , x(s, t) = s2 + t, y(s, t) = s − t2 , z(s, t) = st + s + t Use the chain rule to calculate the given partial derivative at the given point. 8.22. gr (r, θ) at (r, θ) = (2, π), where g(x, y) = xy2 − x2 y, x = r cos θ, y = r sin θ 2 2 2 8.23. hu (u, v) at (u, v) = (2, −3), where h(x, y, z) = e x −y −z , x = 2u + v, y = u − v, z = u + v + 1 8.24. ∂R/∂x at (x, y) = (1, 1), where R(u, v, w) = ln(u2 + v2 + w2 ), u = x + 2y, v = 2x − y, w = 2xy 8.25. Show that the function u(x, y) = aey cos x + bey sin x is a solution of the diﬀerential equation u xx + uyy = 0. 8.26. Show that any function u(x, t) of the form u(x, t) = f (x + at) + g(x − at) is a solution of the diﬀerential equation utt = a2 u xx . Find the linear approximation to the given diﬀerentiable function near the given point. 8.27. f (x, y) = ln(2xy + x + y), (2, 1) 8.28. f (x, y, z) = arctan(xy + xz + 2yz), (3, 2, 1) 8.29. Show that the function f (x, y) = xy is diﬀerentiable at (1, 2) by ﬁnding functions the deﬁnition of diﬀerentiability in §8.4. 8.30. (a) Calculate f x (0, 0) and fy (0, 0) for the function   xy f (x, y) = x2 + y2 0 (b) Show that f (x, y) is not continuous at (0, 0). (c) Show that f (x, y) is not diﬀerentiable at (0, 0). 8.31. Let a, b, c, and d be real numbers.
75
1 (x, y)

and

2 (x, y)

that satisfy

if (x, y)

(0, 0),

if (x, y) = (0, 0).

(a) Calculate f x (0, 0) and fy (0, 0) for the function  3 2 2 3  ax + bx y + cxy + dy f (x, y) = x 2 + y2  0

if (x, y)

(0, 0),

if (x, y) = (0, 0).

(b) Show that f (x, y) is continuous at (0, 0). (c) Show that if a c or b d, then f (x, y) is not diﬀerentiable at (0, 0).

76

LECTURE 9

Directional Derivatives and Gradients
9.1. Directional derivatives Given a function of two variables f (x, y), we know how to compute its rate of change in the x-direction and in the y-direction: the rate of change in the x-direction is given by the partial derivative with respect to x, f (x + h, y) − f (x, y) ∂f (x, y) = lim , f x (x, y) = h→0 ∂x h and the rate of change in the y-direction is given by the partial derivative with respect to y, ∂f f (x, y + h) − f (x, y) fy (x, y) = (x, y) = lim . h→0 ∂y h A natural question then is how to compute the rate of change of f in other directions. This question leads to the notion of a “directional derivative.” Definition. If f is a function of two variables and if u = a, b is a unit vector, the directional derivative of f in the u-direction is f (x + ha, y + hb) − f (x, y) Du f (x, y) = lim , (9.1) h→0 h if the limit exists. If f is a function of three variables and if u = a, b, c is a unit vector, the directional derivative of f in the u-direction is f (x + ha, y + hb, z + hc) − f (x, y, z) Du f (x, y, z) = lim . (9.2) h→0 h Remark. Note that if we introduce the function g(t) = f (x + ta, y + tb), the limit on the right side of (9.1) is also g(h) − g(0) lim = g (0); h→0 h that is, Du f (x, y) = g (0). On the other hand, by the chain rule, g (t) = f x (x + ta, y + tb) · a + fy (x + ta, y + tb) · b, g (0) = f x (x, y)a + fy (x, y)b. This is (essentially) how one proves the following result. Theorem 9.1. If f is a diﬀerentiable function of two variables and u = a, b is a unit vector, then Du f (x, y) = f x (x, y)a + fy (x, y)b. If f is a diﬀerentiable function of three variables and u = a, b, c is a unit vector, then Du f (x, y, z) = f x (x, y, z)a + fy (x, y, z)b + fz (x, y, z)c.
77

2x Example 9.1. Find the directional derivative of f (x, y) = at the point (1, 0) in the direction x−y √ of v = e1 − 3e2 . Solution. First, notice that v is not unit and that the unit vector in the direction of v is √ √ v 1, − 3 3 1 u= = √ ,− . = v 2 2 1+3 The partials of f are f x (x, y) = so the theorem gives √ √ 1 −2y − 3 2x −y − 3x Du f (x, y) = + = , 2 (x − y)2 2 (x − y)2 (x − y)2 Example 9.2. Find the directional derivative of f (x, y, z) = direction of v = 1, 2, 3 . √ Du f (1, 0) = − 3. x at the point (2, 1, 1) in the y+z 2(x − y) − 2x −2y = , 2 (x − y) (x − y)2 fy (x, y) = 2x , (x − y)2

Solution. The unit vector in the direction of v is 1, 2, 3 1 v 2 3 = √ = √ , √ , √ , u= v 1+4+9 14 14 14 and the partial derivatives of f are 1 −x −x f x (x, y, z) = , fy (x, y, z) = , fz (x, y, z) = . 2 y+z (y + z) (y + z)2 Hence, 1 2 1 3 −2 1 1 Du f (2, 1, 1) = √ − √ − √ = √ . 2 14 2 14 2 14 14 9.2. The gradient vector Definition. If f is a function of two variables x and y, its gradient, denoted f or grad f , is the vector function ∂f ∂f f = f x (x, y), fy (x, y) = e1 + e2 . ∂x ∂y If f is a function of three variables x, y, and z, its gradient, f or grad f , is ∂f ∂f ∂f f = f x (x, y, z), fy (x, y, z), fz (x, y, z) = e1 + e2 + e3 . ∂x ∂y ∂z Remark. Note that using the gradient vector, we can restate both formulas in Theorem 9.1 in the form Du f (x) = f (x) · u, (9.3) where x = (x, y) or x = (x, y, z).
78

Theorem 9.2. Let f be a diﬀerentiable function of two or three variables. The maximum value of the directional derivative Du f (x) is f (x) ; it occurs when u has the same direction as f (x). Proof. By (9.3) and the properties of the dot product, Du f (x) = f (x) · u = f (x) u cos θ = f (x) cos θ, where θ is the angle between f (x) and u. Since cos θ ≤ 1, with equality when θ = 0 (i.e., when f (x) and u point in the same direction), we ﬁnd that Du f (x) ≤ f (x) , with an equality when u points in the direction of the gradient. Example 9.3. By Example 9.2, the gradient of the function x f (x, y, z) = y+z −x 1 −x , , . 2 (y + z)2 y + z (y + z) Hence, the largest directional derivative at (2, 1, 1) occurs in the direction of 1 −1 −1 f (2, 1, 1) = , , . 2 2 2 f (x, y, z) = Its value is f (2, 1, 1) =
√ 3 . 2

is

Exercises
Find the directional derivative of the given function in the given direction. 9.1. f (x, y) = x2 + y2 e2xy , in the direction of 1, 2 9.2. f (x, y, z) = ln xy + z2 , in the direction of e2 − 2e3 9.3. Find the directional derivative of F(x, y, z) = 9.4. Find the directional derivative of G(x, y, z) = arcsin at the point P(1, 2, 0) in the direction of the point Q(−1, 3, 1). Find the maximum rate of change of f at the given point and the direction in which it occurs. 9.5. f (x, y) = tan(xy + x + y), P(2, −1) 9.6. f (x, y, z) = xy + yz2 , Q(1, −3, 1) 4x2 2x + y2 + z2 4x2 − y2 − z2 at P(2, −1, 3) in the direction of v = −1, −2, 2 .

79

LECTURE 10

Tangent Planes and Normal Vectors
10.1. Tangent planes and normal vectors to level surfaces We now fulﬁll a promise made in Lecture #8. We start by deﬁning the tangent plane to a level surface. Definition. Let F(x, y, z) be a diﬀerentiable function, and let P(x0 , y0 , z0 ) be a point on the surface Σ with equation F(x, y, z) = k. The tangent plane to Σ at P is the plane, if it exists, that contains P and the tangent vectors to any smooth curve γ on Σ that passes through P. The following theorem establishes a formula for the equation of the tangent plane to a level surface. Theorem 10.1. Let F(x, y, z) be a diﬀerentiable function, and let P(x0 , y0 , z0 ) be a point on the surface Σ with equation F(x, y, z) = k. If F(x0 , y0 , z0 ) 0, then the tangent plane to Σ at the point P exists and is perpendicular to F(x0 , y0 , z0 ). In particular, the tangent plane has equation F x (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − z0 ) = 0. (10.1) Proof. Note that if γ is a curve on Σ given by the vector function r(t) = x(t)e1 + y(t)e2 + z(t)e3 , then the component functions of r(t) must satisfy the equation of Σ: F(x(t), y(t), z(t)) = k. When we diﬀerentiate this equation using the chain rule, we get ∂F dx ∂F dy ∂F dz + + = 0 ⇐⇒ F · r (t) = 0. ∂x dt ∂y dt ∂z dt In other words, the tangent vector to the curve, r (t), is perpendicular to the gradient vector F. First of all, this proves the existence of the tangent plane—it is the plane through P that is perpendicular to F(x0 , y0 , z0 ). Second, we get that (10.1) is the equation of the tangent plane. Example 10.1. Find the equation of the tangent plane at a point P(x0 , y0 , z0 ) of the one-sheet hyperboloid x2 y2 z2 F(x, y, z) = 2 + 2 − 2 = 1. a b c Solution. We have 2x0 2y0 2z0 F x (x0 , y0 , z0 ) = 2 , Fy (x0 , y0 , z0 ) = 2 , Fz (x0 , y0 , z0 ) = − 2 , a b c so the equation of the tangent plane to the hyperboloid at a point P(x0 , y0 , z0 ) is 2x0 2y0 2z0 (x − x0 ) + 2 (y − y0 ) − 2 (z − z0 ) = 0, 2 a b c
81

which is equivalent to
2 2xx0 2yy0 2zz0 2x0 2y2 2z2 + 2 − 2 = 2 + 20 − 20 . a2 b c a b c Since P lies on the hyperboloid, the right side is equal to 2 and we can rewrite the last equation in the form xx0 yy0 zz0 + 2 − 2 = 1. a2 b c

Example 10.2. Find the equation of the tangent plane at a point P(x0 , y0 , z0 ) of the surface z = f (x, y), where f (x, y) is a diﬀerentiable function. Solution. We can rewrite the equation of the surface as f (x, y) − z = 0, which is of the form considered above with F(x, y, z) = f (x, y) − z and k = 0. Since F x (x0 , y0 , z0 ) = f x (x0 , y0 ), Fy (x0 , y0 , z0 ) = fy (x0 , y0 ), Fz (x0 , y0 , z0 ) = −1, we deduce that the equation of the tangent plane to the surface at (x0 , y0 , z0 ) is f x (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) − (z − z0 ) = 0, which is equivalent to the equation stated without proof in §8.3. Therefore, Theorem 8.2 is merely special case of Theorem 10.1. Finally, we deﬁne the normal line to a surface. Definition. Let F(x, y, z) be a diﬀerentiable function and let P(x0 , y0 , z0 ) be a point of the surface Σ with equation F(x, y, z) = k. The normal line to Σ at P is the line that passes through P and is orthogonal to the tangent plane at P. The same reasoning that we used to establish (10.1) shows that the symmetric equations of the normal line to Σ at P is y − y0 z − z0 x − x0 = = , F x (x0 , y0 , z0 ) Fy (x0 , y0 , z0 ) Fz (x0 , y0 , z0 ) with the proper modiﬁcations, if some of the partials vanish at (x0 , y0 , z0 ). 10.2. Tangent planes and normal vectors to parametric surfaces Suppose that Σ is a surface with parametrization r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D) and P0 (x0 , y0 , z0 ) = r(u0 , v0 ) is a point on Σ. We shall focus on curves γ on Σ that are the images of parametric curves u = u(t), v = v(t) (t ∈ I) that lie in D and pass through the point (u0 , v0 ). Definition. The tangent plane to Σ at P is the plane, if it exists, that contains P and the tangent vectors to any smooth curve γ on Σ of the above type.
82

We remark that this deﬁnition, at least at ﬁrst glance, seems more restrictive than the deﬁnition we gave in §8.3 for the tangent plane to a level surface: we look only at some curves on Σ instead of looking at all such curves. However, it can be shown that our deﬁnition and the deﬁnition that refers to all the curves on Σ yield the same notion of tangent plane. The following theorem establishes the existence of the tangent plane at a “smooth” point of a parametric surface. Theorem 10.2. Let r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 , (u, v) ∈ D, be a diﬀerentiable vector function with partials ru = xu e1 + yu e2 + zu e3 , rv = xv e1 + yv e2 + zv e3 ,

and let P(x0 , y0 , z0 ) = r(u0 , v0 ) be a point on the parametric surface Σ given by r(u, v). If n = ru (u0 , v0 ) × rv (u0 , v0 ) 0, then the tangent plane to Σ at the point P exists and is perpendicular to the vector n. Proof. Let γ be a curve on Σ of the type considered in the above deﬁnition. It can be parametrized by means of the vector function f(t) = r(u(t), v(t)) = x(u(t), v(t))e1 + y(u(t), v(t))e2 + z(u(t), v(t))e3 (t ∈ I).

Thus, we can use the chain rule to express its tangent vector f (t) in terms of the derivatives of r(u, v), u(t), and v(t): dy dz df dx = e1 + e2 + e3 dt dt dt dt ∂x du ∂x dv ∂y du = + + e1 + ∂u dt ∂v dt ∂u dt ∂x du ∂y du ∂z du = e1 + e2 + e3 ∂u dt ∂u dt ∂u dt ∂y ∂z dv du ∂x e1 + e2 + e3 + = dt ∂u ∂u ∂u dt

∂y dv ∂z du ∂z dv + e2 + e3 ∂v dt ∂u dt ∂v dt ∂x dv ∂y dv ∂z dv + e1 + e2 + e3 ∂v dt ∂v dt ∂v dt ∂x ∂y ∂z e1 + e2 + e3 = u (t)ru + v (t)rv . ∂v ∂v ∂v

In particular, the tangent vector f (t0 ) at the point P0 of γ is f (t0 ) = u (t0 )ru (u0 , v0 ) + v (t0 )rv (u0 , v0 ). In other words, the tangent vector is the sum of a vector parallel to ru (u0 , v0 ) and a vector parallel to rv (u0 , v0 ). It follows that the cross product n, which is perpendicular to both ru (u0 , v0 ) and rv (u0 , v0 ), will be perpendicular to the tangent vector f (t0 ) to the curve γ. Note that the last vector is independent of the curve and is completely determined by the parametrization of the surface Σ. This establishes the existence of the tangent plane: it is the plane through the point P0 with normal vector n. Example 10.3. Find an equation of the tangent plane to the surface Σ with parametrization r(u, v) = cos u(4 + 2 cos v)e1 + sin u(4 + 2 cos v)e2 + 2 sin ve3 at the point (4, 0, 2) = r(0, π/2).
83

(0 ≤ u, v ≤ 2π)

Solution. We have ru (u, v) = − sin u(4 + 2 cos v)e1 + cos u(4 + 2 cos v)e2 , rv (u, v) = −2 cos u sin ve1 − 2 sin u sin ve2 + 2 cos ve3 , so the normal vector to Σ at (4, 0, 2) is n = (ru × rv )(0, π/2) = 0, 4, 0 × −2, 0, 0 = (4e2 ) × (−2e1 ) = −8(e2 × e1 ) = 8e3 . Thus, the equation of the tangent plane is 8(z − 2) = 0 ⇐⇒ z = 2. Example 10.4. Find the equation of the tangent plane at a point P(x0 , y0 , z0 ) of the surface z = f (x, y), where f (x, y) is a diﬀerentiable function. Solution. We can parametrize the given surface as r(x, y) = xe1 + ye2 + f (x, y)e3 ((x, y) ∈ D), so the normal vector n to it is e1 e2 e3 1 fx 1 0 0 fx e = − f x e1 − fy e2 + e3 . e − e + n = r x × ry = 1 0 f x = 1 fy 1 0 fy 2 0 1 3 0 1 fy Thus, the equation of the tangent plane at (x0 , y0 , z0 ) is − f x (x0 , y0 )(x − x0 ) − fy (x0 , y0 )(y − y0 ) + (z − z0 ) = 0, which is equivalent to the equation we gave earlier in Theorem 8.2 and Example 10.2. 10.3. Smooth and piecewise smooth surfaces Recall that we called the parametric curve γ given by the vector function r(t), t ∈ I, smooth if the tangent vector to the curve, r (t), does not vanish for t ∈ I. We later saw that this condition ensures that γ has a tangent line at each point. In the next deﬁnition, we deﬁne a “smooth surface” so that it has a tangent plane at each point. Definition. The level surface Σ with equation F(x, y, z) = k is called smooth, if the function F(x, y, z) is diﬀerentiable at every point of Σ and F(x, y, z) 0, except possibly at the boundary of Σ. The parametric surface Σ, given by r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D), (10.2) 0, except is called smooth, if the partials ru and rv are deﬁned for all (u, v) ∈ D and ru × rv possibly at the boundary of D.

Note that it is an immediate consequence of this deﬁnition and Theorems 10.1 and 10.2 that a smooth surface (either level or parametric) has a tangent plane at each point. Definition. The level surface Σ with equation F(x, y, z) = k is called piecewise smooth, if the function F(x, y, z) is continuous and we can represent Σ as the union of a ﬁnite number of smooth surfaces, say Σ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn . The parametric surface Σ, given by (10.2) is called piecewise smooth, if r(u, v) is continuous and we can represent Σ as the union of a ﬁnite number of smooth surfaces, say Σ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn .
84

Example 10.5. Determine whether the surface Σ is smooth, piecewise smooth, or neither: (a) Σ is the elliptic paraboloid z = x2 + 4y2 . (b) Σ is the cone z2 = x2 + y2 . First solution. (a) We view Σ as a level surface of F(x, y, z) = z − x2 − 4y2 . Since F(x, y, z) = −2x, −8y, 1 the paraboloid is smooth. (b) We view Σ as a level surface of F(x, y, z) = z2 − x2 − y2 . Since F(x, y, z) = −2x, −2y, 2z = 0 ⇐⇒ x = y = z = 0, 0,

the cone is not smooth. However, we can break Σ into two surfaces: the level surface Σ1 of F(x, y, z) = z2 − x2 − y2 and the level surface Σ2 of F(x, y, z) = z2 − x2 − y2 (z ≤ 0), (z ≥ 0),

each of which is smooth, because the origin is a boundary point for them. Thus, the cone is piecewise smooth. Second solution. (b) We view Σ as a parametric surface given by r(u, v) = v cos ue1 + v sin ue2 + ve3 , where 0 ≤ u ≤ 2π, −∞ < v < ∞. Then ru = −v sin ue1 + v cos ue2 , and ru × rv = (−v sin ue1 + v cos ue2 ) × (cos ue1 + sin ue2 + e3 ) = −v sin2 u(e1 × e2 ) − v sin u(e1 × e3 ) + v cos2 u(e2 × e1 ) + v cos u(e2 × e3 ) = −v sin2 ue3 + v sin ue2 − v cos2 ue3 + v cos ue1 = v cos ue1 + v sin ue2 − ve3 . Hence, ru × rv = 0 when v = 0 and the cone is not smooth. However, we can break Σ into two surfaces: the parametric surface Σ1 given by r(u, v) = v cos ue1 + v sin ue2 + ve3 and the parametric surface Σ2 given by r(u, v) = v cos ue1 + v sin ue2 + ve3 (0 ≤ u ≤ 2π, v ≤ 0), (0 ≤ u ≤ 2π, v ≥ 0), rv = cos ue1 + sin ue2 + e3 ,

each of which is smooth, because the origin is a boundary point for them. Thus, the cone is piecewise smooth.
85

10.4. Orientable surfaces We conclude this lecture with the notion of orientation of a surface, which we shall need in our discussion of surface integrals later in these notes. Intuitively, we think of a surface as a twodimensional object, which is placed in space and thus has an “upper side” and a “lower side.” However, this is not always the case, even for nice, smooth surfaces. We use the concept of orientation of a surface to separate the intuitive, two-sided surfaces from the one-sided ones. However, before giving a formal deﬁnition, we present an example of a surprisingly simple one-sided surface.

C

B

D

A

Figure 10.1. The M¨ bius strip o before gluing

Figure 10.2. after gluing

The M¨ bius strip o

Example 10.6. Take a long strip of paper in the form of a rectangle ABCD, the vertices listed counterclockwise so that AB and CD are the two short sides of the strip (see Figure 10.1). Then twist the paper half a turn and glue the short sides so that A gets glued to C and B to D. The resulting surface (see Figure 10.2) has only one side! Indeed, if you take a paintbrush, started at the cut, and moved along the surface parallel to its edges, you will be able to paint the entire strip without lifting the paintbrush. This surface is known as the M¨ bius strip. You may think at o ﬁrst that the M¨ bius strip is an esoteric example and that “normal” surfaces, given by nice, simple o formulas, don’t behave this way. Think again! The M¨ bius strip is given by a nice, simple formula: o its parametric equations are x = (2 + v cos(u/2)) cos u,
1 where 0 ≤ u ≤ 2π and − 2 ≤ v ≤ 1 . 2

y = (2 + v cos(u/2)) sin u,

z = v sin(u/2),

(10.3)

Definition. Let Σ be a smooth surface in R3 . For each nonboundary point P(x, y, z) on the surface, let n1 (x, y, z) and n2 (x, y, z) be the two unit vectors at P that are normal to Σ. We say that Σ is orientable, if it is possible to select a unit normal vector n(x, y, z) at every nonboundary point of Σ in such a way that n(x, y, z) varies continuously as (x, y, z) varies over Σ. A choice of the unit normal vector n as described above is called an orientation of Σ. Remark. If a surface Σ is orientable, then there are two distinct orientations of that surface. For example, let us choose a distinguished point P0 (x0 , y0 , z0 ) on Σ. Let us select and ﬁx one of the two unit normal vectors at P0 ; we denote it by n0 . Then for each nonboundary point P near P0 only one of the unit normal vectors at P will be close to n0 , the other will be close to −n0 . Thus, the continuity requirement on our choice of normal vectors settles the orientation of all normal vectors near P0 . Of course, once we have decided on the choice of normal vectors at points near P0 , we
86

can repeat the same procedure to settle the orientation of the normal vectors at points that may not be near P0 but are near points that are near P0 ; after that, we can settle the choices at points that may not be near P0 or near points that are near P0 , but are near points that are near points that are near P0 ; etc. Ultimately, after we choose between the two unit normal vectors n0 and −n0 at P0 , we set a domino eﬀect that yields a consistent choice between the unit normal vectors at all other nonboundary points of Σ. r[u_,v_] := {2Cos[u]Sin[v], 2Sin[u]Sin[v], 2Cos[v]} n[u_,v_] := 0.5*r[u,v] p = ParametricPlot3D[r[u,v], {u,0,2Pi}, {v,0,Pi}, Boxed -> False, Axes -> False, Mesh -> None, ViewPoint -> {6,1,4}]; Manipulate[ Show[{p, vector[r[u,v], r[u,v] + n[u,v], Red]}, PlotRange -> All], {{u,0}, 0, 4Pi}, {{v,0}, 0, Pi}] r[u_,v_] := {Cos[u](2 + v*Cos[u/2]), Sin[u](2 + v*Cos[u/2]), v*Sin[u/2]} n[u_,v_] = 0.5*Cross[D[r[u,v], u], D[r[u,v], v]] p = ParametricPlot3D[r[u,v], {u,0,2Pi}, {v,-1/2,1/2}, Boxed -> False, Axes -> False, Mesh -> None, ViewPoint -> {6,1,4}]; Manipulate[ Show[{p, vector[r[u,v], r[u,v] + n[u,v], Red]}, PlotRange -> All], {{u,0}, 0, 4Pi}, {{v,0}, -1/2, 1/2}]

Figure 10.3. Normal vectors of orientable and non-orientable surfaces Example 10.7. A good way to illustrate the concept of orientability is to compare the normal vectors on an orientable surface to those on a non-orientable surface. Such a comparison is most
87

successful, if it is undertaken using graphical software like Mathematica, so that is what we will do. In this example, we will compare the normal vectors to the sphere x2 + y2 + z2 = 4 and the normal vectors to the M¨ bius strip with parametric equations (10.3). o The ﬁrst block of Mathematica code in Figure 10.3 produces a dynamic plot of the sphere (parametrized as in Example 10.9 below) and of the outward unit normal vector at its north pole. One can then use the sliders to move the normal vector on the surface of the sphere. Execution that code1 in Mathematica and experimentation with the sliders will demonstrate that as one moves the normal vector on the sphere, its direction changes but always points away from the center of the sphere. Furthermore, no matter how the normal vector meanders around, every time it returns to a speciﬁc point it points out. This is because the sphere is orientable (see Example 10.9 below); the normal vectors Mathematica displays are the vectors of its “outward orientation.” The second block of code in Figure 10.3 produces a similar dynamic plot of the M¨ bius strip o and one of its normal vectors. However, experimentation with its output will demonstrate that as one moves the normal vector on the M¨ bius strip, the vector may be positioned at the same location o with opposite directions: e.g., when (u, v) = (0, 0) and when (u, v) = (2π, 0) (see the screenshots in Figure 10.3). This is because the M¨ bius strip is non-orientable. o Example 10.8. Let Σ be the graph of a function z = f (x, y), (x, y) ∈ D. In Example 10.4, we used the parametrization r(x, y) = xe1 + ye2 + f (x, y)e3 to show that r x × ry = − f x e1 − fy e2 + e3 is a normal vector to the surface at the point (x, y, z) of the surface. Note that this vector always has a positive third component. Thus, the normalized cross product n= − f x e1 − fy e2 + e3 r x × ry = 1/2 , r x × ry 1 + f x2 + fy2 ((x, y) ∈ D)

will pick consistently one of the two unit normal vectors to the surface: namely, it will always pick the unit normal that points upward. Consequently, n represents an orientation of the surface; this orientation is often referred to as the upward orientation of the surface. Example 10.9. Let Σ be the sphere x2 + y2 + z2 = 4. Its spherical equation is ρ = 2, so we can describe Σ as the set of points (x, y, z) in space whose Cartesian coordinates are related to their spherical coordinates via the equations x = 2 cos θ sin φ, y = 2 sin θ sin φ, z = 2 cos φ,

where 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. This provides a parametrization of Σ: r(θ, φ) = 2 cos θ sin φ e1 + 2 sin θ sin φ e2 + 2 cos φ e3 , where 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. Then rθ = −2 sin θ sin φ e1 + 2 cos θ sin φ e2 ,
1

rφ = 2 cos θ cos φ e1 + 2 sin θ cos φ e2 − 2 sin φ e3 ,

The code uses the vector command described in §2.9, so that code needs to be included as well.
88

and rθ × rφ = (−2 sin θ sin φ e1 + 2 cos θ sin φ e2 ) × (2 cos θ cos φ e1 + 2 sin θ cos φ e2 − 2 sin φ e3 ) = −4 sin2 θ sin φ cos φ(e1 × e2 ) + 4 sin θ sin2 φ(e1 × e3 ) + 4 cos2 θ sin φ cos φ(e2 × e1 ) − 4 cos θ sin2 φ(e2 × e3 ) = −4 cos θ sin2 φ e1 − 4 sin θ sin2 φ e2 − 4 sin2 θ + cos2 θ sin φ cos φ e3 = −4 sin φ cos θ sin φ e1 + sin θ sin φ e2 + cos φ e3 = −2 sin φ r. Hence, the normal vectors at a point of the sphere are parallel to the position vector r of the point. Since the length of r is the distance from the origin to a point on Σ, we have r = 2. Thus, n = 1 r 2 is a unit normal vector at the point P = r(u, v) of Σ. Note that since n has the same direction as − − → the OP, it will consistently be the unit normal vector to the sphere that is pointed away from the origin. Therefore, n yields an orientation of Σ that is often called the outward orientation of the sphere. Exercises
Find the equations of the tangent plane and of the normal line to the given surface at the given point. 10.1. z = xye x+2y , P(2, −1, −2) 10.2. z = log2 xy2 + x2 y , P(1, −2, 1) 10.3. x2 − 2y2 + 2z2 = 10, P(2, −1, −2) 10.4. ln(xyz) = x + yz−1 , P(1, e, e) 10.5. (a) Find the equation of the normal line to the sphere x2 + y2 + z2 = r2 at a point (x0 , y0 , z0 ) of the sphere. (b) Show that the normal line from part (a) passes through the origin. √ √ √ 10.6. (a) Find the equation of the tangent plane to the surface x + y + z = 4 at a point P(x0 , y0 , z0 ) on this surface. (b) Show that the sum of the lengths of the line segments that the tangent plane from part (a) cuts from the coordinate axes is independent of the choice of P. (The line segment that a plane cuts from a coordinate axis is the segment whose endpoints are the origin and the intersection point of the plane and the axis.) 10.7. Let P(x0 , y0 , z0 ) be a point on the surface xyz = 6. Show that the volume of the tetrahedron that the tangent plane at P cuts from the ﬁrst octant is independent of the choice of P. 10.8. Find the points on the hyperboloid x2 − 2y2 − z2 = −2 at which the tangent plane is parallel to the plane 2x − 3y + 2z + 7 = 0. 10.9. Find the points on the hyperbolic paraboloid x = y2 − 4z2 at which the normal line is perpendicular to the plane x + y + z = 1. 10.10. Show that the hyperboloid x2 − 2y2 − z2 = −2 and the ellipsoid x + 4 + y + 1 + 1 z + 1 = 1 are tangent 3 3 4 3 to each other at the point (−1, −1, 1), that is, show that they have the same tangent plane at that point. Find an equation of the tangent plane to the given parametric surface at the given point. 10.11. r(u, v) = uv, u sin v, u cos v , (u, v) = (1, π/2) 10.12. r(y, z) = 4 + y2 + z2 , y, z , (y, z) = (1, 1) 10.13. r(u, v) = uv, u2 + v2 , u + v + 2 , (u, v) = (−1, 0) 10.14. r(u, v) = u2 − v2 , u + v, u2 + 2v , (0, 2, 3) √ 10.15. r(u, v) = u cos 2v, u sin 2v, v , (1, 3, π/6) 10.16. Let Σ be the parametric surface given by r(u, v) = (u + v)e1 + (uv)e2 + (u3 − v3 )e3
89
2 2 2

(−∞ < u, v < ∞).

(a) Find the point P on Σ that corresponds to the values u = 1 and v = 1 of the parameters. (b) Find the equation of the tangent plane to Σ at the point P found in (a). 10.17. Let Σ be the parametric surface given by r(u, v) = (u + v)e1 + (u2 + v2 )e2 + (u3 + v3 )e3 (−∞ < u, v < ∞). Find the point(s) on Σ where the tangent plane is perpendicular to the line x = y = −z. Determine whether the given surface is smooth, piecewise smooth, or neither. 10.18. x2 + 3y2 − 2xyz = 6 10.19. r(u, v) = 3u, u2 − 2v, u3 + v2 , (u, v) ∈ R2 10.20. r(u, v) = u, uv, sin uv , (u, v) ∈ R2 10.21. z2 + sin(xy) = 1 10.22. Find the point(s) where the parametric surface given by r(u, v) = (u + v)e1 + (uv)e2 + (u2 − v2 )e3 is not smooth. (−∞ < u, v < ∞)

90

LECTURE 11

Extremal Values of Multivariable Functions
In this lecture, we shall discuss the local and absolute (or global) maximum and minimum values of a function of two variables f (x, y). 11.1. Review of extremal values of single-variable functions We start with a review of the extrema of single-variable functions. For example, let us classify the local and global extrema of the function f (x) = x4 − 8x3 + 18x2 − 10 in the interval [−1, 2]. First, we ﬁnd the critical points of f : these are the zeros of f (x) = 4x3 − 24x2 + 36x = 4x(x2 − 6x + 9) = 4x(x − 3)2 . We have 4x(x − 3)2 = 0 f (x) = 12x2 − 48x + 36 ⇐⇒ =⇒ x = 0 or x = 3. f (0) = 36 > 0, f (3) = 0, We determine whether the critical points yield local extrema by the second-derivative test: that is, f has a local minimum at x = 0, (0, −10). (We can also show that f has no extremum at x = 3, but that does not follow from the second derivative test.) Finally, to ﬁnd the absolute extrema of f in [−1, 2], we compare the local extrema and the boundary values: f (−1) = 17, f (0) = −10, and f (2) = 14. We ﬁnd that the absolute maximum of f is 17 and its absolute minimum is −10. 11.2. Local extrema of a function of two variables Definition. Let f be a function of two variables. If (a, b) is a point in the domain of f , then f has a local maximum at (a, b) if f (x, y) ≤ f (a, b) for all points (x, y) near (a, b); f has a local minimum at (a, b) if f (x, y) ≥ f (a, b) for all points (x, y) near (a, b). A local maximum or minimum is called also a local extremum. Theorem 11.1. If f has a local extremum at (a, b) and its ﬁrst-order partials f x (a, b) and fy (a, b) exist, then f x (a, b) = fy (a, b) = 0. Proof. If f (x, y) has a local maximum/minimum at (a, b), then the single-variable function g(x) = f (x, b) also has a local maximum/minimum at x = a. Hence, g (a) = 0 =⇒ f x (a, b) = 0. Similarly, fy (a, b) = 0 by considering h(y) = f (a, y).
91

Definition. Let f be a function of two variables. A critical point of f is a point (a, b) such that either f x (a, b) = fy (a, b) = 0, or one of these partial derivatives does not exist. Those critical points (a, b) of f for which f x (a, b) = fy (a, b) = 0 are sometimes called stationary points. Example 11.1. Find all critical points of the function f (x, y) = x3 + y2 − 6xy + 6x + 3y − 2. Solution. This function has ﬁrst-order (and also second-, third-, and higher-order) partial derivatives at any point (x, y) in the R2 , so the only critical points will be the stationary points of f . Since f x (x, y) = 3x2 − 6y + 6 and fy (x, y) = 2y − 6x + 3, the stationary points of f are the solutions of the system 3x2 − 6y + 6 = 0 2y − 6x + 3 = 0 x2 − (6x − 3) + 2 = 0 ⇐⇒ x2 − 2y + 2 = 0 2y = 6x − 3 .

Substituting 2y from the second equation into the ﬁrst, we ﬁnd that x must satisfy the equation ⇐⇒ x2 − 6x + 5 = 0 ⇐⇒ x = 1 or x = 5. When we substitute x = 1 in the second equation of the system, we obtain 2y = 3 so y = 1.5; when we substitute x = 5, we obtain 2y = 27 so y = 13.5. Thus, the critical points of f are (1, 1.5) and (5, 13.5). 11.3. The second-derivative test Note that the theorem in the previous section says that (just as in the case of functions of a single variable) a function of two variables can have a local extremum only at a critical point. However, we know that for a function of one variable a critical point is not necessarily an extremal point. Is this also the case for functions of two variables? And if so, then how can we determine whether a critical point is an extremal point or not? The following theorem is often suﬃcient to answer these questions. It is the two-variable version of the second-derivative test. Theorem 11.2 (Second-derivative test). Suppose that f (x, y) has continuous second-order partial derivatives near the stationary point (a, b) and deﬁne D = D(a, b) = f xx (a, b) fyy (a, b) − f xy (a, b)2 = Then: i) if D > 0 and f xx (a, b) > 0, f has a local minimum at (a, b); ii) if D > 0 and f xx (a, b) < 0, f has a local maximum at (a, b); iii) if D < 0, f does not have a local extremum at (a, b). In case iii), the point (a, b) is called a saddle point, since under those conditions the graph of f near (a, b) resembles a saddle. Example 11.2. Find the local extrema of the function f (x, y) = x3 + y2 − 6xy + 6x + 3y − 2.
92

f xx (a, b) f xy (a, b) . fyx (a, b) fyy (a, b)

Solution. We know from Example 9.1 that the critical points of f are (1, 1.5) and (5, 13.5). The second-order partials of f are: f xx (x, y) = 6x, so f xx (5, 13.5) = 30 and D(1, 1.5) = 6 −6 = −24, −6 2 D(5, 13.5) = 30 −6 = 24. −6 2 fyy (x, y) = 2, f xy (x, y) = −6,

Hence, by the second-derivative test, (1, 1.5) is a saddle point of f and f (5, 13.5) = −29 is a local minimum. Example 11.3. Find the point on the plane 2x − y + 2z − 3 = 0 that is closest to the origin. What is the distance between the plane and the origin? Solution. We want to minimize the function d(x, y, z) = x2 + y2 + z2 for (x, y, z) subject to the constraint 2x − y + 2z − 3 = 0. This is the same as minimizing the function f (x, z) = x2 + (2x + 2z − 3)2 + z2 = 5x2 + 8xz + 5z2 − 12x − 12z + 9 in the xz-plane. The ﬁrst-order partials of f are f x (x, z) = 10x + 8z − 12, so the critical points are the solutions of the system 10x + 8z − 12 = 0 8x + 10z − 12 = 0 ⇐⇒ 5x + 4z = 6 4x + 5z = 6 . fz (x, z) = 8x + 10z − 12,

2 Solving the last system, we ﬁnd that the only critical point of f is ( 3 , 2 ). The second-order partials 3 of f are

f xx (x, z) = fzz (x, z) = 10, so
2 D = D( 3 , 2 ) = 3

f xz (x, z) = fzx (x, z) = 8, 10 8 = 36. 8 10

Since D > 0 and f xx > 0, we conclude that f ( 2 , 2 ) = 4 is a local minimum of f . It is intuitively 3 3 3 clear that this local minimum must actually be an absolute minimum of f , so the minimum distance occurs when the point (x, y, z) on the plane is ( 2 , − 1 , 2 ). The distance from the origin to that point 3 3 3 (which is the distance from the origin to the plane) is (2/3)2 + (−1/3)2 + (2/3)2 = 1.
93

11.4. Absolute extrema of a function of two variables In real life, we are more likely to be interested in absolute (or global) extrema of functions than in local extrema. In the last example, we used practical considerations (i.e., intuition) to make the jump from the local to the global minimum. Is there another (hopefully, more rigorous) way to approach global extrema? The answer is “yes,” if the function is deﬁned on a “closed, bounded set.” First, we must say what does it mean for a set in the plane to be “closed and bounded.” Instead of formal deﬁnitions we will be content with explanations by example. A set R is closed if it contains all its “boundary points.” For example, among the sets on Figure 11.1 (where solid boundary curve belongs to the set and a dotted curve does not), the sets in (a) and (d) are closed and the sets in (b), (c), and (e) are not. A set R is bounded if it is contained inside some disk centered at the origin. For example, of the sets on Figure 11.1, only (d) is not bounded.

(a)

(b)

(c)

(d)

(e)

Figure 11.1. Examples of closed and bounded sets We can now state the algorithm for ﬁnding the absolute extrema of a function of two variables on a closed, bounded set. It is a generalization of the closed interval method for ﬁnding absolute extrema of functions of a single variable on a closed interval. To ﬁnd the absolute extrema of f (x, y) on a closed, bounded set R: 1. Find the critical points of f inside R and the respective values of f . 2. Find the absolute extrema of f on the boundary of R.
94

3. The absolute maximum of f in R is the largest among the numbers found in Steps 1 and 2; the absolute minimum of f in R is the smallest among the numbers found in Steps 1 and 2. Example 11.4. Find the absolute maximum and minimum values of the function f (x, y) = x3 − y3 − 3x + 12y in the closed and bounded region R = (x, y) : 0 ≤ x ≤ 2, −3x ≤ y ≤ 0 . Solution. 1. The ﬁrst-order partials of f are f x (x, y) = 3x2 − 3, 3x2 − 3 = 0 −3y2 + 12 = 0 fy (x, y) = −3y2 + 12, x2 = 1 y2 = 4 x = ±1 y = ±2

so the critical points of f are the solutions of the system ⇐⇒ ⇐⇒ .

That is, f has four critical points: (1, 2), (1, −2), (−1, 2), and (−1, −2). However, of these only P(1, −2) lies inside R. The respective value of f is f (1, −2) = −18. 2. The boundary of R consists of three line segments (see Figure 11.2):
1 2 3

: : :

(x, y) : 0 ≤ x ≤ 2, y = 0 , (x, y) : 0 ≤ x ≤ 2, y = −3x , (x, y) : x = 2, −6 ≤ y ≤ 0 .

2.1. When (x, y) lies on

1,

f (x, y) is really only a function of x: (0 ≤ x ≤ 2).

g1 (x) = f (x, 0) = x3 − 3x

The critical values of g1 are the solutions of 3x2 − 3 = 0. These are x = ±1, but we are interested only in x = 1 (since −1 lies outside [0, 2]). Thus, we need compare g1 (0) = 0, g1 (1) = −2, and
1

g1 (2) = 2. are f (1, 0) = −2 and f (2, 0) = 2.

We ﬁnd that the absolute minimum and maximum of f on y
1

x

P
3 2

Figure 11.2. The boundary of R
95

2.2. When (x, y) lies on

2,

f (x, y) is again a function of x only:

(0 ≤ x ≤ 2). √ The critical values of g2 are the solutions√ 84x2 −39 = 0: x = ± 13/28. Again, we are interested of only in one of these critical points: x = 13/28. We compare g2 (0) = 0, g2 13/28 = −26 13/28 = −17.71 . . . ,
2

g2 (x) = f (x, −3x) = 28x3 − 39x

and

g2 (2) = 146.

We ﬁnd that the absolute minimum and maximum of f on f( 13/28, −3
3,

are f (2, −6) = 146.

13/28) = −17.71 . . .

and

2.3. When (x, y) lies on

f (x, y) is only a function of y: (−6 ≤ y ≤ 0).

g3 (y) = f (2, y) = −y3 + 12y + 2

The critical values of g3 are the solutions of −3y2 + 12 = 0, y = ±2. After dropping y = 2, we compare g3 (0) = 2, g3 (−2) = −14, and g3 (−6) = 146. We ﬁnd that the absolute minimum and maximum of f on 3 are, respectively, f (2, −2) = −14 and f (2, −6) = 146. 3. Comparing all the values found in Steps 1 and 2, we conclude that
(x,y)∈R (x,y)∈R

min f (x, y) = min{−18, −2, 2, −17.71, 146, −14} = −18 = f (1, −2),

max f (x, y) = max{−18, −2, 2, −17.71, 146, −14} = 146 = f (2, −6).

Example 11.5. Find the absolute maximum and minimum values of the function f (x, y) = x2 − 2xy − 2x + 4y in the closed region bounded by the ellipse γ with parametric equations x = 3 cos t, y = 2 sin t (0 ≤ t ≤ 2π). fy (x, y) = −2x + 4, y= x−1=1 x=2

Solution. 1. The ﬁrst-order partials of f are f x (x, y) = 2x − 2y − 2, 2x − 2y − 2 = 0 −2x + 4 = 0

so the critical points of f are the solutions of the system ⇐⇒ x=y+1 x−2=0 ⇐⇒

Hence, P(2, 1) is the only critical point of f (x, y). It lies inside γ (see Figure 11.3) and f (2, 1) = 4. 2. On γ, we have f (x, y) = f (3 cos t, 2 sin t) = 9 cos2 t − 12 cos t sin t − 6 cos t + 8 sin t, so we have to ﬁnd the maximum and the minimum of g(t) = 9 cos2 t − 12 cos t sin t − 6 cos t + 8 sin t
96

(0 ≤ t ≤ 2π).

y γ P x

Figure 11.3. The ellipse γ and the critical point P We have g (t) = −18 cos t sin t + 12 sin2 t − 12 cos2 t + 6 sin t + 8 cos t = 24 sin2 t + 6 sin t − 12 + 2 cos t(4 − 9 sin t). √ Substituting u = sin t and cos t = ± 1 − u2 , we can rewrite the equation g (t) = 0 as √ 24u2 + 6u − 12 ± 2 1 − u2 (4 − 9u) = 0 √ 12u2 + 3u − 6 = ± 1 − u2 (9u − 4) 12u2 + 3u − 6
2

= (1 − u2 )(9u − 4)2

225u4 − 200u2 + 36u + 20 = 0. The last equation has four real roots, whose approximate values are: α1 ≈ −0.980, α2 ≈ −0.245, α3 ≈ 0.547, α4 ≈ 0.678.

These are the values of sin t at the critical numbers of g(t). They yield a total of eight possible critical numbers: four pairs that have the same sines and opposite cosines. In fact, only one of the numbers in each pair is a critical number of g, but we will ignore this and will treat all eight numbers as critical. The values of all these critical (and pseudo-critical) numbers, their sines and cosines, and the respective values of g(t) are listed in Table 11.1. αk −0.980 −0.980 −0.245 −0.245 0.547 0.547 0.678 0.678 t 4.510 4.915 3.389 6.036 0.579 2.563 0.745 2.397 cos t −0.201 0.201 −0.970 0.970 0.837 −0.837 0.735 −0.735 sin t −0.980 −0.980 −0.245 −0.245 0.547 0.547 0.678 0.678 g(t) −8.630 −6.316 9.470 3.533 0.166 21.201 −0.104 20.681

Table 11.1. Values of g(t) at its critical numbers

97

We also have g(0) = g(2π) = 3, so
0≤t≤2π

max g(t) = g(π − arcsin α3 ) ≈ 21.201,

0≤t≤2π

min g(t) = g(π − arcsin α1 ) ≈ −8.630.

3. Comparing all the values found in Steps 1 and 2, we conclude that the absolute minimum of f inside γ is min{4, −8.630, 21.201} = −8.630, and that the absolute maximum of f is max{4, −8.630, 21.201} = 21.201. Exercises
Find and classify the critical points (as local maxima, local minima, or saddle points) of the given function. 11.1. f (x, y) = 9 + x + 2y − x2 + 1 y2 2 11.2. f (x, y) = 2x − 4y − x2 − 2y2 11.3. f (x, y) = x4 + y4 − 4xy + 2 11.4. f (x, y) = xy(x + y + 1) 11.5. f (x, y) = e2y sin(xy) 2 2 11.6. f (x, y) = xye−x −y 11.7. f (x, y) = x2 − 6xy + 4y3 11.8. f (x, y) = x3 − 6xy + 8y3 11.9. f (x, y) = x3 + y4 − 3xy + 4 Find the absolute maxima and minima of the given function on the given set. 11.10. f (x, y) = 1 + 2x + 3y on the triangle with vertices (0, 3), (2, 1), and (5, 3) 11.11. f (x, y) = x2 + 4y2 − xy on the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1 11.12. f (x, y) = x3 + 2xy2 − 2x on the disk x2 + y2 ≤ 4 11.13. f (x, y) = x3 + y2 − 2xy + 3 on the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 11.14. Find the shortest distance from the point (3, 2, −1) to the plane x − y + 2z = 1. 11.15. Find the point(s) on the cone (z − 2)2 = x2 + y2 that is closest to the point (1, 1, 2). 11.16. Among all the rectangular boxes of volume 125 ft3 , ﬁnd the dimensions of that which has the least surface area. 11.17. Find the volume of the largest rectangular box with faces parallel to the coordinate planes that can be inscribed in the ellipsoid x2 + 4y2 + 4z2 = 16.

98

LECTURE 12

Lagrange Multipliers*
12.1. Introduction Recall Example 11.3. In that example, we had to minimize the function f (x, y, z) = x2 + y2 + z2 subject to the constraint 2x − y + 2z − 3 = 0. We used the constraint to eliminate one of the variables and reduced the problem to that of minimizing a related function of two variables (without any constraints). On an abstract level, one expects that the same idea should work if one wants to minimize or maximize any “nice” function f (x, y, z) subject to the constraint that g(x, y, z) = k, where g is another “nice” function and k is a constant. However, while such an approach is theoretically possible, for any practical purpose it is pretty much doomed unless g is a very simple function (such as the linear function that appears in the above example). In this lecture, we describe the method of Lagrange multipliers, which allows us to solve the general problem. 12.2. Basic form of the method of Lagrange multipliers Suppose that P(x0 , y0 , z0 ) is a point at which the function f (x, y, z) is maximal subject to the constraint g(x, y, z) = k. This constraint describes a level surface, which we will denote by Σ. Thus, P is a point on the surface Σ such that f (x0 , y0 , z0 ) is maximal. Let γ be a smooth parametric curve on Σ parametrized by the vector function r(t) = x(t), y(t), z(t) , t ∈ I, and such that r(t0 ) = P (so that, γ passes through P). Since f (x0 , y0 , z0 ) is maximal on Σ and γ lies on Σ, we have f (x0 , y0 , z0 ) ≥ f (x, y, z) Using the parametrization of γ, we get f (x(t0 ), y(t0 ), z(t0 )) ≥ f (x(t), y(t), z(t)) for all t ∈ I, so t = t0 is a local maximum of the single-variable function h(t) = f (x(t), y(t), z(t)), t ∈ I. Therefore, h (t0 ) = 0. On the other hand, by the chain rule, h (t) = f x (x(t), y(t), z(t))x (t) + fy (x(t), y(t), z(t))y (t) + fz (x(t), y(t), z(t))z (t) = f (x(t), y(t), z(t)) · r (t), so the equation h (t0 ) = 0 can be rewritten as f (x0 , y0 , z0 ) · r (t0 ) = 0. That is, the gradient of f at the point P is perpendicular to the tangent vector r (t0 ) to γ at P. Since γ was an arbitrary curve on Σ, this shows that the gradient f (P) is perpendicular to all tangent
99

for all (x, y, z) on γ.

vectors to Σ. Hence, f (P) is parallel to the normal vector to Σ at P—which is g(P). We conclude that if P(x0 , y0 , z0 ) is on Σ and f (x0 , y0 , z0 ) is maximal, then f (x0 , y0 , z0 ) = λ g(x0 , y0 , z0 ) for some number λ. We note that if we had started with a point P(x0 , y0 , z0 ) on Σ such that f (x0 , y0 , z0 ) is minimal, we could have used a similar argument to reach the same conclusion. These observation motivate the following approach towards constrained optimization. To ﬁnd the extrema of a function f (x, y, z) subject to the constraint g(x, y, z) = k (when g 0 on the surface g(x, y, z) = k): 1. Find all the values x, y, z, and λ such that f (x, y, z) = λ g(x, y, z), g(x, y, z) = k. That is, solve this system of four equations in the four unknowns x, y, z, λ. 2. Evaluate f at all the points (x, y, z) found in Step 1. The maximum of f is the largest among these numbers; the minimum of f is the smallest among them. Example 12.1. Find the maximum and minimum values of the function f (x, y, z) = 8x − 4z on the ellipsoid x2 + 10y2 + z2 = 5. Solution. Let g(x, y, z) = x2 + 10y2 + z2 . We have f x (x, y, z) = 8, and g x (x, y, z) = 2x, gy (x, y, z) = 20y, gz (x, y, z) = 2z, so the system in Step 1 of the method of Lagrange multipliers is   8 = 2λx  x = 4/λ     0 = 20λy y = 0 ⇐⇒   −4 = 2λz z = −2/λ    2  2 2 2 x + 10y + z = 5 x + 10y2 + z2 = 5 Here we have used that λ 0, because of the ﬁrst or third equations. Substituting the expressions for x, y, and z in the last equation of the system, we ﬁnd that λ must be a solution of 16/λ2 + 0 + 4/λ2 = 5 ⇐⇒ λ2 = 4 ⇐⇒ λ = ±2. fy (x, y, z) = 0, fz (x, y, z) = −4,

The respective points (x, y, z) are (2, 0, −1) and (−2, 0, 1), so the conditional maximum of f is max{ f (2, 0, −1), f (−2, 0, 1)} = max{20, −20} = 20 = f (2, 0, −1), and the conditional minimum of f is min{ f (2, 0, −1), f (−2, 0, 1)} = min{20, −20} = −20 = f (−2, 0, 1). Example 12.2. Find the maximum and minimum values of the function f (x, y, z) = x3 + y3 + z3 on the sphere x2 + y2 + z2 = 1.
100

Solution. Let g(x, y, z) = x2 + y2 + z2 . We have f x (x, y, z) = 3x2 , and g x (x, y, z) = 2x, gy (x, y, z) = 2y, gz (x, y, z) = 2z,   x = 0, 2λ/3   y = 0, 2λ/3 z = 0, 2λ/3    2 x + y2 + z2 = 1 so the system in Step 1 of the method of Lagrange multipliers is    x(x − 2λ/3) = 0 3x2 = 2λx     2 y(y − 2λ/3) = 0 3y = 2λy ⇐⇒ ⇐⇒ z(z − 2λ/3) = 0 3z2 = 2λz      2  2 x + y2 + z2 = 1 x + y2 + z2 = 1 fy (x, y, z) = 3y2 , fz (x, y, z) = 3z2 ,

For each of the eight choices for x, y, z from the ﬁrst three equations, we now solve the fourth equation for λ: 1. x = y = z = 0. Then the fourth equation fails and we get no solution of the system. 2. x = y = 0, z = 2λ/3. Then the fourth equation becomes 4λ2 /9 = 1 ⇐⇒ λ = ±3/2,

and we obtain the points (0, 0, 1) and (0, 0, −1). 3. x = z = 0, y = 2λ/3. Then the fourth equation becomes 4λ2 /9 = 1 ⇐⇒ λ = ±3/2,

and we obtain the points (0, 1, 0) and (0, −1, 0). 4. y = z = 0, x = 2λ/3. Then the fourth equation becomes 4λ2 /9 = 1 5. ⇐⇒ λ = ±3/2,

6.

7.

8.

and we obtain the points (1, 0, 0) and (−1, 0, 0). x = 0, y = z = 2λ/3. Then the fourth equation becomes √ 8λ2 /9 = 1 ⇐⇒ λ = ±3 2/4, √ √ √ √ and we obtain the points (0, 2/2, 2/2) and (0, − 2/2, − 2/2). y = 0, x = z = 2λ/3. Then the fourth equation becomes √ 8λ2 /9 = 1 ⇐⇒ λ = ±3 2/4, √ √ √ √ and we obtain the points ( 2/2, 0, 2/2) and (− 2/2, 0, − 2/2). z = 0, x = y = 2λ/3. Then the fourth equation becomes √ 8λ2 /9 = 1 ⇐⇒ λ = ±3 2/4, √ √ √ √ and we obtain the points ( 2/2, 2/2, 0) and (− 2/2, − 2/2, 0). x = y = z = 2λ/3. Then the fourth equation becomes √ 12λ2 /9 = 1 ⇐⇒ λ = ± 3/2, √ √ √ √ √ √ and we obtain the points ( 3/3, 3/3, 3/3) and (− 3/3, − 3/3, − 3/3).
101

A list of the values of f at the above points is √ √ √ √ 1, −1, 1/ 2, −1/ 2, 1/ 3, −1/ 3, so the minimum and the maximum of f on the sphere are −1 and 1, respectively. The minimum is attained at the points of intersection of the sphere with the negative directions of the coordinate axes and the maximum is attained at the points of intersection of the sphere with the positive directions of the coordinate axes. 12.3. Lagrange multipliers in the case of two constraints** Sometimes, one needs to ﬁnd the extrema of a function f (x, y, z) subject to multiple constraints: say, g(x, y, z) = k, h(x, y, z) = m. The variant of the method of Lagrange multipliers that addresses this question is as follows: 1. Find all the values x, y, z, λ, and µ such that   f (x, y, z) = λ g(x, y, z) + µ h(x, y, z)  . g(x, y, z) = k  h(x, y, z) = m That is, solve this system of ﬁve equations in the ﬁve unknowns x, y, z, λ, µ. 2. Evaluate f at all the points (x, y, z) found in Step 1. The maximum of f is the largest among these numbers; the minimum of f is the smallest among them. Example 12.3. Find the maximum and minimum values of f (x, y, z) = 3x − y − 3z on the intersection curve of the plane x + y − z = 0 and the elliptic cylinder x2 + 2z2 = 1. Solution. Let g(x, y, z) = x + y − z, The partials of f, g, h are f x = 3, g x = 1, h x = 2x, so we have to solve the system   3 = λ + 2µx   −1 = λ   −3 = −λ + 4µz  x + y − z = 0     2 x + 2z2 = 1 fy = −1, gy = 1, hy = 0,  λ = −1   2µx = 4   4µz = −4  y = z − x     2 x + 2z2 = 1 fz = −3, gz = −1, hz = 4z,  λ = −1    x = 2µ−1   z = −µ−1  y = −3µ−1     2 x + 2z2 = 1 h(x, y, z) = x2 + 2z2 .

⇐⇒

⇐⇒

Substituting the expressions for x and z into the ﬁfth equation, we ﬁnd that µ must satisfy the equation √ 4µ−2 + 2µ−2 = 1 ⇐⇒ µ2 = 6 ⇐⇒ µ = ± 6.
102

√ √ √ √ √ √ These give the points (2/ 6, −3/ 6, −1/ 6) and (−2/ 6, 3/ 6, 1/ 6). Since √ √ 2 −3 −1 −2 3 1 12 f √ , √ , √ and f √ , √ , √ = √ =2 6 = −2 6, 6 6 6 6 6 6 6 √ √ we conclude that the maximum of f on the given curve is 2 6 and that its minimum is −2 6. Exercises
Use the method of Lagrange multipliers to ﬁnd the maximum and minimum values of the function f subject to the given constraint. 12.1. f (x, y) = x2 y; x2 + y4 = 1 12.2. f (x, y, z) = 3x − 2y + z; 3x2 + 2y2 + z2 = 24 12.3. f (x, y, z) = xyz; x2 + 2y2 + z2 = 3 12.4. Use the method of Lagrange multipliers to ﬁnd the minimum of the function f (x, y, z) = xy + yz + zx subject to the constraints xy = 1, z = x2 + y2 + 1, and x > 0. What can you say about the maximum of f subject to the same constraints? 12.5. Use the method of Lagrange multipliers to ﬁnd the point(s) on the cone (z − 2)2 = x2 + y2 that is closest to the point (1, 1, 2). 12.6. Use the method of Lagrange multipliers to ﬁnd three positive numbers whose sum is S , where S > 0, and whose product is maximal. 12.7. Maximize n i=1

xi yi subject to the constraints

n i=1

xi2 = 1 and

n 2 i=1 yi

= 1.

103

LECTURE 13

Double Integrals over Rectangles
In this lecture, we generalize the idea of the deﬁnite integral (see Appendix B) to functions of two variables. 13.1. Deﬁnition Let f (x, y) be a bounded function on the rectangle R, R = [a, b] × [c, d] = (x, y) : a ≤ x ≤ b, c ≤ y ≤ d . For each n ≥ 1, let x0 = a, y0 = c, x1 = a + ∆x, y1 = c + ∆y, x2 = a + 2∆x, y2 = c + 2∆y, ..., ..., xn = a + n∆x = b, yn = c + n∆y = d, ∆x = (b − a)/n, ∆y = (d − c)/n, be the points that divide the interval [a, b] into n subintervals of equal lengths; and let be the points that divide the interval [c, d] into n subintervals of equal lengths. Also, pick n2 sample points (xi∗j , y∗j ) so that i xi−1 ≤ xi∗j ≤ xi , y j−1 ≤ y∗j ≤ y j i n n

(1 ≤ i, j ≤ n).

Then the double integral of f over R is deﬁned by f (x, y) dA = lim
R n→∞

f (xi∗j , y∗j )∆A, i i=1 j=1

∆A = ∆x∆y,

(13.1)

provided that the limit exists. The sums on the right side of (13.1) are called Riemann sums of f on R. Remarks. 1. Note that the vertical lines x = xi and y = y j split the rectangle R into n2 congruent subrectangles, which we may label according to their upper right corner points: if 1 ≤ i, j ≤ n, then Ri j is the subrectangle whose upper right corner is the point (xi , y j ). With this labelling of the subrectangles, we can report that each sample point (xi∗j , y∗j ) lies in the subrectangle Ri j with i the same indices and there is exactly one sample point per subrectangle (see the left side of Figure 13.1). Furthermore, the quantity ∆A in (13.1) is equal to the common area of the rectangles Ri j : one n2 th part of the area of R. Thus, each term in the Riemann sum of f is the product of a sample value of f in one of the rectangles Ri j and the area of that rectangle. As the latter product is the volume of a rectangular box with base Ri j and height f (xi∗j , y∗j ), the entire Riemann sum equals the i volume of a solid built by sticking together n2 rectangular boxes (see the right side of Figure 13.1). 2. The limit in (13.1) is supposed to exist and be independent of the choice of the sample points. This is not necessarily true for any old function, but it does hold for functions that are piecewise continuous in R (see §6.7 for the deﬁnition of piecewise continuity).
105

y d yj

(xi∗j , y∗j ) i

c a xi b x

Figure 13.1. The sample points and solid volume of a Riemann sum 3. If f (x, y) ≥ 0 on R, then R f (x, y) dA represents the volume of the solid whose upper boundary is the surface z = f (x, y) and whose lower boundary is R. That is, f (x, y) dA = vol. (x, y, z) : (x, y) ∈ R, 0 ≤ z ≤ f (x, y)
R

= vol. (x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z ≤ f (x, y) . 4. From now on, we shall use the notation |D| to denote the area of a plane region D. Let us divide both sides of (13.1) by |R| = (b − a)(c − d). Since ∆A = |R|/n2 , we have ∆A/|R| = 1/n2 , so we get n n 1 1 f (x, y) dA = lim 2 f (xi∗j , y∗j ). i n→∞ n |R| R i=1 j=1 Since n12 i j f (xi∗j , y∗j ) is the average of the sample values f (xi∗j , y∗j ), we interpret the last equai i tion in terms of the average value of f : 1 f (x, y) dA. (13.2) average f (x, y) = |R| R (x,y)∈R Example 13.1. Evaluate the double integral of the function f (x, y) = over the rectangle R = [0, 3] × [0, 2]. Solution. We shall use the interpretation of the double integral as volume. By Remark 3 above, the value of the integral f (x, y) dA is the volume of the solid E that lies between R and the part of the surface z = f (x, y) with (x, y) ∈ R. The graph of f (x, y) (before the restriction to R) consists of the part of the plane z = 5 − 2x − 3y with 5−2x−3y ≥ 0 and the part of the xy-plane z = 0 with 5−2x−3y ≤ 0. The line with equation 5 5 − 2x − 3y = 0 splits the rectangle R in two parts: a triangle with vertices (0, 0), ( 2 , 0), and (0, 5 ), 3
106
R

5 − 2x − 3y if 5 − 2x − 3y ≥ 0, 0 if 5 − 2x − 3y ≤ 0,

(0, 2) 5 (0, 3 )

(3, 2)

(0, 0)

( 5 , 0) 2

(3, 0)

Figure 13.2. The pyramid E and its base
5 where 5 − 2x − 3y ≥ 0; and a pentagon with vertices ( 5 , 0), (3, 0), (3, 2), (0, 2), and (0, 3 ), where 2 5 − 2x − 3y ≤ 0 (see Figure 13.2). For points (x, y) lying in the pentagon, the graph z = f (x, y) and the xy-plane coincide, and for (x, y) in the triangle, the graph z = f (x, y) is a triangle in the plane 5 z = 5 − 2x − 3y. Thus, the solid E is the pyramid with vertices O(0, 0, 0), A( 2 , 0, 0), B(0, 5 , 0), and 3 C(0, 0, 5). Its volume is 1 1 1 1 1 5 5 125 |OC| · area( OAB) = |OC| · |OA| |OB| = 5 · = . 3 3 2 3 223 36 Hence, 125 f (x, y) dA = . 36 R

13.2. Iterated integrals In practice, we want to reverse the order of things in the above example, so that we are able to compute volumes, averages, etc. by interpreting them as double integrals and then evaluating the latter using some “standard” machinery independent of our interpretation. One such method is the method of iterated integrals. An iterated integral is an expression of one of the forms d c a b b d

f (x, y) dx dy Sometimes, we write these integrals as d c a b

or a c

f (x, y) dy dx.

b

d

f (x, y) dxdy

and a c

f (x, y) dydx,

respectively. Note that these are deﬁnite integrals of functions deﬁned by means of deﬁnite integrals and not double integrals. Thus, we can evaluate such integrals using the familiar methods from single-variable calculus. Example 13.2. Evaluate
1 0 0 3 3 1

x3 y dydx Solution. We have
1 0 0 3 1

and
0 0

x3 y dxdy.

x y dydx =
3 0

x3 y2 2

3

9 dx = 2 0
107

1 0

9x4 x dx = 8
3

1 0

9 = , 8

and
0

3 0

1

3

x3 y dxdy =
0

x4 y 4

1

dy =
0

1 4

3

y dy =
0

y2 8

3 0

9 = . 8

13.3. Fubini’s theorem Notice that in the last example, we evaluated the two possible iterated integrals of x3 y over [0, 1] × [0, 3] and obtained the same value for both despite following diﬀerent routes. Is this always the case? The answer to this question is in the aﬃrmative. Moreover, the common value of the two iterated integrals is the value of the double integral of f over the rectangle. This fact is known as Fubini’s theorem. Theorem 13.1 (Fubini). Let f (x, y) be piecewise continuous in the rectangle R = [a, b] × [c, d]. Then b d d b

f (x, y) dA =
R a c

f (x, y) dydx = c a

f (x, y) dxdy.

Example 13.3. Find the volume of the solid bounded above by the graph of f (x, y) = x3 y and below by the rectangle R = [0, 1] × [0, 3]. Solution. We can express this volume by the double integral x3 y dA.
R

By Fubini’s theorem and by Example 13.2,
1 3 3 1 0

x y dA =
3 R 0 0

x y dydx =
3 0

9 x3 y dxdy = , 8

so the volume of the solid is 9/8. Example 13.4. Evaluate √ ey x + ey dA,
R

where R = [0, 4] × [0, 1].
4 0 0 1

Solution. By Fubini’s theorem, √ ey x + ey dA =
R

√ ey x + ey dydx.

By the substitution u = x + ey , du = ey dy, the inner integral is e+x 1 e+x √ u3/2 2 y 1/2 y dy = e x+e u du = = (x + e)3/2 − (x + 1)3/2 . 3/2 1+x 3 0 1+x Hence, √ 2 4 (x + e)3/2 − (x + 1)3/2 dx ey x + ey dA = 3 0 R 22 4 = (x + e)5/2 − (x + 1)5/2 0 35 4 = (e + 4)5/2 − e5/2 − 55/2 + 1 ≈ 13.308. 15
108

Second solution. Let us try to see how things change, if we use the other iterated integral. We have 4+ey 4 √ 4+ey √ 2 u3/2 = (4 + ey )3/2 − e3y/2 . x + ey dx = u du = 3/2 ey 3 0 ey Hence, Fubini’s theorem yields 1 4 1 4 √ √ √ ey x + ey dA = ey x + ey dxdy = ey x + ey dx dy
R 0

=

2 2 e ey (4 + ey )3/2 − e3y/2 dy = (4 + u)3/2 − u3/2 du 3 0 3 1 4 22 e (4 + u)5/2 − u5/2 1 = (4 + e)5/2 − e5/2 − 55/2 + 1 . = 35 15 Exercises

0 1

0

0

13.1. Use a Riemann sum with n = 3 to estimate the value of sample points to be: (a) the lower right corners of the subrectangles;

R

xy + y2 dA, where R = [1, 2] × [−1, 1]. Take the

(b) the centers of the subrectangles.

Evaluate the given double integral by identifying it as the volume of a solid and calculating that volume using geometric means. 13.2.
R

2 dA, R = [−1, 2] × [0, 3] (3 − x) dA, R = [0, 1] × [0, 1]
R

13.3.

Evaluate the given iterated integral.
2 3 1

13.4.
1 4 3

1 + 2xy2 dxdy x + y ln x dydx y 3 x+2y dydx

13.5.
1 1 1 1

13.6.
0 0

Evaluate the given double integral. 13.7.
R

xy + y2 dA, R = [1, 2] × [−1, 1]
R

13.8. 13.9.
R

x2

xy dA, R = [0, 1] × [0, 1] + y2 + 1

1 sin(3x + 2y) dA, R = [−π, 0] × [− 2 π, 1 π] 4

13.10.
R

2+y dA, R = [0, 1] × [−2, 3] 1 + x2 (x + y)e2xy dA, R = [1, 2] × [0, 1] √ √

13.11.
R

13.12.
R

xy cos x2 + y2 dA, R = 0,

π × 0,

π

109

LECTURE 14

Double Integrals over General Regions
To deﬁne the double integral of a bounded function f (x, y) on a bounded region D in the plane, we use that the deﬁnition given in §13.1 applies to any piecewise continuous function F(x, y) deﬁned on a rectangle (recall Remark 2 in §13.1). 14.1. Deﬁnition Let D be the closed region in R2 bounded by a piecewise smooth curve γ and let f (x, y) be piecewise continuous inside D. Deﬁne the function F(x, y) = f (x, y) if (x, y) ∈ D, 0 otherwise.

Then the double integral of f over D is deﬁned by f (x, y) dA =
D R

F(x, y) dA,

where R is any rectangle containing D. Remark. Note that with the assumption that f (x, y) is piecewise continuous in D, F(x, y) is continuous everywhere except possibly at the points of γ and the (possible) points inside D where f (x, y) is discontinuous. Thus, if f (x, y) is actually continuous in D, the only discontinuities of F(x, y) lie on γ. If f (x, y) is a genuine piecewise continuous function, with discontinuities on some smooth curves γ1 , . . . , γn inside D, then the discontinuities of F(x, y) lie on the curves γ, γ1 , . . . , γn . In either case, F(x, y) is piecewise continuous and R F(x, y) dA is deﬁned by (13.1). 14.2. Properties of the double integral We now state several properties of the double integral which are two-dimensional versions of the properties of the deﬁnite integral listed in Appendix B. Theorem 14.1. Let f (x, y), g(x, y) be piecewise continuous functions and a, b denote constants. Then i)
D

[a f (x, y) + bg(x, y)] dA = a
D

f (x, y) dA + b
D

g(x, y) dA.

ii) If f (x, y) ≤ g(x, y), then f (x, y) dA ≤
D D

g(x, y) dA.

111

iii) If D = D1 ∪ D2 , D1 ∩ D2 = ∅, then f (x, y) dA =
D D1

f (x, y) dA +
D2

f (x, y) dA.

iv)
D

1 dA = |D|, |D| being the area of D.

Furthermore, similar to the double integral over a rectangle, the double integral over a general region D can be interpreted as the volume of a solid or the average value of the function. When f (x, y) ≥ 0, f (x, y) dA = vol. (x, y, z) : (x, y) ∈ D, 0 ≤ z ≤ f (x, y) .
D

Also, for any function f (x, y) that is piecewise continuous on D, we have 1 f (x, y) dA. average f (x, y) = |D| D (x,y)∈D 14.3. Evaluation of double integrals

(14.1)

We now state two formulas which follow from Fubini’s theorem and which we shall use to evaluate double integrals. First, consider a region D of the form D = (x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x) , where g1 and g2 are continuous functions of x. We call such regions type I regions. If f is a continuous function on D, we have b f (x, y) dA =
D a

g2 (x)

f (x, y) dy dx. g1 (x)

(14.2)

We also deﬁne type II regions, which a regions of the form D = (x, y) : c ≤ y ≤ d, g1 (y) ≤ x ≤ g2 (y) , where g1 and g2 are continuous functions of y. If f is a continuous function on such a region, we have d f (x, y) dA =
D c

g2 (y)

f (x, y) dx dy. g1 (y)

(14.3)

Example 14.1. Evaluate ye x dA, where D = (x, y) : 0 ≤ x ≤ 1,
1

√

D

x≤y≤1 .
1 1 √ 1

Solution. Since D is a type I region, (14.2) yields
1 √

ye dA = x D 0

ye dydx = x x 1 0 0 1

y2 e x 2

dx = x 1 2

(1 − x)e x dx
0 1 0

=

1 2

(1 − x)e x

+
0

e x dx

=

1 − 1 + ex 2

=

e−2 . 2

112

(0, 1) D y= (0, 0) √ x

(1, 1)

Figure 14.1. The region D Second solution. Alternatively, we can try to describe D as a type II region and then use (14.3). We start by sketching D: see Figure 14.1. In our type I description of D, we ﬁrst let x vary in the √ range 0 ≤ x ≤ 1, and then for each ﬁxed x, we restrict y to the range x ≤ y ≤ 1. To obtain a type II description, we interchange the roles of x and y: ﬁrst, we let y vary in 0 ≤ y ≤ 1, and then for each ﬁxed y, we restrict x to the range 0 ≤ x ≤ y2 . Here, the limit y2 comes from the description √ of the curve y = x in the form x = x(y): √ y = x ⇐⇒ x = y2 . Thus, D = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y2 , and (14.3) yields
1 y2 1

f (x, y) dA =
D 0 1 0

ye x dxdy =
0 1 du 2

y ex = u y2 0

1

dy =
0 1

y ey − 1 dy

2

=
0

eu − 1

e −u 2

=
0

e−2 . 2

Example 14.2. Evaluate the double integral ey dA,
D
2

where D = (x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1 . Solution. Our initial impulse is to write ey dA =
D 0 x
2

1

1

ey dydx, ey dy is not solvable in elementary
2

2

but this approach is doomed, because the indeﬁnite integral functions. On the other hand, if we observe that

D = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y ,
113

we have ey dA =
D 0 1 0
2

1

y

ey dxdy = eu
0 1 du 2

2

1 0 1 0

ey x 0 dy = = e−1 . 2

2

y

1 0

ey y dy

2

=

=

e 2

u

Example 14.3. Evaluate the double integral 2xy dA,
D

where D is the triangle with vertices (0, 0), (0, 3), and (1, 2). (0, 3)

y=3−x D (1, 2) y = 2x

(0, 0) Figure 14.2. The triangle D Solution. The sides of the triangle D are the lines in the plane determined by the pairs of points (0, 0) and (0, 3), (0, 0) and (1, 2), and (0, 3) and (1, 2). These are the lines with equations x = 0, y = 2x, and y = 3 − x, respectively (see Figure 14.2). Hence, we can express D as a type I region: D = (x, y) : 0 ≤ x ≤ 1, 2x ≤ y ≤ 3 − x . Thus, the given double integral is
1 3−x 1

2xy dA =
D 0 1 2x

2xy dydx =
0

x (3 − x)2 − 4x2 dx 3x4 9x2 − 2x3 − 2 4
1 0

=
0

9x − 6x2 − 3x3 dx =

7 = . 4

Example 14.4. Find the volume of the solid E in the ﬁrst octant that is bounded by the cylinder y2 + z2 = 4 and by the plane x = 2y. Solution. We start by sketching the solid E: the left side of Figure 14.3 shows a joint display of the parts of the cylinder y2 + z2 = 4 and the plane x = 2y that lie in the ﬁrst octant and how they intersect. From that display, we see that E is bounded from below by a triangle D in the xy-plane and from above by the part of the cylinder that lies above D. The triangle D (see the right side of
114

(0, 2) D (0, 0) Figure 14.3. The solid E and its base x = 2y

(4, 2)

Figure 14.3) is the region in the xy-plane bounded by the y-axis, by the line x = 2y, and by the line y = 2, along which the cylinder intersects tha xy-plane. Thus, we see that D = (x, y) : 0 ≤ x ≤ 4,
1 x 2

≤ y ≤ 2 = (x, y) : 0 ≤ y ≤ 2, 0 ≤ x ≤ 2y . 4 − y2 , it follows that 4 − y2 .
2 2y

Since the surface of the cylinder is the graph of z = Hence, by Remark 3 and Fubini’s theorem,
4 2

E = (x, y, z) : (x, y) ∈ D, 0 ≤ z ≤

vol.(E) =
D

4 − y2 dA =
0 x/2

4 − y2 dydx =
0 0

4 − y2 dxdy.

We shall use the latter iterated integral, because its evaluation is less technical:
2 2y 2 0

vol.(E) =
0 0

4−

y2

dxdy =
0

2y

4−

y2

dy =
4

√

u3/2 u (−du) = 3/2

4

=
0

16 . 3

Example 14.5. Reverse the order of integration in the iterated integral
1 0 0 arctan x

f (x, y) dydx. First solution. We have
1 0 0 arctan x

f (x, y) dydx =
D

f (x, y) dA,

where D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ arctan x is the region displayed on Figure 14.4.

y = arctan x D (0, 0)

(1, π/4)

(1, 0)

Figure 14.4. The region D
115

Since y = arctan x ⇐⇒ x = tan y, a reversal of the roles of x and y yields a representation of D as a type II region: D = (x, y) : 0 ≤ y ≤ π/4, tan y ≤ x ≤ 1 . Thus, f (x, y) dA =
D 0 tan y π/4 1

f (x, y) dxdy.

Second solution. We now demonstrate an alternative way for changing D into a type II set. Initially, D is described by the inequalities 0 ≤ x ≤ 1, 0 ≤ y ≤ arctan x. Since arctan x ≤ π/4 for all x, 0 ≤ x ≤ 1, we can add to the above inequalities the redundant inequality y ≤ π/4. Thus, we now have D described by 0 ≤ x ≤ 1, or equivalently, by 0 ≤ x ≤ 1, We observe that y ≤ arctan x ⇐⇒ tan y ≤ x so we can now replace the deﬁnition of D by 0 ≤ x ≤ 1, 0 ≤ y ≤ π/4, ⇐⇒ x ≥ tan y, 0 ≤ y ≤ π/4, y ≤ arctan x. 0 ≤ y ≤ arctan x, y ≤ π/4,

x ≥ tan y.

Since tan y ≥ 0 for all y in the range 0 ≤ y < π/2, we ﬁnd that the condition x ≥ 0 is superﬂuous and can be dropped. This yields D = (x, y) : 0 ≤ y ≤ π/4, tan y ≤ x ≤ 1 , whence
0 1 0 arctan x π/4 1

f (x, y) dydx =
0 tan y

f (x, y) dxdy.

Example 14.6. Reverse the order of integration in the iterated integral
2 0 3−x

f (x, y) dydx. x/2 First solution. We have
2 0 3−x

f (x, y) dydx = x/2 D

f (x, y) dA,

where D = (x, y) : 0 ≤ x ≤ 2, x/2 ≤ y ≤ 3 − x is the region displayed on Figure 14.5. We can easily convert the equations of the lines to the form x = g(y) (see the ﬁgure). However, this time we have to be careful, because the function x = g(y) that bounds D from the right is neither x = 3 − y nor x = 2y. The precise representation of D as a type II region is D = (x, y) : 0 ≤ y ≤ 3, 0 ≤ x ≤ g(y) ,
116

(0, 3) y=3−x↔ x=3−y D (0, 0) (2, 1) y = x/2 ↔ x = 2y

Figure 14.5. The region D where g(y) = 2y if 0 ≤ y ≤ 1, 3 − y if 1 ≤ y ≤ 3.

However, this function is not particularly convenient to work with. To obtain a “friendlier” answer, we split D in two using the horizontal line y = 1. Then each of the two subregions we get is bounded from the right by a single line, say: D1 = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y , Thus,
2 0 3−x 1 2y 3 3−y

D2 = (x, y) : 1 ≤ y ≤ 3, 0 ≤ x ≤ 3 − y . f (x, y) dxdy + f (x, y) dxdy.
1 0

f (x, y) dydx = x/2 0 0

Second solution. Initially, D is described by the inequalities 0 ≤ x ≤ 2, x/2 ≤ y ≤ 3 − x. On noting that 3 − x ≤ 3 and x/2 ≥ 0, we can add to the above inequalities the redundant condition 0 ≤ y ≤ 3, that is, we have D described by 0 ≤ x ≤ 2, Next, we observe that x/2 ≤ y ⇐⇒ x ≤ 2y, 0 ≤ x ≤ 2, y≤3−x x ≤ 2y, ⇐⇒ x ≤ 3 − y, so we can further transform the deﬁnition of D to 0 ≤ y ≤ 3, or equivalently to 0 ≤ y ≤ 3, 0 ≤ x ≤ min(2, 2y, 3 − y). This is the equivalent of the description of D using g(y) in our ﬁrst solution. By comparing 2, 2y, and 3 − y, we ﬁnd that the minimum equals 2y or 3 − y according as y ≤ 1 or y ≥ 1. Therefore, D is the union of the region D1 described by 0 ≤ y ≤ 1, and the region D2 described by 1 ≤ y ≤ 3, 0 ≤ x ≤ 3 − y.
117

0 ≤ y ≤ 3,

x/2 ≤ y ≤ 3 − x.

x ≤ 3 − y,

0 ≤ x ≤ 2y

Thus,
2 0 3−x 1 2y 3 3−y

f (x, y) dydx = x/2 0 0

f (x, y) dxdy +
1 0

f (x, y) dxdy.

Exercises
Evaluate the given iterated integral.
2 4−y

14.1.
0 2 y 2x 0

y2 − 3x dxdy √ 3 4 − x2 dydx

14.2.
1 1 y

14.3.
0 0

y cos(x − y) dxdy

Evaluate the given double integral. 14.4.
D

yx2 dA, D = (x, y) : 1 ≤ y ≤ 3, y ≤ x ≤ y2 3x sin 2y dA, D = (x, y) : 0 ≤ x ≤ π, x/2 ≤ y ≤ x
D D

14.5. 14.6. 14.7.
D

√ 2y dA, D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x x2 + 1 e x/y dA, D = (x, y) : 0 ≤ y ≤ 1, y2 ≤ x ≤ y x sin y dA, D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x2

14.8.
D

14.9.
D

x2 + y3 dA, D is the region bounded by y = 2x and y = x2 (3x − 4y) dA, D is the triangle with vertices (1, −1), (3, 1), and (1, 2)
D

14.10. 14.11.
D

xy2 dA, D is the right half of the disk x2 + y2 ≤ 4 x2 dA, D is the triangle with vertices (0, −1), (1, 1), and (0, 2)
D

14.12. 14.13.
D

6x2 dA, D = (x, y) : 1 ≤ y ≤ 2, 0 ≤ x ≤ 2y y4 + 2
1 √ x √ 4−x2

Sketch the region of integration and reverse the order of integration.
1

14.14.
0 2

f (x, y) dydx f (x, y) dydx

14.15.
0 2

√ − 4−x2 4−x x2 /2 4−y

14.16.
0 2

f (x, y) dydx f (x, y) dxdy
0 2 2 ln y

14.17. 14.18.
1 π/2 0 1

f (x, y) dxdy f (x, y) dydx
0 sin x

14.19.

118

Evaluate the given integral by reversing the order of integration.
1 1 y2 2 √ 3 4 y 2 x

14.20.
0 8

ye−x dxdy dxdy x4 + 1 ey dydx
3

2

14.21.
0

14.22.
0

√

14.23. Find the average value of the function f (x, y) = y3 over the triangle with vertices (0, 2), (1, 1), and (3, 2). 14.24. Find the volume of the solid lying below the hyperbolic paraboloid z = 4 + x2 − y2 and above the rectangle R = [0, 1] × [0, 2]. 14.25. Find the volume of the solid in the ﬁrst octant bounded by the cylinder x2 + z2 = 1, the plane y = 3, and the coordinate planes. 14.26. Find the volume of the solid lying below the plane 4x−y+2z = 6 and above the region bounded by the parabola x = y2 − 1 and the right half of the circle x2 + y2 = 1. 14.27. Find the volume of the solid lying below the paraboloid z = 3xy + 2y + 1 and above the region bounded by the parabola y = x2 and the line y = 2x + 3. 14.28. Find the volume of the solid lying below the parabolic cylinder z = y2 and above the region bounded by the parabola y = x2 and the line y = x + 2. 14.29. Find the volume of the solid bounded by the cylinders x2 + y2 = 1 and x2 + z2 = 1. 14.30. Evaluate D 1 − x2 − y2 dA, where D is the unit disk x2 + y2 ≤ 1, by identifying it as the volume of a solid and calculating that volume using geometric means. 14.31. Evaluate
D

x8 sin x + y9 + 1 dA, where D is the disk x2 + y2 ≤ 4.

119

LECTURE 15

Double Integrals in Polar Coordinates
15.1. The formula Theorem 15.1. Let f (x, y) be piecewise continuous in a region D in the (Cartesian) xy-plane and let R be the same region in polar coordinates, that is, R = (r, θ) : (r cos θ, r sin θ) ∈ D . Then f (x, y) dA =
D R

f (r cos θ, r sin θ)r dA.

(15.1)

The formal proof of (15.1) lies beyond the scope of these notes, but we should give at least some motivation for this formula. Let a, b, α, β be non-negative numbers such that a < b and 0 ≤ α < β ≤ 2π. Consider the case when D is the region bounded by the circles x2 + y2 = a2 and x2 + y2 = b2 and by the half-lines through the origin forming angles α and β with the positive x-axis (see Figure 15.1). Note that D contains the points (x, y) in the plane whose polar coordinates (r, θ) satisfy the inequalities a ≤ r ≤ b and α ≤ θ ≤ β. We recall that the area of D is given by the formula b+a 1 . (15.2) area(D) = 2 (β − α) b2 − a2 = (β − α)(b − a) 2 For each integer n ≥ 1, let r0 = a, θ0 = α, r1 = a + h, θ1 = α + δ, θ=β β θ = θj (xi∗j , y∗j ) i r=a θ=α θj (ri∗ , θ∗ ) j r2 = a + 2h, θ2 = α + 2δ, ..., ..., rn = a + nh = b, θn = α + nδ = β, θ h = (b − a)/n, δ = (β − α)/n, be the points that divide the interval [a, b] into n subintervals of equal lengths; and let

r=b r = ri

α a ri b r

Figure 15.1. The regions D and R, with partitions and sample points
121

be the points that divide the interval [α, β] into n subintervals of equal lengths. We now subdivide the region D into n2 subregions of similar shapes bounded by the circles of radii a, r1 , r2 , . . . , rn−1 , b centered at the origin and the half-lines through the origin forming angles α, θ1 , θ2 , . . . , θn−1 , β with the positive x-axis (see Figure 15.1). Let Di j denote the subregion bounded by the circles of radii ri−1 and ri and by the half-lines at angles θ j−1 and θ j with the positive x-axis: in polar coordinates, Di j is given by the inequalities ri−1 ≤ r ≤ ri , θ j−1 ≤ θ ≤ θ j . We now pick a sample point (xi∗j , y∗j ) in each subregion Di j : we choose (xi∗j , y∗j ) to be the point in i i the plane whose polar coordinates are ri∗ = (ri−1 +ri )/2 and θ∗ = (θ j−1 +θ j )/2. Using the subregions j Di j and the sample points (xi∗j , y∗j ), we form the sum i n n

f (xi∗j , y∗j )|Di j |, i i=1 j=1

|Di j | = area(Di j ).

Note that this sum bears a strong resemblance to a Riemann sum for f (x, y). Indeed, it can be proved that if f (x, y) is continuous, then n n

f (x, y) dA = lim
D

n→∞

f (xi∗j , y∗j )|Di j |. i i=1 j=1

(15.3)

We now express the double sum in (15.3) in terms of the ri ’s and the θ j ’s. By (15.2) with a = ri−1 , b = ri , α = θ j−1 and β = θ j , we have ri + ri−1 = δhri∗ . |Di j | = (θ j − θ j−1 )(ri − ri−1 ) 2 Furthermore, since the polar coordinates of (xi∗j , y∗j ) are (ri∗ , θ∗ ), we have i j xi∗j = ri∗ cos θ∗ , j n n n n

y∗j = ri∗ sin θ∗ , i j

whence f (xi∗j , y∗j ) = f (ri∗ cos θ∗ , ri∗ sin θ∗ ). Therefore, i j j f (xi∗j , y∗j )|Di j | i i=1 j=1

= i=1 j=1

f (ri∗ cos θ∗ , ri∗ sin θ∗ )ri∗ δh. j j

Substituting this expression into the right side of (15.3), we obtain n n

f (x, y) dA = lim
D

n→∞

f (ri∗ cos θ∗ , ri∗ sin θ∗ )ri∗ δh. j j i=1 j=1

(15.4)

The double sum on the right side of (15.4) is a Riemann sum for the function f (r cos θ, r sin θ)r on the rectangle R = [a, b] × [α, β] when the sample points (ri∗ , θ∗ ) are always chosen at the centers of j the subrectangles (see Figure 15.1). Hence, n n→∞ n

lim

f (ri∗ cos θ∗ , ri∗ sin θ∗ )ri∗ δh = j j i=1 j=1 R

f (r cos θ, r sin θ)r dA.

The latter equality and (15.4) justify the (15.1) in the special case when D is a sector of the above kind.
122

15.2. Examples Example 15.1. Evaluate the integral (x + y) dA,
D

where D is the region lying between the circles x2 + y2 = 1 and x2 + y2 = 4 and to the left of the y-axis. Solution. The polar form of D is R = (r, θ) : 1 ≤ r ≤ 2, π/2 ≤ θ ≤ 3π/2 . Hence,
2 3π/2 π/2 2 2 1

(x + y) dA =
D R 2

(r cos θ + r sin θ)r dA =
1 3π/2 dr π/2

r2 cos θ + r2 sin θ) dθ dr 14 . 3

=
1

r2 sin θ − cos θ

= −2

r dr = −

Example 15.2. Find the area enclosed by the curve r = 4 + 3 cos θ. Solution. Since 4 + 3 cos θ ≥ 1 for all θ, the given curve “loops” around the origin, enclosing the region (see Figure 15.2) R = (r, θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4 + 3 cos θ . If D is the region R in Cartesian coordinates, then we know from property (iv) in Theorem 14.1 that its area is |D| =
D

1 dA.

O

D

Figure 15.2. The region r ≤ 4 + 3 cos θ
123

Figure 15.3. The solid in Example 15.3

Hence, by (15.1),
2π 4+3 cos θ

|D| =
D

1 dA =
R 2π

r dA =
0

r dr dθ 1 2
0 2π

1 = 2 =
0

(4 + 3 cos θ)2 dθ =
0 2π

16 + 24 cos θ + 9 cos2 θ dθ
0 2π

41θ 9 9 + 12 sin θ + sin 2θ 8 + 12 cos θ + (1 + cos 2θ) dθ = 4 4 8

=
0

41π . 2

Example 15.3. Find the volume of the solid that lies inside the sphere x2 + y2 + z2 = 16 and outside the cylinder x2 + y2 = 4. Solution. The given solid is symmetric with respect to the xy-plane, so it is twice as large as the solid E that lies inside the sphere, outside the cylinder, and above the xy-plane (see Figure 15.3). This solid is the part of the space between the upper hemisphere and the annular region D in the xy-plane bound by the sphere and the cylinder: D = (x, y) : 4 ≤ x2 + y2 ≤ 16 . Since the upper hemisphere of x2 + y2 + z2 = 16 can be described as the graph of the function z = 16 − x2 − y2 , we can express E as E = (x, y, z) : (x, y) ∈ D, 0 ≤ z ≤ Hence, vol.(E) =
D

16 − x2 − y2 .

16 − x2 − y2 dA.

The polar form of D is R = (r, θ) : 2 ≤ r ≤ 4, 0 ≤ θ ≤ 2π , and 16 − (r cos θ)2 − (r sin θ)2 = Hence, (15.1) yields vol.(E) =
D 4

16 − r2 (cos2 θ + sin2 θ) = √ r 16 − r2 dA
R 4 2 2

√ 16 − r2 .

16 − x2 − y2 dA =
2π 0

=
2

√ r 16 − r2 dθ dr = 2π

√ r 16 − r2 dr.

We evaluate the last integral using the substitution u = 16 − r . Then du = −2r dr, so
0 12

u(2) = 12,

u(4) = 0, 2π 3/2 u 3
12 0

vol.(E) = 2π
12

u1/2 − 1 du = π 2 √

u1/2 du =
0

√ = 16 3 π.

Thus, the volume of the original solid is 32 3 π.
124

Exercises
Evaluate the given integral by changing to polar coordinates. 15.1.
D

(x + 3y) dA, D is the disk x2 + y2 ≤ 4 ex
D
2

15.2. 15.3.
D

+y2

dA, D is the right half of the disk x2 + y2 ≤ 1

16 − x2 − y2 dA, D is the part of the annulus 1 ≤ x2 + y2 ≤ 9 lying in the ﬁrst quadrant
D

15.4. 15.5.
D

xy dA, D is the part of the disk x2 + y2 ≤ 9 lying below the line y = x and above the x-axis 4 + x2 + y2 xye−x
2

−y2

dA, D is the larger region bounded by the y-axis, the line x+y = 0, and the semicircle x =

4 − y2

15.6. Find the volume of the solid enclosed by the paraboloid z = 4 − x2 − y2 and the xy-plane. 15.7. Find the volume of the solid enclosed by the cone z2 = x2 + y2 and the plane z = 3. 15.8. Find the volume of the solid lying inside both the cylinder y2 + z2 = 4 and the sphere x2 + y2 + z2 = 9. 15.9. Find the volume lying above the paraboloid z =
1 2

x2 + y2 and inside the ellipsoid x2 + y2 + 4z2 = 2. 9 − x2 − y2 and above the region D in the xy-plane where

15.10. Find the volume lying below the hemisphere z = x2 + y2 ≤ 2x.

15.11. Find the volume of the solid lying below the plane z = x + y and above the disk x2 + y2 ≤ x + y. 15.12. Find the volume of the solid lying inside all three cylinders x2 + y2 = 1, x2 + z2 = 1, and y2 + z2 = 1.

125

LECTURE 16

Triple Integrals
16.1. Deﬁnition Let f (x, y, z) be a bounded function on the rectangular box B, [a1 , b1 ] × [a2 , b2 ] × [a3 , b3 ] = (x, y, z) : a1 ≤ x ≤ b1 , a2 ≤ y ≤ b2 , a3 ≤ z ≤ b3 . For each n ≥ 1, let x0 = a1 , y0 = a2 , z0 = a3 , x1 = a1 + ∆x, y1 = a2 + ∆y, z1 = a3 + ∆z, x2 = a1 + 2∆x, y2 = a2 + 2∆y, z2 = a3 + 2∆z, ..., xn = a1 + n∆x = b1 , yn = a2 + n∆y = b2 , zn = a3 + n∆z = b3 , ∆x = (b1 − a1 )/n, ∆y = (b2 − a2 )/n, ∆z = (b3 − a3 )/n,

be the points that divide the interval [a1 , b1 ] into n subintervals of equal lengths; let ...,

be the points that divide the interval [a2 , b2 ] into n subintervals of equal lengths; and let ...,

be the points that divide the interval [a3 , b3 ] into n subintervals of equal lengths. Also, pick n3 sample points (xi∗jk , y∗jk , z∗jk ) so that i i xi−1 ≤ xi∗jk ≤ xi , y j−1 ≤ y∗jk ≤ y j , i n n n

zk−1 ≤ z∗jk ≤ zk i

(1 ≤ i, j, k ≤ n).

Then the triple integral of f over B is deﬁned by f (x, y, z) dV = lim
B

n→∞

f (xi∗jk , y∗jk , z∗jk )∆V, i i i=1 j=1 k=1

∆V = ∆x∆y∆z,

(16.1)

provided that the limit exists. The sums on the right side of (16.1) are called Riemann sums of f on B. More generally, if S is a solid region in R3 bounded by a piecewise smooth surface Σ and f is a piecewise continuous function on S , we deﬁne the triple integral of f over S as f (x, y, z) dV =
S B

F(x, y, z) dV,

where B is any rectangular box containing S and F is the function F(x, y, z) = f (x, y, z) if (x, y, z) ∈ S , 0 otherwise.

Remark. As in the cases of deﬁnite and double integrals, the above deﬁnitions deﬁne the triple integral for all piecewise continuous functions over solids bounded by piecewise smooth surfaces.
127

Furthermore, similarly to the deﬁnite and double integrals, the triple integral of a function is related to its average in the region of integration: 1 f (x, y, z) dV. (16.2) average f (x, y, z) = vol.(S ) (x,y,z)∈S S 16.2. Properties of the triple integral The following theorem lists the three-dimensional versions of the properties of double integrals from Theorem 14.1. Theorem 16.1. Let f (x, y, z), g(x, y, z) be piecewise continuous functions and a, b denote constants. Then i)
S

[a f (x, y, z) + bg(x, y, z)] dV = a
S

f (x, y, z) dV + b
S

g(x, y, z) dV.

ii) If f (x, y, z) ≤ g(x, y, z), then f (x, y, z) dV ≤
S

g(x, y, z) dV.
S

iii) If S = S 1 ∪ S 2 , S 1 ∩ S 2 = ∅, then f (x, y, z) dV =
S S1

f (x, y, z) dV +
S2

f (x, y, z) dV.

iv)
S

1 dV = vol.(S ).

There is also a three-dimensional version of Fubini’s theorem, which we shall use to evaluate triple integrals. Before stating it, we need to introduce some terminology. We say that a solid region S in R3 is of type I, if it can be written in the form S = (x, y, z) : (x, y) ∈ D, g1 (x, y) ≤ z ≤ g2 (x, y) for some region D in the xy-plane and some continuous functions g1 and g2 on D. We say that S is a type II region, if it can be written in the form S = (x, y, z) : (y, z) ∈ D, g1 (y, z) ≤ x ≤ g2 (y, z) for some region D in the yz-plane. We say that S is a type III region, if it can be written in the form S = (x, y, z) : (x, z) ∈ D, g1 (x, z) ≤ y ≤ g2 (x, z) for some region D in the xz-plane. Theorem 16.2. Let f (x, y, z) be a piecewise continuous function on a solid region S in R3 . i) If S is of type I, then f (x, y, z) dV =
S D g2 (x,y)

f (x, y, z) dz dA; g1 (x,y)

(16.3)

ii) If S is of type II, then f (x, y, z) dV =
S D g2 (y,z)

f (x, y, z) dx dA; g1 (y,z)

(16.4)

128

iii) If S is of type III, then f (x, y, z) dV =
S D g2 (x,z)

f (x, y, z) dy dA. g1 (x,z)

(16.5)

16.3. Evaluation of triple integrals Example 16.1. Evaluate the triple integral yz cos(x5 ) dV,
S

where S = (x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x, x ≤ z ≤ 2x . Solution. Let D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x . By (16.3),
2x

yz cos(x5 ) dV =
S D x

yz cos(x5 ) dz dA y cos(x5 ) (2x)2 − x2 dA =
D 1 0 1 0 x

1 = 2 3 = 2 = 3 20

3 2
1 0

yx2 cos(x5 ) dA
D

yx2 cos(x5 ) dy dx = cos u du =
0

3 4

x4 cos(x5 ) dx

3 sin u 20

1 0

=

3 sin 1 ≈ 0.126. 20

Example 16.2. Evaluate the triple integral x2 + z2 dV,
S

where S is the solid bounded by the paraboloid y = x2 + z2 and the plane y = 4. Solution. We can write S as a type III region: S = (x, y, z) : x2 + z2 ≤ 4, x2 + z2 ≤ y ≤ 4 . Hence, by (16.5),
4

x2 + z2 dV =
S x2 +z2 ≤4

x2 +z2

x2 + z2 dy dA = x2 +z2 ≤4

4 − x 2 − z2

x2 + z2 dA.

We can easily compute the last integral by changing to polar coordinates in the xz-plane:
2 2π

4−x −z x2 +z2 ≤4

2

2

x2

+

z2

dA =
0≤r≤2

4 − r r r dA =
2 0 2 0

r2 4 − r2 dθdr
2

= 2π
0

4r − r

2

4

4r3 r5 dr = 2π − 3 5

=
0

128π . 15

129

Example 16.3. Compute the volume of the solid T described by the inequalities x ≥ 0, z ≥ 0, y ≥ 2x, x + 2y + 2z ≤ 2.
2 Solution. In fact, T is the tetrahedron with vertices (0, 0, 0), (0, 1, 0), (0, 0, 1), and ( 5 , 4 , 0), but 5 we shall ignore this fact and make a point of using triple integrals and inequalities. By solving the inequlity x + 2y + 2z ≤ 2 for z, we can rewrite the deﬁnition of T as

x ≥ 0,

y ≥ 2x,

0 ≤ z ≤ 1 − x/2 − y.

(16.6)

Note that the implicit in the inequality for z is the condition 1 − x/2 − y ≥ 0 ⇐⇒ y ≤ 1 − x/2.

Adding the latter inequality to those in (16.6), we get x ≥ 0, 2x ≤ y ≤ 1 − x/2, 0 ≤ z ≤ 1 − x/2 − y
2 We now note that the condition x ≤ 5 is implicit in the inequalities for y. Making this condition explicit, we ﬁnally have T described as a type I set:

T = (x, y, z) : 0 ≤ x ≤ 2 , 2x ≤ y ≤ 1 − x/2, 0 ≤ z ≤ 1 − x/2 − y . 5 We can now compute the volume of T :
2/5 1−x/2 2x 0 2/5 1−x/2−y

vol.(T ) =
T 2/5

1 dV =
0 1−x/2

1 dzdydx (1 − x/2)y − 1 y2 2
1−x/2 2x

=
0 2/5 2x

(1 − x/2 − y) dydx =
0

dx

=
0 2/5

(1 − x/2)2 − 1 (1 − x/2)2 − 2x(1 − x/2) + 2x2 dx 2
1 1 2

=
0

−

5 x 2

+

25 2 x 8

dx =
0

1 2

− u + 1 u2 2

2 du 5

=

1 . 15

16.4. Double and triple integrals in Mathematica Mathematica’s built-in commands Integrate and NIntegrate can evaluate not only deﬁnitite integrals in one variable, but also double, triple, and even higher-dimensional integrals. There are two main ways to use these commands to evaluate double and triple integrals. Suppose ﬁrst that the integral can be easily converted to an iterated one: say, b a g2 (x) b g2 (z) g1 (z) h2 (y,z)

f (x, y) dydx g1 (x)

or a h1 (y,z)

f (x, y, z) dxdydz.

If the integrations are within Mathematica’s reach, we can use Integrate as follows Integrate[f[x,y], {x,a,b}, {y,g1[x],g2[x]}] Integrate[f[x,y,z], {z,a,b}, {y,g1[z],g2[z]}, {x,h1[y,z],h2[y,z]}] If the functions are such that Mathematica cannot evaluate the integrals exactly, we can use NIntegrate in a similar fashion to obtain a numeric approximation. Here are some examples:
130

In[1]:= Integrate[x*yˆ2, {x,0,1}, {y,0,1}] Out[1]:= 1/6 In[2]:= Integrate[x*yˆ2, {x,0,1}, {y,0,x}] Out[2]:= 1/15 In[3]:= Integrate[y*z*Cos[xˆ5], {x,0,1}, {y,0,x}, {z,x,2x}] Out[3]:= 3Sin[1]/20 In[4]:= Integrate[Sin[x+y]/(x+y), {x,1,2}, {y,1,2}] Out[4]:= Cos[2] - 2Cos[3] + Cos[4] + 2SinIntegral[2] - 6SinIntegral[3] + 4SinIntegral[4] In[5]:= NIntegrate[Sin[x+y]/(x+y), {x,1,2}, {y,1,2}] Out[5]:= 0.0619179 Next, suppose that we are interested in an integral over two- or three-dimensional domain D that is described by one or more inequalities. In such situations, we use a familiar trick: we replace the original function f by another which is zero outside D. Mathematica provides two commands to execute this trick: If[] and Boole[]. For example, suppose that we want to integrate the function x2 − xy + y2 over the region D above the x-axis and the line y = −x and inside the ellipse 2x2 + y2 = 3 (see Figure 16.1). To deﬁne the function f (x, y) = x2 − xy + y2 0 if (x, y) ∈ D, if (x, y) D,

we can use either of the following two Mathematica syntaxes: f[x_,y_] := If[2xˆ2+yˆ2=0 && y>=0, xˆ2-x*y+yˆ2, 0] f[x_,y_] := (xˆ2-x*y+yˆ2)Boole[2xˆ2+yˆ2=0 && y>=0] Thus, to evaluate D (x2 − xy + y2 ) dA, we can use (the condition y ≥ 0 is incorporated in the range of y) Integrate[(xˆ2-x*y+yˆ2)Boole[2xˆ2+yˆ2=0], {x,-1,2}, {y,0,2}]

1.5

1.0

0.5

0.0 1.0 0.5 0.0 0.5 1.0

Figure 16.1. The region above y = 0 and y = −x and insider 2x2 + y2 = 3
131

Here are some examples: In[6]:= Integrate[(x*y+yˆ2)Boole[xˆ2+yˆ2 0 and v > 0. Show that if (x, y) is a solution of the system of equations xy = u, 2x2 − y2 = v, then x must be a solution of the equation 2x4 − vx2 − u2 = 0. Show that the latter equation has a unique positive real solution, and deduce that T is a one-to-one transformation of R onto D. (c) Find the Jacobian of T . (d) Use the transformation T to evaluate R (2x2 + y2 ) dA. [Hint: Switch the usual roles of x, y and u, v.] 19.12. In this exercise, we will ﬁnd the volume of the solid E enclosed by the surface Σ : x2/3 + y2/3 + z2/3 = 4. (a) Let x = u3 , y = v3 , and z = w3 . Observe that the point (x, y, z) is on/inside Σ if and only if the point (u, v, w) is on/inside the sphere u2 + v2 + w2 = 4. Use this observation and spherical coordinates in the uvw-space to show that the formulas x = ρ3 cos3 θ sin3 φ, y = ρ3 sin3 θ sin3 φ, z = ρ3 cos3 φ deﬁne a transformation T of B onto E, where B is the rectangular box 0 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. Show that T is one-to-one inside B. (b) Find the Jacobian of the transformation T from part (a). (c) Express the volume of E as a triple integral and use the transformation T above to evaluate that triple integral. 19.13. Let p, q, r, s be non-negative integers, and let S be the triangular pyramid (called also simplex) deﬁned by the inequalities x ≥ 0, y ≥ 0, z ≥ 0, and x + y + z ≤ 1. (a) Let U denote the cube of side length 1 in the uvw-space deﬁned by the inequalities 0 ≤ u, v, w ≤ 1. Show that the equations x + y + z = u, y + z = uv, z = uvw deﬁne a transformation T of U onto S and that this transformation is one-to-one inside U. Is T one-to-one on U; if not, then at which boundary points does T fail to be one-to-one? (b) Find the Jacobian of the transformation T from part (a). (c) Use Theorem 19.2 and the transformation T above to evaluate x p yq zr (1 − x − y − z) s dV.
S

[Hint: You may want ﬁrst to evaluate

1 a 0 t (1

− t)b dt, where a, b are non-negative integers.]

156

LECTURE 20

Vector Fields
20.1. Deﬁnition If D is a region in R2 , a vector ﬁeld on D is a function F that assigns to each point (x, y) in D a two-dimensional vector F(x, y). If D is a solid region in R3 , a vector ﬁeld on D is a function F that assigns to each point (x, y, z) ∈ D a three-dimensional vector F(x, y, z). Note that we can express each two-dimensional vector ﬁeld F in terms of a pair of functions of two variables: F(x, y) = P(x, y), Q(x, y) = P(x, y)e1 + Q(x, y)e2 . Similarly, we can express each three-dimensional vector ﬁeld F in terms of a triple of functions of three variables: F(x, y, z) = P(x, y, z), Q(x, y, z), R(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 . Example 20.1. The function F(x, y, z) = x2 e1 + y2 e2 + z2 e3 is a vector ﬁeld on R3 . The function F(x, y) = cos(xy)e1 + sin(xy)e2 is a vector ﬁeld on R2 . The standard way to visualize a vector ﬁeld is to draw the arrows representing F(x, y) (or F(x, y, z)) applied at the points (x, y) (or (x, y, z)) for a number of choices for (x, y) (or (x, y, z)) from D. Example 20.2. Sketch twenty vectors of the vector ﬁeld F(x, y) = 1 e1 − xe2 . 2 Answer. See Figure 20.1. y

1 1 O x

1 Figure 20.1. The vector ﬁeld F(x, y) = 2 e1 − xe2

157

20.2. Gradient ﬁelds and potential functions We are already familiar with an important class of vector ﬁelds. Suppose that f (x, y) is a diﬀerentiable function on D. In Lecture #9, we deﬁned the gradient f of f : f (x, y) = f x (x, y)e1 + fy (x, y)e2 . We now see that f is a two-dimensional vector ﬁeld on D. Similarly, if f (x, y, z) is a diﬀerentiable function on a solid D in R3 , then f (x, y, z) is a three-dimensional vector ﬁeld on D. From now on, we will refer to the gradient of a function f as the gradient vector ﬁeld of f . Example 20.3. Find the gradient vector ﬁeld of the function f (x, y) = x2 + y2 . Solution. We have f (x, y) = f x (x, y)e1 + fy (x, y)e2 = 2xe1 + 2ye2 . In the subsequent lectures, we will see that vector ﬁelds F which are the gradient ﬁelds of some function play an important role in vector calculus. Such vector ﬁelds are known as conservative, that is, F is conservative if there is a function f such that F = f . If F = f , then f is called a potential function of F, or simply a potential. Example 20.4. By Example 20.3, the vector ﬁeld F(x, y) = 2xe1 +2ye2 is conservative: F = f , where f (x, y) = x2 + y2 . Similarly, F(x, y, z) = 2ye1 + 2xe2 + e3 is conservative, because F = g, where g(x, y, z) = 2xy + z. Not every vector ﬁeld is conservative, and it is not very diﬃcult to give an example of a vector ﬁeld that is not conservative. Example 20.5. Show that the vector ﬁeld F(x, y) = (x2 + y)e1 + y3 e2 is not conservative. Solution. Suppose that F is conservative, that is, there is some function f (x, y) such that f x (x, y) = x2 + y, Then f xy (x, y) = ∂y (x2 + y) = 1, fyx (x, y) = ∂ x (y3 ) = 0. Since both mixed partials are constants, they are continuous everywhere. Thus, by Theorem 8.1, we must have f xy = fyx . However, they are not equal, a contradiction. Therefore, our assumption that F is conservative must be false. We shall use this example as a starting point of our discussion of conservative vector ﬁelds. However, before delving into that, we need to ﬁx some terminology. 20.3. Terminology Recall that in §11.4 we deﬁned a closed region in the plane to be a region containing all of its boundary. We call a region D in the plane or in space open, if it contains none of the points on its boundary. Some examples of open and non-open sets in the plane are shown on Figure 20.2. On the ﬁgure, solid and dashed curves indicate boundary that does and does not belong to the set, respectively. The ﬁrst set is open, because it does not contain any of its boundary; the second set is not open, because it contains all of its boundary (it is closed); and the last set is not open, because it contains a part of its boundary.
158

fy (x, y) = y3 .

not open

open Figure 20.2. Open and non-open sets

not open

We say that a set D, in the plane or in space, is connected, if we can connect every two points of D with a continuous path that lies entirely in D. Geometrically, this means that D “has only one piece”. Several examples of connected and disconnected plane sets are displayed on Figure 20.3.

disconnected

connected

connected

disconnected

disconnected

connected

Figure 20.3. Connected and disconnected sets We say that a set D in the plane is simply-connected, if it is connected and every curve γ that lies entirely in D can be shrunk to a point in D without leaving D. We say that a set D in space is simply-connected, if it is connected and every surface Σ that lies entirely in D can be shrunk to a point in D without leaving D. In both cases, it is “closed” curves and surfaces that we really need to test. Both in two and in three dimensions, simple-connectedness means that D “has only one piece and no holes”. Also, in both cases, we can check whether a connected set is simply-connected by looking at the compliment of D to the whole plane or the whole space. If we denote that set by D , then we must be able to connect each point of D to any other point of D and to “inﬁnity” by a path that lies entirely in D . Some examples of simply-connected and not simply-connected plane sets are displayed on Figure 20.4.

simply-connected

not simply-connected not simply-connected

Figure 20.4. Simply-connected and not simply-connected sets
159

20.4. Conservative vector ﬁelds in R2 We can easily generalize the solution of Example 20.5. Let F(x, y) = P(x, y)e1 + Q(x, y)e2 be a conservative vector ﬁeld and let f (x, y) be its potential, that is, P(x, y) = f x (x, y), Then Py = f xy and Q x = fyx =⇒ This is the basic idea underlying the following theorem. Py = Q x . Q(x, y) = fy (x, y).

Theorem 20.1. If F(x, y) = P(x, y)e1 + Q(x, y)e2 is a conservative vector ﬁeld and P(x, y) and Q(x, y) have continuous ﬁrst order partials in a domain D, then throughout D we have ∂y P = ∂ x Q. (20.1) Note that we can use this theorem to show that a given vector ﬁeld is not conservative. However, we would rather have a tool for establishing that a vector ﬁeld is conservative. The next theorem says that equation (20.1) implies that F is conservative, provided that the region D has a special property. Theorem 20.2. If F(x, y) = P(x, y)e1 + Q(x, y)e2 is a vector ﬁeld that satisﬁes (20.1) on an open, simply-connected region D, then F is conservative. Example 20.6. Determine whether or not the vector ﬁeld is conservative: (a) F(x, y) = (2x + y2 )e1 + (2y + x2 )e2 ; (b) F(x, y) = (x3 + 3xy2 )e1 + (y3 + 3x2 y)e2 ; −y x (c) F(x, y) = 2 e + 2 e2 . 2 1 x +y x + y2 Solution. (a) Let P(x, y) = 2x + y2 and Q(x, y) = 2y + x2 . Then Py = 2y 2x = Q x , so the vector ﬁeld is not conservative. (b) Let P(x, y) = x3 + 3xy2 and Q(x, y) = y3 + 3x2 y. Then Py = 6xy = Q x for all (x, y) ∈ R2 . Since R2 is open and simply-connected, it follows that the vector ﬁeld is conservative. (c) We have −y (−1)(x2 + y2 ) − (−y)(2y) y2 − x2 ∂ = = 2 , ∂y x2 + y2 (x2 + y2 )2 (x + y2 )2 ∂ x (1)(x2 + y2 ) − (x)(2x) y2 − x2 = = 2 , ∂x x2 + y2 (x2 + y2 )2 (x + y2 )2 for all (x, y) (0, 0). However, the set of all (x, y) (0, 0) is not simply-connected. Thus, Theorem 20.2 says nothing about F. On the other hand, if we consider F on the set D obtained by cutting the xy-plane along a half-line containing the origin (e.g., we can cut the plane along the positive x-axis, or the negative y-axis, or the part of the line y = 2x with x ≤ 1), then the theorem does say that F is conservative on D (which is open and simply-connected). Later we shall show that this vector ﬁeld is not conservative on the set of all (x, y) 0.
160

20.5. Curl and conservative vector ﬁelds in R3 Definition. If F(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 is a vector ﬁeld on R3 such that P, Q, R are diﬀerentiable, then the curl of F is the vector ﬁeld on R3 deﬁned by ∂P ∂R ∂Q ∂P ∂R ∂Q − e1 + − e2 + − e3 . curl F = ∂y ∂z ∂z ∂x ∂x ∂y Remark. A formula for curl F that is easier to remember is that e1 e2 e3 curl F = × F = ∂ x ∂y ∂z . P Q R In reality, this is a terrible abuse of notation, but it helps remember the formula, just as (2.6) helps us remember the deﬁnition of the cross product. Example 20.7. Find curl F for F(x, y, z) = xez e1 + yez e2 + (x2 z + y2 z)e3 . Solution. We have e1 e2 e3 ∂z curl F = ∂ x ∂y xez yez x2 z + y2 z = ∂y ∂z ∂x ∂z ∂ x ∂y e z 2 2 e1 − z 2 2 e2 + ye x z + y z xe x z + y z xez yez 3

= ∂y (x2 z + y2 z) − ∂z (yez ) e1 − ∂ x (x2 z + y2 z) − ∂z (xez ) e2 + ∂ x (yez ) − ∂y (xez ) e3 = (2yz − yez )e1 − (2xz − xez )e2 + (0 − 0)e3 = y(2z − ez )e1 + x(ez − 2z)e2 . We can use the curl of a vector ﬁeld to state three-dimensional versions of the theorems in the previous section. Theorem 20.3. If f (x, y, z) has continuous second-order partials, then curl( f ) = 0. In particular, if F is a conservative three-dimensional vector ﬁeld, then curl F = 0. Proof. The proof is straightforward: e1 e2 e3 ∂ ∂z ∂ ∂z ∂ ∂y curl( f ) = ∂ x ∂y ∂z = y e1 − x e2 + x e ∂y f ∂z f ∂ x f ∂z f ∂ x f ∂y f 3 ∂ x f ∂y f ∂z f = ∂2 f ∂2 f − e1 − ∂y∂z ∂z∂y ∂2 f ∂2 f − e2 + ∂x∂z ∂z∂x ∂2 f ∂2 f − e3 = 0. ∂x∂y ∂y∂x

The converse theorem is not universally true, but as in the case of two-dimensional ﬁelds, it can be proved for special sets. Theorem 20.4. If F is a three-dimensional vector ﬁeld that has continuous partials and satisﬁes curl F = 0 on an open, simply-connected set in R3 , then F is conservative.
161

Example 20.8. Determine whether the vector ﬁeld F = ye2xy+z e1 + xe2xy+z e2 + ze2xy+z e3 is conservative.
2 2 2

Solution. We have e1 e2 e3 ∂x ∂y ∂z curl F = 2xy+z2 2xy+z2 2xy+z2 ze xe ye = xe ∂y 2xy+z2 ze ∂z 2xy+z2 e1 −
2

ye

∂x 2xy+z2

ze

∂z 2xy+z2

e2 +
2

ye

∂x 2xy+z2

xe

∂y 2xy+z2

e3
2

= (z(2x) − x(2z))e2xy+z e1 − (z(2y) − y(2z))e2xy+z e2 + (x(2y) − y(2x))e2xy+z e3 = 0. Since the partials of F are continuous everywhere in R3 (a simply-connected set), it follows that F is conservative. Remark. Note that when F(x, y, z) = P(x, y)e1 + Q(x, y)e2 + 0e3 (that is, when F(x, y, z) is a two-dimensional vector ﬁeld that “pretends” to be three-dimensional), then ∂Q ∂P curl F = − e3 , ∂x ∂y and the condition curl F = 0 is equivalent to the condition discussed in §20.4. 20.6. Finding potentials Example 20.9. Find a potential for the vector ﬁeld F(x, y) = (x3 + 3xy2 )e1 + (y3 + 3x2 y)e2 , if it exists. Solution. We know that this ﬁeld is conservative from Example 20.6, so a potential exists. If we denote it by f (x, y), then f must satisfy the conditions f x (x, y) = x3 + 3xy2 , fy (x, y) = y3 + 3x2 y. In other words, for a ﬁxed y, f (x, y) is an antiderivative of x3 + 3xy2 , and for a ﬁxed x, f (x, y) is an antiderivative of y3 + 3x2 y. Hence, x4 3x2 y2 + + g(y). 4 2 Note that our answer infolves an arbtrary function of y instead of an arbitrary constant. In order to ﬁnd out more about g(y), we diﬀerentiate the above expression for f (x, y) with respect to y: f (x, y) = x3 + 3xy2 dx = fy (x, y) = ∂y Comparing this with the equality fy = y + 3x2 y, we ﬁnd that 3x2 y + g (y) = y3 + 3x2 y ⇐⇒ g (y) = y3 ⇐⇒ g(y) = y4 + C. 4
1 4 x 4 3 3 + 2 x2 y2 + g(y) = 3x2 y + g (y).

1 1 We conclude that f (x, y) = 4 x4 + 3 x2 y2 + 4 y4 + C. 2

Note that in the above example we can easily check our answer: direct diﬀerentiation gives f (x, y) = F(x, y), so f (x, y) is potential for F(x, y) by the deﬁnition of potential. x y Example 20.10. Find a potential for the vector ﬁeld F(x, y) = 2 e1 + 2 e2 , if it exists. x + y2 x + y2
162

Solution. The potential f (x, y), if it exists, must satisfy the conditions y x , fy (x, y) = 2 . f x (x, y) = 2 x + y2 x + y2 Thus, x 1 d(x2 + y2 ) ln(x2 + y2 ) f (x, y) = dx = = + g(y). x2 + y2 2 x2 + y2 2 To ﬁnd g(y), we use fy (x, y): 1 2y y fy (x, y) = + g (y) = 2 ⇐⇒ g (y) = 0 ⇐⇒ g(y) = C. 2 + y2 2 x x + y2 We conclude that f (x, y) = 1 ln(x2 + y2 ) + C. 2 Remark. Note that the function f (x, y) = 1 ln(x2 + y2 ) +C in the previous example is a potential 2 for the vector ﬁeld F(x, y) for all (x, y) (0, 0). Therefore, F is conservative in the set R2 − {(0, 0)}. On the other hand, if we tried to apply Theorem 20.2 to it, we would run into the same trouble as in the solution of Example 20.6, part (c): this set is not simply-connected. Example 20.11. Find a potential for the vector ﬁeld F(x, y, z) = (2xz+y2 )e1 +2xye2 +(x2 +3z2 )e3 , if it exists. Solution. The potential f (x, y, z), if it exists, must satisfy the conditions f x (x, y, z) = 2xz + y2 , f (x, y, z) = fy (x, y, z) = 2xy, fz (x, y, z) = x2 + 3z2 . Thus, arguing similarly to the last two examples, we ﬁnd that (2xz + y2 ) dx = x2 z + y2 x + g(y, z), ⇐⇒ ⇐⇒ gy (y, z) = 0 h (z) = 3z2 ⇐⇒ ⇐⇒ g(y, z) = h(z), h(z) = z3 + C.

fy (x, y, z) = 2yx + gy (y, z) = 2xy fz (x, y, z) = x2 + h (z) = x2 + 3z2

We conclude that f (x, y, z) = x2 z + y2 x + z3 + C. Example 20.12. Find a potential for the vector ﬁeld F(x, y, z) = e2xy e1 + e2xy e2 + z2 e3 , if it exists. Solution. Any potential f (x, y, z) must satisfy f x (x, y, z) = e2xy , From the ﬁrst of these equations, we ﬁnd e2yx + g(y, z). 2y Next, we diﬀerentiate the last function with respect to y and get f (x, y, z) = e2xy dx = 2xe2xy e2xy x 1 − 2 + gy (y, z) = e2xy ⇐⇒ gy (y, z) = e2xy 1 − + 2 . 2y 2y y 2y We seem to be running in circles, and we indeed are, because the ﬁeld is non-conservative. For example, ∂ x e2xy = 2ye2xy 2xe2xy = ∂y e2xy , so the third component of curl F is non-zero. fy (x, y, z) =
163

fy (x, y, z) = e2xy ,

fz (x, y, z) = z2 .

20.7. The divergence of a vector ﬁeld Definition. If F(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 is a vector ﬁeld on R3 such that P, Q, R are diﬀerentiable, then the divergence of F is the function deﬁned by ∂P ∂Q ∂R + + . div F = ∂x ∂y ∂z Example 20.13. Find div F for F = ez sin(xy)e1 + ez cos(xy)e2 + xyze3 . Solution. We have div F = ∂ x (ez sin(xy)) + ∂y (ez cos(xy)) + ∂z (xyz) = yez cos(xy) − xez sin(xy) + xy. Theorem 20.5. If F = Pe1 + Qe2 + Re3 has continuous second-order partials, then div curl F = 0. Proof. The proof is straightforward: ∂ ∂R ∂Q ∂ ∂P ∂R ∂ ∂Q ∂P div curl F = − + − + − ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y 2 2 2 2 2 2 ∂R ∂ Q ∂P ∂R ∂Q ∂P = − + − + − = 0. ∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y Exercises
20.1. Sketch the vector ﬁeld F(x, y) = (x − y)e1 + xe2 by drawing arrows in a similar fashion to Example 20.2. Find the gradient vector ﬁeld of the given function. 20.2. f (x, y) = e2y sin(xy) 20.3. f (x, y, z) = x2 + y2 + z2 20.4. f (x, y, z) = ln x2 + yz Determine whether or not the given set is open, connected, and simply-connected. 20.5. (x, y) : x > 0, y > 0 20.6. (x, y) : x > 0, y ≥ 0 20.7. (x, y) : 1 < x2 + y2 < 2 20.8. (x, y) : x2 + y2 ≤ 1 20.9. (x, y) : x2 + y2 < 1 or x2 + y2 > 2 20.10. (x, y) : xy ≤ 0 Determine whether the vector ﬁeld F(x, y) is conservative. If it is, ﬁnd a potential function for it. 20.11. F(x, y) = (x − y)e1 + (x + y)e2 20.12. F(x, y) = ye1 + (x + 3y)e2 20.13. F(x, y) = ey e1 + xey e2 y2 20.14. F(x, y) = e1 + (2y arctan x)e2 1 + x2 20.15. F(x, y) = (4x + sin y)e1 + (sin y + x cos y)e2 Find the curl and the divergence of the given vector ﬁeld.

164

20.16. F(x, y, z) = x2 yze1 − 4xyz2 e3 20.17. F(x, y, z) = 2 arcsin ye1 +

e2 − 3z2 e3 1 − y2 20.18. F(x, y, z) = cos xze1 + 2 sin xye2 + ze2xy e3 Determine whether the vector ﬁeld F(x, y, z) is conservative. If it is, ﬁnd a potential function for it. 20.19. F(x, y, z) = (yz + x2 )e1 + (xz + 4y)e2 + (xy − 3z2 )e3 20.20. F(x, y, z) = y2 cos ze1 + 2xy cos ze2 − xy2 sin ze3 20.21. F(x, y, z) = xey e1 + ye x e2 + xyze3 2x 20.22. F(x, y, z) = 2 arcsin ye1 + e2 − 3z2 e3 1 − y2 20.23. Is there a vector ﬁeld F on R3 such that curl F = xy2 e1 + yz2 e2 + zx2 e3 ? Explain.

2x

165

LECTURE 21

Line Integrals
21.1. Line integrals of scalar functions We start with a smooth parametric curve γ, given by a vector function r(t) = x(t)e1 + y(t)e2 + z(t)e3 (a ≤ t ≤ b). Recall that the smoothness of γ means that r (t) = x (t)e1 + y (t)e2 + z (t)e3 is continuous and r (t) 0 when a < t < b (see §5.3). Further, suppose that f (x, y, z) is a continuous function on a set S in R3 that contains the curve γ. In particular, f (x, y, z) is deﬁned at each point (x, y, z) on γ. For each n ≥ 1, set ∆n = (b−a)/n and deﬁne the numbers that partition [a, b] into n subintervals of equal lengths: t0 = a, t1 = a + ∆ n , t2 = a + 2∆n , ..., tn = a + n∆n = b. We write xi = x(ti ), yi = y(ti ), zi = z(ti ) for the coordinates of the point Pi (xi , yi , zi ) that corresponds to the value t = ti of the parameter. The points P0 , P1 , P2 , . . . , Pn partition γ into n arcs: γ1 , γ2 , . . . , γn , where γi has Pi−1 and Pi as its endpoints. Further, we pick n values of the parameter ∗ ∗ ∗ t1 , t2 , . . . , tn so that ∗ ∗ ∗ t0 ≤ t1 ≤ t1 , t1 ≤ t2 ≤ t2 , . . . , tn−1 ≤ tn ≤ tn . These yield n sample points P∗ (xi∗ , y∗ , z∗ ) on the curve with coordinates xi∗ = x(ti∗ ), y∗ = y(ti∗ ), i i i i z∗ = z(ti∗ ). Then the line integral of f along γ is deﬁned by i n f (x, y, z) ds = lim γ n→∞

f (xi∗ , y∗ , z∗ )∆si , i i i=1 ∆si = length(γi ),

(21.1)

provided that the limit exists. Recall that, by Theorem 5.1, ∆si = ti ti−1

x (t)2 + y (t)2 + z (t)2 dt.

Using this formula, the above deﬁnition, the properties of continuous functions, and the deﬁnition of the deﬁnite integral, it can be shown that for continuous functions f the limit (21.1) always exists and the line integral can be expressed as a deﬁnite integral. Theorem 21.1. Let γ be a smooth parametric curve, given by the vector function r(t) = x(t)e1 + y(t)e2 + z(t)e3 If f (x, y, z) is continuous on γ, then b (a ≤ t ≤ b).

f (x, y, z) ds = γ a

f (x(t), y(t), z(t))
167

x (t)2 + y (t)2 + z (t)2 dt.

(21.2)

Remarks. 1. Using more extensively the language of vector functions, we can write (21.2) in the form b f (x, y, z) ds = γ a

f (r(t)) r (t) dt.

(21.3)

2. As in the cases of deﬁnite and multiple integrals, the line integral of a continuous function along a curve is related to its average value on the points of the curve: average f (x, y, z) =
(x,y,z)∈γ

1 length(γ)
−1

f (x, y, z) ds. γ Example 21.1. Evaluate x2 + y2 + z2 γ ds

where γ is the helix x = cos t, Solution. We have x2 + y2 + z2 γ −1 T

y = sin t,

z=t

(0 ≤ t ≤ T ).

ds =
0 T

(cos t)2 + (sin t)2 + t2 √ (1 + t ) 1 + 1 dt =
2 −1

−1

(− sin t)2 + (cos t)2 + 12 dt √ √ 2 dt = 2 arctan T. 2 1+t

T 0

=
0

We have already seen on several instances that properties of plane curves or two-dimensional vectors often appear as special cases of analogous properties of space curves or three-dimensional vectors. This general principle also applies to line integrals. We just deﬁned the line integral of a function of three variables along a space curve. Is it possible to deﬁne the line integral of a function of two variables along a plane curve? The answer to this question is in the aﬃrmative. One way to develop the theory of such line integrals is by repeating the above deﬁnitions and reproving the above theorems with one less variable. Alternatively, we can deﬁne two-dimensional line integrals as a special case of three-dimensional line integrals. We take the latter route. Observe that a two-variable function f (x, y) is also a three-variable function f (x, y, z) (which does not really depend on z) and that a plane curve, given by r(t) = x(t)e1 + y(t)e2 is also a space curve, given by r0 (t) = x(t)e1 + y(t)e2 + 0e3 (a ≤ t ≤ b). (21.5) (a ≤ t ≤ b), (21.4)

If γ is the plane curve (21.4) and f (x, y) is continuous on γ, we deﬁne the line integral of f along γ by f (x, y) ds = γ γ0

g(x, y, z) ds,

168

where γ0 is the space curve (21.5) and g(x, y, z) = f (x, y) for all (x, y) in the domain of f and all z. In particular, by (21.2), we have b f (x, y) ds = γ a b

g(x(t), y(t), 0) f (x(t), y(t)) a b

x (t)2 + y (t)2 + 02 dt x (t)2 + y (t)2 dt.

= That is, we have the formula b f (x, y) ds = γ a

f (r(t)) r (t) dt = a f (x(t), y(t))

x (t)2 + y (t)2 dt.

(21.6)

Note that the vector form of (21.6) and the vector form (21.3) of (21.2) look identical; the only diﬀerence is that the vector function r(t) is two-dimensional in (21.6) and three-dimensional in (21.3). Example 21.2. Evaluate xey ds, γ where γ is the arc of the curve x = ey from (1, 0) to (e, 1). Solution. The curve can be parametrized by the equations x = et , Hence, by (21.6),
1 1

y=t

(0 ≤ t ≤ 1). √ e2t e2t + 1 dt.

xey ds = γ 2t 0

et et

(et )2 + 12 dt =
0

Using the substitution u = e + 1, we ﬁnd that du = 2e2t dt, and
1

u(0) = 2, u1/2
1 du 2

u(1) = e2 + 1, = (e2 + 1)3/2 − 23/2 ≈ 7.157. 3

e
0

2t

√ e2t

+ 1 dt =
2

e2 +1

We can easily extend the deﬁntion of line integrals to piecewise smooth curves. If γ is a piecewise smooth parametric curve that is the union of the smooth curves γ1 , γ2 , . . . , γn , γ = γ1 ∪ γ2 ∪ · · · ∪ γn , we deﬁne f (x, y, z) ds = γ γ1

f (x, y, z) ds + γ2 f (x, y, z) ds + · · · + γn f (x, y, z) ds.

Note that all the line integrals on the right side of the formula are along smooth curves and can be evaluated via the limit in (21.1) or via the integral formulas (21.2), (21.3), or (21.6). Example 21.3. Evaluate the line integral xy2 ds γ where γ is the triangle with vertices (0, 1, 2), (1, 0, 3), and (0, −1, 0).
169

Solution. We split γ into three line segments: • γ1 is the segment connecting (0, 1, 2) and (1, 0, 3): r(t) = t 0, 1, 2 + (1 − t) 1, 0, 3 = 1 − t, t, 3 − t • γ2 is the segment connecting (0, 1, 2) and (0, −1, 0): r(t) = t 0, 1, 2 + (1 − t) 0, −1, 0 = 0, 2t − 1, 2t • γ3 is the segment connecting (1, 0, 3) and (0, −1, 0): r(t) = t 1, 0, 3 + (1 − t) 0, −1, 0 = t, t − 1, 3t By (21.2),
1

(0 ≤ t ≤ 1);

(0 ≤ t ≤ 1);

(0 ≤ t ≤ 1). √ t2 − t3 dt =
0

xy2 ds = γ1 0 1

(1 − t)t2

√ (−1)2 + 12 + (−1)2 dt = 3

1

3 ; 12

xy2 ds = γ2 0 1

√ 0(2t − 1)2 02 + 22 + 22 dt = 0; t(t − 1) √ 2 12 + 12 + 32 dt = √ 11
0 1

√ t3 − 2t2 + t dt = 11 . 12

xy2 ds = γ3 0

Hence,

√ √ 3 11 xy ds = xy ds + xy ds + xy ds = + ≈ 0.421. 12 12 γ γ1 γ2 γ3
2 2 2 2

21.2. Line integrals of vector ﬁelds Let F be a continuous vector ﬁeld deﬁned on a piecewise smooth curve γ given by a vector function r(t), a ≤ t ≤ b. We deﬁne the line integral of F along γ by F · dr = γ γ

(F · T) ds,

where T is the unit tangent vector to the curve γ. Recall from §5.4 that the unit tangent vector T(t) at the point of γ corresponding to the value t of the parameter can be expressed in terms of r(t) by the formula r (t) T(t) = . r (t) Therefore, by (21.3) or (21.6) (depending on the dimension of F and γ), b (F · T) ds = γ a

F(r(t)) ·

r (t) r (t)

b

r (t) dt = a F(r(t)) · r (t) dt.

That is, we have b F · dr = γ γ

(F · T) ds = a F(r(t)) · r (t) dt.

(21.7)

170

What is the meaning of this kind of line integral? Based on the interpretation that we gave to the line integral of a function, we can say that 1 1 F · dr = (F · T) ds = average (F · T). length(γ) γ length(γ) γ γ Recall from §2.7 that the dot product F · T is the size of the projection of F onto the direction of T, taken with positive or negative sign according as the projection and T point in the same direction or in opposite directions. Since T(t) is the direction of the curve at the point corresponding to the value t of the parameter, we conclude that 1 F · dr length(γ) γ is the “average projection of F in the direction of the curve”. Example 21.4. Evaluate ye1 + ze2 + xe3 · dr, γ where γ is the helix r(t) = 2 cos t, 2 sin t, 3t , 0 ≤ t ≤ 2π. Solution. By (21.7),
2π

ye1 + ze2 + xe3 · dr = γ 0 2π

2 sin t, 3t, 2 cos t · (2 cos t) , (2 sin t) , (3t) dt 2 sin t, 3t, 2 cos t · −2 sin t, 2 cos t, 3 dt
0 2π

= =
0

− 4 sin2 t + 6t cos t + 6 cos t dt
2π 2π

= −2
0

(1 − cos 2t) dt + 6
0 2π 0

(t + 1) d(sin t)
2π 0 2π

= − 2t + sin 2t

+ 6(t + 1) sin t

−6
0

sin t dt = −4π.

In the above example, the parametrization of γ was given explicitly. However, we often have only a geometric description of the curve and have to obtain the parametrization ourselves. In such situations, it is important to remember that (21.7) is sensitive to the orientation of the curve, that is, the direction in which r(t) traces the curve as t increases. Indeed, if γ1 and γ2 are two parametric curves that represent the same geometric curve γ but with opposite orientations, then F · dr = − γ1 γ2

F · dr.

(21.8)

That is, if we change the orientation of the curve, the line integral changes sign. This fact goes back to the property of deﬁnite integrals that a b

f (x) dx = − b a

f (x) dx.

Because of (21.8), a geometric description of the curve must always specify its orientation.
171

Example 21.5. Evaluate x2 e1 + y2 e2 · dr, γ where γ is the right half of the ellipse x + 4y2 = 4 oriented in the counterclockwise direction.
2

Solution. The point (a, b) is on the given ellipse exactly when the point (a, 2b) is on the circle x + y2 = 4. Thus, we can parametrize γ by
2

x = 2 cos t,

y = sin t

(−π/2 ≤ t ≤ π/2).

Note that as t increases from −π/2 to π/2, the point (x(t), y(t)) moves along γ in the counterclockwise direction. Applying (21.7) to the above parametrization, we get π/2 x e1 + y e2 · dr =
2 2 γ −π/2 π/2

(2 cos t)2 e1 + (sin t)2 e2 · (2 cos t) e1 + (sin t) e2 dt 4 cos2 te1 + sin2 te2 · − 2 sin te1 + cos te2 dt
−π/2 π/2

= =
−π/2

− 8 cos2 t sin t + sin2 t cos t dt π/2 π/2

= −8 =0+2
0

cos2 t sin t dt +
−π/2 π/2 −π/2

sin2 t cos t dt
1

sin2 t cos t dt = 2
0

u2 du =

2 3

(u = sin t).

Suppose that F(x, y, z) = f (x, y, z)e1 . Then (21.7) yields b b

F · dr = γ a

f (r(t))e1 · x (t)e1 + y (t)e2 + z (t)e3 dt = a γ

f (r(t))x (t) dt.

Another notation for this integral is

f (x, y, z) dx, that is, b f (x, y, z) dx = γ γ

( f e1 ) · dr = a f (r(t))x (t) dt.

(21.9)

Similarly, applying (21.7) to the vector ﬁelds F(x, y, z) = f (x, y, z)e2 and F(x, y, z) = f (x, y, z)e3 , we obtain b f (x, y, z) dy = γ γ

( f e2 ) · dr = a b

f (r(t))y (t) dt

(21.10)

and f (x, y, z) dz = γ γ

( f e3 ) · dr = a f (r(t))z (t) dt.

(21.11)

These three integrals are called the line integrals of f with respect to x, y, and z, respectively. Since they are derived from (21.7), these integrals are also sensitive to the orientation of the curve γ
172

and change signs when we reverse the orientation of the curve. For example, if γ1 and γ2 are two parametric curves that represent the same geometric curve with opposite orientations, then f (x, y, z) dx = − γ1 γ2

f (x, y, z) dx.

Note that if F(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 , then we can use (21.9)–(21.11) to write F · dr = γ γ

P(x, y, z) dx + γ Q(x, y, z) dy + γ R(x, y, z) dz.

In other words, the line integral of a vector ﬁeld decomposes into a sum of the line integrals of its components with respect to the corresponding variables. The right side of the last identity is often written as γ P dx + Q dy + R dz. With this notation, we can rewrite that identity in the form (Pe1 + Qe2 + Re3 ) · dr = γ γ

P dx + Q dy + R dz.

(21.12)

Example 21.6. Evaluate y dx + x dy + xyz dz, γ where γ is the space curve x = et + e−t , Solution. By (21.9)–(21.11),
1

y = et − e−t ,

z = t2

(0 ≤ t ≤ 1).

y dx + x dy + xyz dz = γ 0 1

et − e−t et + e−t et − e−t
0 1 2

+ et + e−t et − e−t
2

+ e2t − e−2t t2 (2t) dt dt

= =
0 1

+ et + e−t

+ 2t3 e2t − e−2t
1

e2t − 2 + e−2t + e2t + 2 + e−2t dt +
0

t3 d e2t + e−2t
1

=2
0

e +e
2t 1 0

−2t

dt + t e + e
3 2t

−2t

1 0

−
0

e2t + e−2t (3t2 ) dt

= e2t − e−2t

+ e2 + e−2 −

1 3 = 2e2 − t2 e2t − e−2t e2t − e−2t (2t) dt 2 0 1 2 −2 e + 3e 3 = + t d e2t + e−2t 2 2 0 e2 + 3e−2 3 3 1 2t 1 = + t e2t + e−2t 0 − e + e−2t dt 2 2 2 0 3 5e2 + 15e−2 1 . = 2e2 + 3e−2 − e2t − e−2t 0 = 4 4

3 2 3 1 + 0 2

1

t2 d e2t − e−2t
0

173

Example 21.7. Evaluate y2 − xy dx, γ where γ is the curve y = ln x from (1, 0) towards (e, 1). Solution. We can parametrize γ as x = t, Hence, (21.9) gives y2 − xy dx = γ 1 e e e

y = ln t e (1 ≤ t ≤ e). ln2 t − t ln t dt.

Using integration by parts, we ﬁnd that e ln2 t dt = t ln2 t
1

e 1

−
1 e 1

t d ln2 t = e −
1 e

t 2t−1 ln t dt = e − 2
1

ln t dt

= e − 2 t ln t e e

+2
1

t t−1 dt = −e + 2(e − 1) = e − 2;
1 2 t 2

t ln t dt =
1 1

ln t d

1 2 t 2

=

ln t

e 1 1 4

e

e 1 2 t 2

−
1 2

t−1 dt = 1 e2 − 2

1 2

t dt =
1

1 4

e2 + 1 .

Thus, the given line integral equals e − 2 −

e + 1 ≈ −1.379.

Exercises
Evaluate the given line integral. 21.1. γ x3 y ds; γ is given by r(t) = te1 + t2 e2 , 0 ≤ t ≤ 1 x2 y ds; γ is the upper half of the circle x2 + y2 = 4 γ 21.2. 21.3. γ (xy + ln x) dy; γ is the arc of the curve y = x3/2 from (1, 1) to (0, 0) xy dx − x2 dy; γ is the arc of the circle x2 + y2 = 4 from (2, 0) to (0, 2) γ 21.4. 21.5. γ xy + z2 ds; γ is given by r(t) = cos 2te1 + sin 2te2 + 3te3 , 0 ≤ t ≤ π. z2 dx + y dz; γ is given by r(t) = (t − sin t)e1 + (1 − cos t)e2 + sin(t/2)e3 , 0 ≤ t ≤ 2π γ 21.6. 21.7. γ yz dx + 2xz dy − xy dz; γ consists of the line segments from (1, 0, 0) to (1, 0, 2) to (2, 2, 2) y2 dx + x2 dy + xyz dz; γ consists of the upper half of the circle x2 + y2 = 1 in the xy-plane taken clockwise γ 21.8.

and the line segment from (1, 0, 0) to (0, 1, 1) Evaluate the line integral γ F · dr for the given vector ﬁeld F and the given curve γ.

21.9. F(x, y, z) = x2 z e2 − (x + y)e3 ; γ is the ellipse r(t) = 2e1 + 3 sin te2 + cos te3 ,√ ≤ t ≤ 2π 0 21.10. F(x, y, z) = sin x e1 + cos y e2 − 2xyze3 ; γ is given by r(t) = t3 e1 + 2t2 e2 + te3 , 0 ≤ t ≤ 1 21.11. F(x, y) = x sin ye1 + ye2 ; γ is the arc of the parabola x = y2 from (1, −1) to (4, 2)

174

LECTURE 22

The Fundamental Theorem for Line Integrals
22.1. The main theorem So far, we have been reducing line integrals to deﬁnite integrals. In the case of conservative vector ﬁelds, there is another, faster way to evaluate their line integrals. That alternative approach is based on the following theorem, known as the fundamental theorem for line integrals. Theorem 22.1. Let γ be a piecewise smooth curve given by a vector function r(t), a ≤ t ≤ b, and let f (x, y, z) be a diﬀerentiable function, whose gradient vector ﬁeld f is continuous on γ. Then f · dr = f (r(b)) − f (r(a)). (22.1) γ Proof. We have f · dr = γ γ b

f x dx + fy dy + fz dz f x (r(t))x (t) + fy (r(t))y (t) + fz (r(t))z (t) dt a b

by (21.12) by (21.9)–(21.11) by the chain rule by the FTC.

= =

d f (r(t)) dt a dt = f (r(b)) − f (r(a))

Suppose that F(x, y, z) is a conservative vector ﬁeld with potential function f (x, y, z), that is, F = f . Theorem 22.1 then gives F · dr = γ γ

f · dr = f (r(b)) − f (r(a)).

This demonstrates the importance of conservative vector ﬁelds. If F is conservative and its potential f is known, the evaluation of the line integral of F is rather easy: all we need to do is calculate the net change of f along the curve. The next example illustrates this. Example 22.1. Evaluate F · dr, γ where F(x, y, z) = 2ye1 + 2xe2 + e3 and γ is the curve given by r(t) = sin 2t, cos 2t, ln(t + 1) , 0 ≤ t ≤ π. Solution. We observe that F = g, where g(x, y, z) = 2xy + z. Hence, by (22.1), F · dr = γ γ

g · dr = g(r(π)) − g(r(0)) = g(0, 1, ln(π + 1)) − g(0, 1, 0) = ln(π + 1).
175

Observe that the answer will stay the same, if we replace γ by any other piecewise smooth curve connecting the points (0, 1, 0) and (0, 1, ln(π + 1))! For example, the line integral of F along the curve displayed on Figure 22.1 (I have no idea what its parametrization may be!) will also equal ln(π + 1). (0, 1, ln(π + 1)) γ

(0, 1, 0) Figure 22.1. A curve with γ (2ye1

+ 2xe2 + e3 ) · dr = ln(π + 1)

22.2. Path independence We just saw that if F(x, y, z) is a conservative vector ﬁeld, Theorem 22.1 implies that the line integral γ F·dr depends only on the endpoints of γ. Thus, if γ1 and γ2 are two curves in the domain of F that share the same endpoints, then F · dr = γ1 γ2

F · dr,

that is, γ F · dr is path independent. The next theorem says that the converse of this observation is also true: if the line integrals of F are path independent, then F is conservative. Theorem 22.2. Let F be a vector ﬁeld that is continuous in a open, connected region D. If the line integral γ F · dr is path independent in D, then F is conservative in D. Note that unlike Theorems 20.2 and 20.4, this theorem requires less of the set D: only that it be open and connected (as opposed to simply connected). On the other hand, the test that it provides is much more elusive: we must determine that all (inﬁnitely many) line integrals of F are path independent. This hardly seems to be a practical test. And yet, if you look up the proofs of Theorems 20.2 and 20.4 in an advanced calculus text, you will discover that those results are deduced from Theorem 22.2. Thus, every time we show that some vector ﬁeld is conservative by means of Theorems 20.2 or 20.4, we implicitly make use of Theorem 22.2 too. We now return to the vector ﬁeld x −y e + 2 e2 , 2 1 +y x + y2 which we studied in Example 20.6(c). Recall that this ﬁeld is deﬁned and diﬀerentiable everywhere but at the origin. Furthermore, we have F(x, y) = x2 ∂ −y 2 + y2 ∂y x = x ∂ 2 + y2 ∂x x
176

=

y2 − x2 , (x2 + y2 )2

also for all (x, y) (0, 0). Back in Lecture #20, we showed that this vector ﬁeld is conservative on the plane cut along a half-line passing through the origin, but were unable to decide whether F is conservative in its entire domain. The next example shows that, in fact, this vector ﬁeld is not conservative in its full domain. −y x Example 22.2. The vector ﬁeld F(x, y) = 2 e + 2 e2 is not conservative. 2 1 x +y x + y2 Solution. Let γ be the unit circle x = cos t, By (21.9) and (21.10), F · dr = γ γ

y = sin t

(0 ≤ t ≤ 2π).

−y x dx + 2 dy x2 + y2 x + y2
2π

=
0 2π

=
0

(cos t)(sin t) (− sin t)(cos t) + 2 cos2 t + sin t cos2 t + sin2 t (− sin t)2 + (cos t)2 dt = 2π. cos2 t + sin2 t

dt

On the other hand, if F was conservative, with potential function, say f (x, y), Theorem 22.1 would give F · dr = f (cos 2π, sin 2π) − f (cos 0, sin 0) = f (1, 0) − f (1, 0) = 0. γ Since this value diﬀers from the value of the integral we got via direct calculation, F must be non-conservative. Example 22.3. Evaluate γ x2

−y x dx + 2 dy, 2 +y x + y2

where γ is the circle (x − 2)2 + (y + 1)2 = 25, oriented counterclockwise. First solution (not recommended). We just showed that the vector ﬁeld F(x, y) = x2 x −y e + 2 e2 2 1 +y x + y2

is conservative on the plane with a half-line removed and non-conservative on the plane with just the origin removed. Thus, F is conservative on curves that don’t loop around the origin and nonconservative on curves that do. In particular, F is not conservative on γ, since the origin lies inside it. This means that we can’t use Theorem 22.1 to evaluate the given integral, so we go back to the methods from the last lecture. A point (a, b) is on γ exactly when (a − 2, b + 1) is on the circle x2 + y2 = 25. Thus, we can derive the parametrization of γ from that of the latter circle: the parametric equations of γ are x = 2 + 5 cos t, y = −1 + 5 sin t
177

(0 ≤ t ≤ 2π).

Denote the given line integral by I. By (21.9) and (21.10),
2π

I=
0 2π

(1 − 5 sin t)(2 + 5 cos t) (2 + 5 cos t)(−1 + 5 sin t) + 2 + (−1 + 5 sin t)2 (2 + 5 cos t) (2 + 5 cos t)2 + (−1 + 5 sin t)2

dt

(1 − 5 sin t)(−5 sin t) + (2 + 5 cos t)(5 cos t) dt 4 + 20 cos t + 25 cos2 t + 1 − 10 sin t + 25 sin2 t 0 2π 2π 5 − sin t + 2 cos t 25 − 5 sin t + 10 cos t dt = dt. = 30 − 10 sin t + 20 cos t 6 − 2 sin t + 4 cos t 0 0 Most likely, you have not seen integrals like the last in single-variable calculus. It can be reduced to an integral of a rational function by the substitution u = tan(t/2), but we must be extremely careful, because this substitution is not “well-behaved”. We must even be careful, if we use mathematical software or an integral table. For example, Mathematica reports that = 5 − sin t + 2 cos t t dt = + arctan 6 − 2 sin t + 4 cos t 2 which would suggest that
2π 0

2 1 − tan(t/2) 2 1 − tan(t/2)

+ C,

t 5 − sin t + 2 cos t dt = + arctan 6 − 2 sin t + 4 cos t 2
2π

2π

= π,
0

... if it wasn’t for the fact that Mathematica also proudly proclaims that 5 − sin t + 2 cos t dt = 2π. 6 − 2 sin t + 4 cos t 0 In fact, both answers that Mathematica returns are correct and our calculation between them is ﬂawed, but this comes to show that we may want to avoid this approach... Second solution (recommended). Recall that in the solution of the last example, we showed −y x dx + 2 dy = 2π, 2 +y x + y2 γ0 where γ0 is the unit circle taken in the counterclockwise direction. It turns out that, although we cannot appeal to Theorem 22.1 to evaluate the given line integral directly, we can use it to show that x −y x −y dx + 2 dy = dx + 2 dy. 2 + y2 2 2 + y2 x +y x + y2 γ x γ0 x Let γ+ be the part of γ that lies above the x-axis and let τ+ be the curve that has the same √ √ endpoints as γ+ : (2 ± 24, 0), but consists of the line segment from (2 + √24, 0) to (1, 0), the upper half γ+ of the unit circle γ0 , and the line segment from (−1, 0) to (2 − 24, 0) (see Figure 22.2). 0 Also, let D+ be the plane cut along the negative y-axis (dashed line). Since γ+ and τ+ lie in D+ and F is conservative in D+ , Theorem 22.2 gives −y x −y x dx + 2 dy = dx + 2 dy. 2 + y2 2 2 + y2 x +y x + y2 γ+ x τ+ x x2 Next, let γ− be the part of γ that lies below the x-axis and let τ− be the curve that consists of √ the line segment from (2 − 24, 0) to (−1, 0), the lower half γ− of the unit circle γ0 , and the line 0
178

that

y γ+ x τ
+

y τ− x

γ− Figure 22.3. The curves γ− and τ− in D−

Figure 22.2. The curves γ+ and τ+ in D+

√ segment from (1, 0) to (2 + 24, 0) (see Figure 22.3). Also, let D− be the plane cut along the positive y-axis (dashed line). Since γ− and τ− lie in D− and F is conservative in D− , Theorem 22.2 gives −y x x −y dx + 2 dy = dx + 2 dy. 2 + y2 2 2 + y2 x +y x + y2 γ− x τ− x Combining this identity with the one before it, we get F · dr = γ γ+

F · dr + γ− F · dr =

τ+

F · dr +

F · dr. τ− We break the integrals over τ+ and τ− into six integrals over γ+ , γ− , and the two horizontal edges. 0 0 Each horizontal edge appears in the sum twice: once oriented from left to right and once oriented from right to left. Thus, those contributions cancel out (recall (21.8)) and we get F · dr + F · dr = F · dr + F · dr = γ0 τ+

τ−

γ+ 0

F · dr.

γ− 0

It follows that, as promised, x2 x −y dx + 2 dy = 2 +y x + y2 x2 −y x dx + 2 dy = 2π. 2 +y x + y2

γ

γ0

Exercises
Find a potential for the given vector ﬁeld and then use the fundamental theorem to evaluate γ F · dr.

22.1. F(x, y) = ey e1 + xey e2 ; γ is a smooth curve connecting (1, 0) and (2, −1) 22.2. F(x, y) = (4x + sin y)e1 + (sin y + x cos y)e2 ; γ is given by r(t) = (t − sin t)e1 + (1 − cos t)e2 , 0 ≤ t ≤ 2π √ 22.3. F(x, y) = ye1 + (x + 3y)e2 ; γ follows the ellipse x2 + 2y2 = 5 from (1, 2) to (− 5, 0) and then the line segment √ from (− 5, 0) to (1, −2) 22.4. F(x, y, z) = (yz + x2 )e1 + (xz + 4y)e2 + (xy − 3z2 )e3 ; γ consists of the line segments from (1, 2, 3) to (0, 2, 1) to (−1, 3, 0) to (3, 2, 1) 22.5. F(x, y, z) = y2 cos ze1 + 2xy cos ze2 − xy2 sin ze3 ; γ is given by r(t) = t2 e1 + t3 e2 − 2t2 e3 , 0 ≤ t ≤ 2
179

y γ −y x dx + 2 dy, 2 +y x + y2 γ where γ is the curve from the ﬁgure to the right. x2 22.6. Evaluate

x

180

LECTURE 23

Green’s Theorem
In this lecture we discuss the ﬁrst of the three major theorems of vector calculus. It is known as Green’s theorem and provides an alternative technique for evaluation of certain line integrals. 23.1. Some deﬁnitions First, we need to ﬁx some terminology. A curve γ, given by r(t), a ≤ t ≤ b, is called closed, if r(a) = r(b). In other words, the curve is closed, if its initital point is also its terminal point (see Figure 23.1). A curve γ is called simple, if r(t1 ) r(t2 ) whenever t1 t2 , except that we

γ

r(a) = r(b) Figure 23.1. A closed, positively oriented curve allow r(a) = r(b). In other words, a simple curve does not self-intersect and does not touch itself, but can be closed. For example, among the curves displayed on Figure 23.2, those in (a) and (e) are simple and those in (b)–(d) are not: the curve in (b) self-intersects at point A; the curve in (c) passes through its initial point before reaching its terminal point; and the curve in (d) touches itself at point B. It is intuitively clear (though quite diﬃcult to prove) that a piecewise smooth, simple, closed plane curve γ divides the plane into two sets, which we can call intuitively the interior and the exterior of γ, the interior being the bounded set and the exterior the unbounded. This fact is known as Jordan’s theorem, named after the French mathematician Camille Jordan for his work on plane curves, including this result. However, the ﬁrst full proof of Jordan’s theorem was given by an American: Oswald Veblen. Let γ be a piecewise smooth, simple, closed curve. We say that γ is positively oriented, if its interior lies always to the left of the point r(t) as it traces the curve in the direction of increasing of t. For “round” curves, such as a circle or the curve on Figure 23.2(e), the positive orientation of the curve is the counterclockwise orientation. However, this is not the case for all curves: the positive orientation goes clockwise for the part of the curve on Figure 23.1 that follows immediately the initial point r(a).
181

A r(a) (a) B (b) (c)

(d)

(e)

Figure 23.2. Simple and non-simple curves 23.2. Green’s theorem Theorem 23.1 (Green). Let P(x, y) and Q(x, y) have continuous ﬁrst-order partials on an open set U. Let γ be a positively oriented, piecewise smooth, simple, closed plane curve such that γ and all of its interior D are contained inside U. Then P dx + Q dy = γ D

∂Q ∂P − ∂x ∂y

dA.

(23.1)

In other words, Green’s theorem relates a line integral along a simple, closed plane curves to a double integral over the interior of the curve. Since we know how to evaluate double integrals this yields another possible integration technique for line integrals along closed curves. However, there is a catch: we absolutely must check the conditions of Green’s theorem before we use it. For example, we showed in Example 20.6(c) that the functions −y x P(x, y) = 2 , Q(x, y) = 2 2 x +y x + y2 are diﬀerentiable and satisfy the condition ∂P/∂y = ∂Q/∂x everywhere except at the origin. Thus, if γ is a simple, closed curve whose interior D does not contain the origin, Green’s theorem gives P dx + Q dy = γ D

∂Q ∂P − ∂x ∂y

dA = 0.

Note that for such a curve the application of Green’s theorem is legitimate. On the other hand, if the origin is in the interior of the curve γ, the conditions of Green’s theorem are violated (at one point) and the theorem is not applicable. Were we to apply it anyways, the result would again be that the line integral along γ is 0. For example, such an illegitimate application of Green’s theorem yields x −y dx + 2 dy = 0 2 + y2 x + y2 γ0 x
182

when γ0 is the unit circle x2 + y2 = 1. However, we know from Example 22.2 that the last integral equals 2π, not 0. We see that if the functions P(x, y) and Q(x, y) or some of their derivatives are not deﬁned at even a single point in the interior of the curve, Green’s theorem is not applicable and should not be used. Now, let us take a look at some of its legitimate uses. Example 23.1. Evaluate e4x sin x + 2y dx + x2 + arctan y dy, γ where γ is the rectangle with vertices (1, 2), (5, 2), (5, 4), (1, 4). Solution. We have P(x, y) = e4x sin x + 2y and Q(x, y) = x2 + arctan y. The partials of these functions are: 1 . P x (x, y) = (4 sin x + cos x)e4x , Py (x, y) = 2, Q x (x, y) = 2x, Qy (x, y) = 1 + y2 Since these are all continuous everywhere, the open set U in the statement of Green’s theorem can be taken to be R2 . We can therefore apply Green’s theorem. We get
5 4

e4x sin x + 2y dx + x2 + arctan y dy =
C D 5

(2x − 2) dA =
1 2

(2x − 2) dydx
5 1

=
1

2(2x − 2) dx = 2 x2 − 2x

= 32.

Here, D denotes the rectangle [1, 5] × [2, 4]. Example 23.2. Evaluate e4x sin x + 2y dx + x2 + arctan y dy, γ where γ is the boundary of the semiannular region shown on Figure 23.3 (the radii of the two semicircles are 2 and 3). y (0, 2) γ x

(0, −3) Figure 23.3. The region D Solution. The integrand is the same as in the previous problem, but the curve is diﬀerent. We can appeal to Green’s theorem for the same reasons as in the previous example. We get e4x sin x + 2y dx + x2 + arctan y dy = γ D

(2x − 2) dA,

183

where D is the semiannular region shown on Figure 23.3. Note that this region is conveniently described in polar coordinates: D = (r, θ) : 2 ≤ r ≤ 3, −π/2 ≤ θ ≤ π/2 . Thus, by converting the double integral to polar coordinates, we obtain
3 π/2

e sin x + 2y dx + x + arctan y dy =
4x 2 γ 2 3 −π/2

(2r cos θ − 2)r dθdr (2r sin θ − 2θ)r
2 3 π/2 −π/2

= =
2

dr

4r2 − 2πr dr = 25 1 − 5π. 3

In the above examples, we used Green’s theorem to evaluate line integrals by means of double integrals. However, Green’s theorem is a relation between the line integral and the double integral appearing in (23.1) that can be used in either direction. Sometimes, we want to use Green’s theorem to evaluate a double integral by means of a line integral. For example, under the assumption of Theorem 23.1, we have x dy = γ D

∂ x (x) − ∂y (0) dA =
D

1 dA = area(D), 1 dA = area(D).
D

(−y) dx = γ D

∂ x (0) − ∂y (−y) dA =

In other words, we have the following new formulas for the area enclosed by a simple, closed curve. Theorem 23.2. Let D be a plane region, bounded by a piecewise smooth, simple, closed curve γ. Then 1 x dy − y dx. (23.2) area(D) = x dy = (−y) dx = 2 γ γ γ x2 y2 Example 23.3. Find the area enclosed by the ellipse 2 + 2 = 1. a b Solution. We will use formula (23.2) and the parametric representation of the ellipse: x = a cos t, By (23.2), area(D) = 1 2 1 = 2 x dy − y dx =
C 2π 0 2

y = b sin t
2π

(0 ≤ θ ≤ 2π).

1 2

(a cos t)(b cos t) − (b sin t)(−a sin t) dt
0 2π 2

1 ab cos t + sin t dt = 2
184

ab dt = πab.
0

23.3. Vector forms of Green’s theorem* We can use the curl and divergence of a vector ﬁeld to give alternative formulations of Green’s theorem. Suppose that γ is a positively oriented, piecewise smooth, simple, closed curve with parametrization r(t) = x(t)e1 + y(t)e2 + 0e3 (a ≤ t ≤ b), that is, γ is a plane curve viewed as a space curve. Furhter, consider the vector ﬁeld F(x, y, z) = P(x, y)e1 + Q(x, y)e2 + 0e3 . By (21.12), F · dr = γ γ

P(x, y) dx + Q(x, y) dy. ∂Q ∂P − e3 = ∂x ∂y ∂Q ∂P − e3 , ∂x ∂y

Also, by the deﬁnition of curl, curl F = ∂0 ∂Q − e1 + ∂y ∂z ∂P ∂0 − e2 + ∂z ∂x

because P(x, y) and Q(x, y) do not depend on z. Thus, we can rewrite (23.1) as F · dr = γ D

(curl F) · e3 dA.

(23.3)

Another formulation of Green’s theorem involves the vector ﬁeld G(x, y, z) = Q(x, y)e1 − P(x, y)e2 + 0e3 , which is orthogonal to the vector ﬁeld F that we used in (23.3): F · G = P(x, y)Q(x, y) + Q(x, y)(−P(x, y)) + 0 = 0. Note that div F = Also, by (21.7), F·r= γ γ

∂Q ∂P − . ∂x ∂y (F · T) ds,

where T is the unit tangent vector to γ. It is a nice exercise to try to use the properties of the dot product to convince yourself that n(t) = y (t)e1 − x (t)e2 x (t)2 + y (t)2

is a unit vector that is normal to the curve γ and points towards the exterior of the curve; it is called the outward normal vector of γ. Since F · T = G · n, we get F·r= γ γ

(G · n) ds,

and we can restate (23.1) as (G · n) ds = γ D

div G dA.

(23.4)

185

Exercises
23.1. (a) Let γ be the square with vertices (1, 1), (2, 1), (2, 2), and (1, 2). Parametrize γ and use the formulas from Lecture #21 to evaluate x2 y dx + xy2 dy. γ (b) Evaluate the integral from part (a) by means of Green’s theorem. Use Green’s theorem to evaluate the given line integral (all the curves are assumed positively oriented). 23.2. γ y3 dx − x3 dy; γ is the circle x2 + y2 = 4 (xy + tan x) dx + x2 + ecos y dy; γ is the boundary of the region bounded by y = x and y = x2 γ 23.3. 23.4. γ xy3 dx + 4x2 y2 dy; γ consists of the semicircle x =

9 − y2 and the line segment from (0, 3) to (0, −3)

23.5. xy2 + sin x2 dx + (x2 + y2 )3/2 + arctan y dy; γ is the boundary of region bounded by the semicircle γ √ √ y = 4 − x2 and the lines y = x and y = − 3x 23.6. γ e x sin y dx + e x cos y dy; γ is the ellipse (x + 1)2 + 5(y − 2)2 = 7 F · dr for the given ﬁeld F(x, y) and the given curve γ. γ Use Green’s theorem to evaluate the line integral

23.7. F(x, y) = y3/2 + x e1 + x2 + y e2 ; γ consists of the graph of y = sin2 x from (0, 0) to (π, 0) and the line segment from (π, 0) to (0, 0) 23.8. F(x, y) = x3 + y e1 + 2xy + y4 e2 ; γ is the cardioid r = 2 + cos θ, positively oriented 23.9. F(x, y) = xe1 + x3 + 3xy2 e2 ; γ consists of the graph of y = 4 − x2 from (−2, 0) to (2, 0) and the line segment from (2, 0) to (−2, 0) 23.10. When the circle x2 + (y − 1)2 = 1 is rolled along the x-axis in the positive direction, the point whose original position is the origin traces a curve called a cycloid (see Figure 23.4). This curve can be parametrized by the vector function r(t) = (t − sin t)e1 + (1 − cos t)e2 (t ≥ 0).

Use Theorem 23.2 to ﬁnd the area bounded by the x-axis and the ﬁrst arc of the cycloid (that is, the part of the cycloid with 0 ≤ t ≤ 2π).

Figure 23.4. Cycloid 23.11. When a circle of radius 1 is rolled along the outside of the circle x2 + y2 = 4 a ﬁxed point on the smaller circle traces a curve called an epicycloid (see Figure 23.5). This curve can be parametrized by the vector function r(t) = (3 cos t − cos 3t)e1 + (3 sin t − sin 3t)e2 Use Theorem 23.2 to ﬁnd the area enclosed by this epicycloid.
186

(0 ≤ t ≤ 2π).

Figure 23.5. Epicycloid

Figure 23.6. Hypocycloid

Figure 23.7. Astroid

23.12. When a circle of radius 1 is rolled along the inside of the circle x2 + y2 = 9 a ﬁxed point on the smaller circle traces a curve called a hypocycloid (see Figure 23.6). This curve can be parametrized by the vector function r(t) = (2 cos t + cos 2t)e1 + (2 sin t − sin 2t)e2 Use Theorem 23.2 to ﬁnd the area enclosed by this hypocycloid. 23.13. When a circle of radius 1 is rolled along the inside of the circle x2 + y2 = 16 a ﬁxed point on the smaller circle traces a diﬀerent kind of hypocycloid called also an astroid (see Figure 23.7). This curve can be parametrized by the vector function r(t) = (3 cos t + cos 3t)e1 + (3 sin t − sin 3t)e2 (0 ≤ t ≤ 2π). Use Theorem 23.2 to ﬁnd the area enclosed by this astroid. (0 ≤ t ≤ 2π).

187

LECTURE 24

Surface Integrals
24.1. Surface integrals of scalar functions We start with a smooth parametric surface Σ, given by the vector function r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D). Recall that the smoothness of Σ means that ru (u, v) and rv (u, v) are continuous and (ru ×rv )(u, v) 0 in D (see §10.3). Further, suppose that f (x, y, z) is a continuous function on a set S in R3 that contains the surface Σ. In particular, f (x, y, z) is deﬁned at each point (x, y, z) on Σ. For simplicity, we shall assume that D is a rectangle in the uv-plane: D = [a, b] × [c, d]. We can treat the general case similarly, but there are more technical details to pay attention to and little extra insight to gain. For each n ≥ 1, let a = u0 < u1 < · · · < un = b be the points that divide the interval [a, b] into n subintervals of equal lengths; let c = v0 < v1 < · · · < vn = d be the points that divide the interval [c, d] into n subintervals of equal lengths. The points (un , v0 ) (un , v1 ) · · · . . . . . . (ui , v0 ) (ui , v1 ) · · · . . . . . . (u1 , v0 ) (u1 , v1 ) · · · (u0 , v0 ) (u0 , v1 ) · · · (un , v j ) · · · . . . (ui , v j ) · · · . . . (u1 , v j ) · · · (u0 , v j ) · · · (un , vn ), . . . (ui , vn ), . . . (u1 , vn ), (u0 , vn ),

deﬁne a partition of the rectangle D into n2 congruent subrectangles. The vector function r maps each of those small rectangles to a small “patch” on Σ and each grid point (ui , v j ) to a point Pi j on the surface. The patches form a partition of Σ into n2 pieces, which we label as Σi j , 1 ≤ i, j ≤ n, so that Σi j is the image of the rectangle that has (ui , v j ) as its upper right corner point. Further, we pick n2 sample points (u∗j , v∗j ) so that i i ui−1 ≤ u∗j ≤ ui , i v j−1 ≤ v∗j ≤ v j . i These yield n2 sample points P∗j on the surface, with coordinates xi∗j = x(u∗j , v∗j ), y∗j = y(u∗j , v∗j ), i i i i i i z∗j = z(u∗j , v∗j ). Then the surface integral of f over Σ is deﬁned by i i i n Σ n

f (x, y, z) dS = lim

n→∞

f (xi∗j , y∗j , z∗j )∆S i j , i i i=1 j=1

∆S i j = area(Σi j ),

(24.1)

provided that the limit exists. Recall that, by Theorem 18.1, ∆S i j =
Di j

ru × rv dA,
189

where Di j is the rectangle in the uv-plane whose image is Σi j . Using this formula, the above deﬁnition, the properties of continuous functions, and the deﬁnition of the double integral, it can be shown that for continuous functions f the limit (24.1) always exists and the surface integral can be expressed as a double integral. Theorem 24.1. Let Σ be a smooth parametric surface, given by the vector function r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 If f (x, y, z) is continuous on Σ, then
Σ

((u, v) ∈ D).

f (x, y, z) dS =
D

f (r(u, v)) ru × rv dA.

(24.2)

Similarly to line integrals, we can extend the above deﬁnition of the surface integral to piecewise smooth surfaces. If Σ is a piecewise smooth surface that is the union of the smooth surfaces Σ1 , Σ2 , . . . , Σn , Σ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn , then we deﬁne
Σ

f (x, y, z) dS =

Σ1

f (x, y, z) dS +

Σ2

f (x, y, z) dS + · · · +

f (x, y, z) dS .
Σn

Remark. Like all other types of integrals we have encountered so far, the surface integral can be interpreted in terms of averages. In this case, we are dealing with the average value of the function on the surface: average f (x, y, z) =
(x,y,z)∈Σ

1 area(Σ)

f (x, y, z) dS .
Σ

(24.3)

Example 24.1. Evaluate xy dS ,
Σ

where Σ is the graph of z =

2 3

x3/2 + y3/2 over the rectangle 0 ≤ x, y ≤ 1.

Solution. The surface can be parametrized by the vector function r(u, v) = ue1 + ve2 + Hence, ru (u, v) = e1 + 0e2 + u1/2 e3 , and e1 e2 e3 0 u1/2 1 u1/2 1 0 ru × rv = 1 0 u1/2 = e1 − e = −u1/2 e1 − v1/2 e2 + e3 . 1/2 1/2 e2 + 1 v 0 v 0 1 3 1/2 0 1 v Applying (24.2), we get xy dS =
S 0≤u,v≤1 2 3

u3/2 + v3/2 e3

(0 ≤ u, v ≤ 1).

rv (u, v) = 0e1 + e2 + v1/2 e3 ,

√ uv u + v + 1 dA =
0

1

1

v
0

√ u u + v + 1 du dv.

(24.4)

190

Using the integration by parts, we ﬁnd that
1 0

√ u u + v + 1 du = =

1 2 3

u d(u + v + 1)3/2
0 2 u(u 3

+v+

1 1)w3/2 0

1

−

2 3

(u + v + 1)3/2 du
0

= 2 (v + 2)3/2 − 3 = 2 (v + 2)3/2 − 3
1

1 2 2 (u + v + 1)5/2 0 3 5 4 4 (v + 2)5/2 + 15 (v + 15

1)5/2 .

Inserting the result back into (24.4) and using another integration by parts, we obtain xy dS =
S 0 1 2 v 3 (v + 2)3/2 − 4 (v 15 4 (v 15

+ 2)5/2 +

4 (v 15

+ 1)5/2 dv + 1)7/2
1 0

=
0

vd

+ 2)5/2 −

8 (v 105

+ 2)7/2 +

8 (v 105

= v

4 (v + 15 1

2)5/2 −
4 (v 15

8 (v 105

+ 2)7/2 +
8 (v 105

8 (v 105

+ 1)7/2
8 (v 105

−
0

+ 2)5/2 −

+ 2)7/2 +

+ 1)7/2 dv
4 (v 63

= =

4 15 4 105

3

4 2 4 − 2 37/2 + 2 27/2 − 15 7 (v + 2)7/2 − 63 (v + 2)9/2 + 7 7 √ √ √ √ 8 9 3 + 16 2 − 105 1 81 3 − 8 2 − 2 ≈ 0.372. 9 5/2

+ 1)9/2

1 0

24.2. Surface integrals of vector ﬁelds Definition. Let F be a continuous vector ﬁeld deﬁned on an oriented surface Σ with unit normal vector n. Then the surface integral of F over Σ is
Σ

F · dS =

(F · n) dS .
Σ

(24.5)

Remark. Since F · n is the size of the projection of F in the direction of n, we can interpret the surface integral of F as the average size of its projection onto the direction of orientation of Σ. Suppose that the surface is given by the vector function r(u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 and that it is oriented using the unit normal vector n(u, v) = Combining (24.5) and (24.2), we get
Σ

((u, v) ∈ D),

ru × rv . ru × rv ru × rv ru × rv

F · dS = =

Σ

(F · n) dS = F· ru × rv ru × rv

F·
Σ

dS F · (ru × rv ) dA.
D

ru × rv dA =
191

(24.6)

D

In particular, when Σ is the graph z = f (x, y), (x, y) ∈ D, with the upward orientation (recall Example 10.8), we obtain
Σ

(Pe1 + Qe2 + Re3 ) · dS =
D

− P(x, y) f x (x, y) − Q(x, y) fy (x, y) + R(x, y) dA.

(24.7)

Example 24.2. Evaluate
Σ

(xe1 − 2ye2 + ze3 ) · dS,

where Σ is the elliptic paraboloid z = 4 − 2x2 − 2y2 , z ≥ 0, with the upward orientation. Solution. Condition z ≥ 0 restricts x, y to the circle 4 − 2x2 − 2y2 ≥ 0 ⇐⇒ x2 + y2 ≤ 2.

Since we are using the upward orientation, we can appeal to (24.7). We have ∂ x 4 − 2x2 − 2y2 = −4x, so (24.7) gives
Σ

∂y 4 − 2x2 − 2y2 = −4y,

(xe1 − 2ye2 + ze3 ) · dS = x2 +y2 ≤2

4x2 − 8y2 + z(x, y) dA 4x2 − 8y2 + 4 − 2x2 − 2y2 dA x2 +y2 ≤2

= = x2 +y2 ≤2

4 + 2x2 − 10y2 dA.

Converting the double integral to polar coordinates, we obtain
√ 2π Σ 2

(xe1 − 2ye2 + ze3 ) · dS =
0 2π 0

4 + 2r2 cos2 θ − 10r2 sin2 θ r drdθ
√ 1 2r2 + 2 r4 cos2 θ − 5 r4 sin2 θ 2 2π 2 0

=
0

dθ

=
0 2π

4 + 2 cos2 θ − 10 sin2 θ dθ 4 + (1 + cos 2θ) − 5(1 − cos 2θ) dθ = 0.
0

=

Example 24.3. Evaluate
Σ

(xze1 + xe2 + ye3 ) · dS,

where Σ is the hemisphere x2 + y2 + z2 = 25, y ≥ 0, oriented in the direction of the positive y-axis.
192

Solution. It is best to represent the hemisphere in spherical coordinates: r(θ, φ) = 5 cos θ sin φe1 + 5 sin θ sin φe2 + 5 cos φe3 Then rθ = −5 sin θ sin φe1 + 5 cos θ sin φe2 , rφ = 5 cos θ cos φe1 + 5 sin θ cos φe2 − 5 sin φe3 ; rθ × rφ = −25 sin2 θ sin φ cos φ(e1 × e2 ) + 25 sin θ sin2 φ(e1 × e3 ) + 25 cos2 θ sin φ cos φ(e2 × e1 ) − 25 cos θ sin2 φ(e2 × e3 ) = −25 cos θ sin2 φe1 − 25 sin θ sin2 φe2 − 25 sin φ cos φe3 = −25 sin φ(cos θ sin φe1 + sin θ sin φe2 + cos φe3 ). Note that the vector rθ × rφ has a negative second component for all θ, φ with 0 ≤ θ, φ ≤ π. Therefore, we can orient Σ in the direction of the positive y-axis by taking n= With this choice, we have F(r(θ, φ)) = 25 cos θ sin φ cos φe1 + 5 cos θ sin φe2 + 5 sin θ sin φe3 = 5 sin φ(5 cos θ cos φe1 + cos θe2 + sin θe3 ), F(r(θ, φ)) · rφ × rθ = 125 sin2 φ 5 cos2 θ sin φ cos φ + sin θ cos θ sin φ + sin θ cos φ . Thus, (24.6) yields
Σ

(0 ≤ θ, φ ≤ π).

rφ × rθ −(rθ × rφ ) = . rθ × rφ rφ × rθ

F · dS =
0≤θ,φ≤π

125 sin2 φ(5 cos2 θ sin φ cos φ + sin θ cos θ sin φ + sin θ cos φ) dA π π

= 125
0 0

5 cos2 θ sin3 φ cos φ + sin θ cos θ sin3 φ + sin θ sin2 φ cos φ dθdφ. π We now use that
0

π cos θ dθ = , 2
2

π

π

sin θ cos θ dθ = 0,
0 0 π

sin θ dθ = 2.

Substituting these values in the iterated integral above, we obtain F · dS = 125
S 0 0

(2.5π sin3 φ cos φ + 2 sin2 φ cos φ) dφ (2.5πu3 + 2u2 ) du = 0.
0

= 125

Example 24.4. The surface Σ with vector equation r(u, v) = u cos ve1 + u sin ve2 + ve3 ,
193

Figure 24.1. Two views of the helicoid where 0 ≤ u ≤ 1, 0 ≤ v ≤ π, is called helicoid (see Figure 24.1). Evaluate the surface integral of F(x, y, z) = x over Σ with upward orientation. Solution. We have ru = cos ve1 + sin ve2 , rv = −u sin ve1 + u cos ve2 + e3 ; ru × rv = u cos2 v(e1 × e2 ) + cos v(e1 × e3 ) − u sin2 v(e2 × e1 ) + sin v(e2 × e3 ) = sin ve1 − cos ve2 + ue3 . Note that the vector ru × rv has a positive third component for all u with 0 ≤ u ≤ 1. Therefore, we can orient Σ in the upward direction by the unit normal vector ru × rv n= . ru × rv With this choice, we have √ √ F(r(u, v)) = u cos v 1 + u2 e1 + u sin v 1 + u2 e2 + uv(cos v + sin v)e3 F(r(u, v)) · (ru × rv ) = u2 v(cos v + sin v). Thus, (24.6) yields π Σ 1

1 + x2 + y2 e1 + y

1 + x2 + y2 e2 + (x + y)ze3

F · dS =
0 0

u2 v(cos v + sin v) dudv

1 π 1 π = v(cos v + sin v) dv = v d(sin v − cos v) 3 0 3 0 1 1 π π 2 π = v(sin v − cos v) 0 − (sin v − cos v) dv = − . 3 3 0 3 3
194

Exercises
24.1. Evaluate the surface integral
Σ

x2 yz dS , where Σ is the surface 0 ≤ x ≤ 3, 0 ≤ y ≤ 2,

z = 1 + 2x + 3y,

that is, the part of the plane z = 1 + 2x + 3y that lies above the rectangle [0, 3] × [0, 2]. 24.2. Evaluate the surface integral
Σ

x2 z2 dS , where Σ is the surface z= x2 + y2 , 1 ≤ x2 + y2 ≤ 4,

that is, the part of the cone z2 = x2 + y2 that lies between the planes z = 1 and z = 2. 24.3. Evaluate the surface integral parametric surface given by
Σ

x2 z + y2 z dS , where Σ is the hemisphere x2 + y2 + z2 = 4, z ≥ 0, that is, the 0 ≤ u ≤ 2π, 0 ≤ v ≤ π/2.

r(u, v) = 2 cos u sin ve1 + 2 sin u sin ve2 + 2 cos ve3 , 24.4. Evaluate the surface integral
Σ

x2 y + z2 dS , where Σ is the part of the cylinder x2 + y2 = 9 that lies between 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2.

the planes z = 0 and z = 2, that is, the parametric surface given by r(u, v) = 3 cos ue1 + 3 sin ue2 + ve3 , 24.5. Evaluate the surface integral
2 2 Σ

F · dS, where F(x, y, z) = xye1 + yze2 + zxe3 and Σ is the part of the paraboloid

z = 4 − x − y that lies above the square 0 ≤ x, y ≤ 1 and is oriented upward. 24.6. Evaluate the surface integral
Σ

F · dS, where F(x, y, z) = xzey e1 − xzey e2 + ze3 and Σ is the part of the plane

x + y + z = 1 in the ﬁrst octant with the upward orientation. 24.7. Evaluate the surface integral
Σ

F · dS, where F(x, y, z) = xze1 + 2yze3 and Σ is the paraboloid y = x2 + z2 ,

0 ≤ y ≤ 1, oriented in the direction of the positive y-axis. 24.8. Evaluate the surface integral
Σ

F · dS, where F(x, y, z) = xe1 + ye2 + xyze3 and Σ is the helicoid given by 0 ≤ u ≤ 1, 0 ≤ v ≤ π,

r(u, v) = u cos ve1 + u sin ve2 + ve3 , and oriented by n = ru × rv .

195

LECTURE 25

Stokes’ Theorem
In the last two lectures, we discuss the other two major integral theorems of vector calculus: Stokes’ theorem and Gauss’ divergence theorem. Stokes’ theorem is a generalization of Green’s theorem to closed curves on surfaces in space. Before we state it, we introduce the notion of orientation of curves on an oriented surface. Definition. Suppose that Σ is an oriented surface with unit normal vector n. We say that a closed curve γ on Σ is positively oriented, if the vector T × n points towards the interior of the curve; here T is the unit tangent vector to γ. Theorem 25.1 (Stokes). Let F be a vector ﬁeld that has continuous partial derivatives on an open set U in R3 . Let Σ be an oriented, piecewise smooth surface that is contained in U and is bounded by a simple, closed, piecewise smooth, positively oriented curve γ. Then F · dr = γ curl F · dS.
Σ

(25.1)

We said above that Stokes’ theorem is a generalization of Green’s. It would be nice to corroborate that. Consider the parametric surface Σ given by r(x, y) = xe1 + ye2 + 0e3 ((x, y) ∈ D), that is, Σ is the region D in the xy-plane turned into a “three-dimensional surface”. If we give this surface an upward orientation, then the unit normal vector n is e3 , and Stokes’ theorem gives F · dr = γ Σ

curl F · dS =

Σ

(curl F) · e3 dS .

Converting the latter surface integral into a double integral, we ﬁnd r x (x, y) = e1 , and hence, F · dr = γ Σ

ry (x, y) = e2 ,

r x × ry = e3 , (curl F) · e3 dA.
D

(curl F) · e3 dS =

This is exactly one of the vector forms of Green’s theorem (recall (23.3)). We now move to describe the main applications of Stokes’ theorem. They are similar to the applications of Green’s theorem. First, we can apply Stokes’s theorem to convert a line integral over a closed space curve γ to a surface integral over a surface Σ that has γ as its boundary. Example 25.1. Evaluate e x +x + z2 e1 + arctan y + x2 e2 + ln(z2 + 1) + y2 e3 · dr,
3

γ

197

Figure 25.1. The spherical patch x2 + y2 + z2 = 9, x, y, z ≥ 0, and its boundary where γ is the positively oriented boundary of the part of the sphere x2 + y2 + z2 = 9 that lies in the ﬁrst octant, with downward orientation (see Figure 25.1. Solution. All the partials of F will be continuous everywhere, so we can apply Stokes’ theorem. We get F · dr = γ curl F · dS,
Σ

where Σ is the given surface and F(x, y, z) = e x +x + z2 e1 + arctan y + x2 e2 + ln(z2 + 1) + y2 e3 .
3

As we do usually when dealing with spheres, we parametrize Σ using spherical coordinates: r(θ, φ) = 3 cos θ sin φe1 + 3 sin θ sin φe2 + 3 cos φe3 Then, similarly to Example 24.3, rθ = −3 sin θ sin φe1 + 3 cos θ sin φe2 , rφ = 3 cos θ cos φe1 + 3 sin θ cos φe2 − 3 sin φe3 ; rθ × rφ = −9 sin φ(cos θ sin φe1 + sin θ sin φe2 + cos φe3 ). Since −9 sin φ cos φ ≤ 0 for all φ, 0 ≤ φ ≤ π/2, the downward orientation of Σ is given by n= Furthermore, we have curl F = (2y − 0)e1 − (0 − 2z)e2 + (2x − 0)e3 = 2ye1 + 2ze2 + 2xe1 ; curl F(r(θ, φ)) = 6 sin θ sin φe1 + 6 cos φe2 + 6 cos θ sin φe3 , curl F(r(θ, φ)) · (rθ × rφ ) = −54 sin2 φ sin θ cos θ sin φ + (sin θ + cos θ) cos φ .
198

(0 ≤ θ, φ ≤ π/2).

rθ × rφ . rθ × rφ

We conclude that F · dr = γ 0≤θ,φ≤π/2 π/2 π/2

curl F(r(θ, φ)) · (rθ × rφ ) dA sin2 φ sin θ cos θ sin φ + (sin θ + cos θ) cos φ dθdφ
0 π/2 0

= −54 = −54
0 π/2

sin2 φ sin2 φ
0 π/2

1 2 1 2

sin2 θ sin φ + (− cos θ + sin θ) cos φ sin φ + 2 cos φ dφ π/2 π/2 0

dφ

= −54 = −27
0 1

1 − cos2 φ sin φ dφ − 108
0 1

sin2 φ cos φ dφ

= −27
0

1 − u2 du − 108
0

u2 du = −54.

Example 25.2. Evaluate (yze1 + 2xze2 + 3xye3 ) · dr, γ where γ is the intersection curve of the ellipsoid x2 + 2y2 + z2 = 4 and the plane x + z = 2, oriented positively relative to the outward orientation of the ellipsoid. Solution. The intersection curve γ of the ellipsoid and the plane is displayed on Figure 25.2. It can be seen from the ﬁgure that the positive orientation of γ relative to the outward orientation of the ellipsoid is in the counterclockwise direction. Note that this is also the positive orientation of γ relative to the upward orientation of the plane x + z = 2. The equations of γ are x2 + 2y2 + z2 = 4, x+z=2 ⇐⇒ x2 + 2y2 + (2 − x)2 = 4, z = 2 − x.

Figure 25.2. The intersection curve of x2 + 2y2 + z2 = 4 and x + z = 2
199

Some basic algebra shows that x2 + 2y2 + (2 − x)2 = 4 ⇐⇒ ⇐⇒ so γ is described by the equations (x − 1)2 + y2 = 1, z = 2 − x. That is, γ contains the points (x, y, z) on the plane z = 2 − x for which the point (x, y) lies on the circle (x − 1)2 + y2 = 1 in the xy-plane. It is possible to parametrize γ and to evaluate the given line integral directly (see the exercises), but we want to use Stokes’ theorem. Let Σ be the portion of the plane z = 2 − x that lies within γ, with the upward orientation. Then, by Stokes’ theorem, (yze1 + 2xze2 + 3xye3 ) · dr = γ Σ

x2 + 2y2 + 4 − 4x + x2 = 4 x2 − 2x + y2 = 0 ⇐⇒

⇐⇒

2x2 − 4x + 2y2 = 0

(x − 1)2 + y2 = 1,

curl(yze1 + 2xze2 + 3xye3 ) · dS.

We have e1 e2 e3 ∂ ∂z ∂ ∂y ∂ ∂z ∂ x ∂y ∂z = y e − x e + x e curl(yze1 + 2xze2 + 3xye3 ) = 2xz 3xy 1 yz 3xy 2 yz 2xz 3 yz 2xz 3xy = (3x − 2x)e1 − (3y − y)e2 + (2z − z)e3 = xe1 − 2ye2 + ze3 . Further, Σ is the graph of z = 2 − x for (x, y) in the disk D with equation (x − 1)2 + y2 ≤ 1. Thus, (24.7) gives
Σ

curl(yze1 + 2xze2 + 3xye3 ) · dS = =

Σ

(xe1 − 2ye2 + ze3 ) · dS (−x∂ x (2 − x) − (−2y)∂y (2 − x) + z(x, y)) dA

D

=
D

(x + 0 + (2 − x)) dA =
D

2 dA.

The last double integral represents twice the area of D, which is a disk of radius 1. Consequently, (yze1 + 2xze2 + 3xye3 ) · dr = γ D

2 dA = 2area(D) = 2π.

In the last two examples, we used Stokes’ theorem to reduce a line integral of a vector ﬁeld to a surface integral of its curl. In the next couple of examples, we describe the other main application of Stokes’ theorem: to simplify the evaluation of surface integrals. To illustrate the ﬁrst ﬂavor of such applications, we take a look at a surface integral that we evaluated in Example 24.2 by writing it as a double integral. We now give an alternative solution, which uses Stokes’ theorem. Example 25.3. Evaluate
Σ

(xe1 − 2ye2 + ze3 ) · dS,

where Σ is the part of the paraboloid z = 4−2x2 −2y2 that lies above the xy-plane, oriented upward.
200

Figure 25.3. The paraboloid z = 4 − 2x2 − 2y2 , z ≥ 0 and its boundary Solution. The main idea is to replace the given surface integral (which is a two-dimensional object) by a line integral along the surface boundary (which is a one-dimensional object). Recall from the previous example that xe1 − 2ye2 + ze3 = curl F, F = yze1 + 2xze2 + 3xye3 .

Hence, when we apply Stokes’ theorem to F on Σ, we get
Σ

(xe1 − 2ye2 + ze3 ) · dS =

Σ

curl F · dS = γ F · dr,

where γ is the boundary of the given surface, positively oriented. The curve γ is the intersection curve of the paraboloid and the xy-plane, that is, the circle x2 + y2 = 2, z = 0. We can parametrize γ using √ √ r(t) = 2 cos te1 + 2 sin te2 + 0e3 (0 ≤ t ≤ 2π). Note that with this parametrization, γ is positively oriented relative to the upward orientation of Σ (see Figure 25.3). Thus,
2π

F · dr = γ 0 2π

F(r(t)) · r (t) dt 0e1 + 0e2 + 6 cos t sin te3 · −
0

=

√ √ 2 sin te1 + 2 cos te2 dt = 0.

In the above solution, the vector ﬁeld F resulted from a “happy coincidence”. Had we not just worked out Example 25.2, we would have started the above solution of Example 25.3 by “observing” that xe1 − 2ye2 + ze3 = curl yze1 + 2xze2 + 3xye3 . Such an “observation” falls just short of reliance upon “divine intervention”. In general, given a surface integral Σ F·dS, in order to argue similarly to above, we need to ﬁnd a vector ﬁeld G such that curl G = F is some given vector ﬁeld. This is a diﬃcult problem, which limits the applicability of the method. Still, the above example is not a complete waste of time: sometimes, one has to calculate the surface integral of the curl of a given ﬁeld, and in those situations one can argue as above.
201

Figure 25.4. The surface Σ from Example 25.4 Example 25.4. Evaluate
Σ

curl yze1 + 2xze2 + 3xye3 · dS,

where Σ is the piecewise smooth surface displayed on Figure 25.41, oriented upward (the boundary curve of Σ is the square formed by the lines x = ±2 and y = ±2 in the xy-plane). Solution. The main idea of this solution is to replace the given surface integral by another surface integral that is easier to evaluate. By Stokes’ theorem,
Σ

curl yze1 + 2xze2 + 3xye3 · dS = γ yze1 + 2xze2 + 3xye3 · dr,

where γ is the boundary square of Σ, oriented positively relative to the upward orientation of the surface (that is, γ is oriented counterclockwise in the xy-plane). We now observe that γ is also the positively oriented boundary of the surface Σ0 with parametrization r(x, y) = xe1 + ye2 + 0e3 yze1 + 2xze2 + 3xye3 · dr = γ (−2 ≤ x, y ≤ 2)

and with the upward orientation. Hence, another application of Stokes’ theorem gives
Σ0

curl yze1 + 2xze2 + 3xye3 · dS.

We have curl yze1 + 2xze2 + 3xye3 = xe1 − 2ye2 + ze3 , r x × ry = e1 × e2 = e3 , Hence, by (24.7),
Σ

curl yze1 + 2xze2 + 3xye3 · dS = =

Σ0

xe1 − 2ye2 + ze3 · dS xe1 − 2ye2 + 0e3 · e3 dA = 0.

−2≤x,y≤2

In fact, the surface is the graph of the function z = min purposes of the example.
1

√

4 − x2 ,

4 − y2 , but that is unimportant for the

202

Exercises
Use Stokes’ theorem to evaluate γ F · dr for the given ﬁeld F(x, y, z) and the given curve γ.

25.1. F(x, y, z) = 2ze1 + xe2 +3ye3 ; γ is the ellipse that is the intersection of the plane z = x and the cylinder x2 +y2 = 4, oriented clockwise 25.2. F(x, y, z) = (z − y)e1 + (z − x)e2 + (x + y)e3 ; γ is the triangle with vertices (0, 1, 0), (0, 2, 2) and (1, 2, 2), oriented positively as viewed from above (i.e., relative to the upward orientation of its plane) 25.3. F(x, y, z) = x2 e1 −2xye2 +yz2 e3 ; γ is the boundary of the surface z = x2 y, 0 ≤ x, y ≤ 1, oriented counterclockwise 1 25.4. F(x, y, z) = x2 ye1 + 3 x3 e2 + xye3 ; γ is the curve of intersection of the hyperbolic paraboloid z = y2 − x2 and the 2 2 cylinder x + y = 1, oriented counterclockwise as viewed from above Use Stokes’ theorem to evaluate
Σ

curl F · dS for the given ﬁeld F(x, y, z) and the given surface Σ.

25.5. F(x, y, z) = 3ye1 − xze2 − yz2 e3 ; Σ is the part of the elliptic paraboloid 2z = x2 + y2 below z = 2, oriented upward 25.6. F(x, y, z) = yze1 + xze2 + xye3 ; Σ is the part of the paraboloid z = 9 − x2 − y2 above z = 8, oriented downward 25.7. F(x, y, z) = x2 e1 − 2xye2 + yz2 e3 ; Σ is the the surface z = x2 y, 0 ≤ x, y ≤ 1, oriented upward 25.8. F(x, y, z) = ye1 + xze2 + 2xe3 ; Σ is the hemisphere x2 + y2 + z2 = 1, y ≥ 0, oriented in the direction of the positive y-axis

203

LECTURE 26

Gauss’ Divergence Theorem
We now turn to another generalization of Green’s theorem, known as Gauss’ divergence theorem. It relates the surface integral over the boundary of a simply-connected solid to a triple integral over the entire solid, similarly to how Green’s theorem relates a line integral over the boundary of a simply-connected plane region to the double integral over the entire region. Like Green’s theorem, the divergence theorem requires that we pay attention to terminology. Recall the deﬁnition of a simply-connected region from §20.3. By a simply-connected solid, we mean a simply-connected region in space that is closed (contains all its boundary points). For such a solid, its boundary consists of the points on its “outer” (visible) surface. The boundary of a simply-connected solid is the rightful three-dimensional analog of the simple, closed curve appearing in Green’s theorem. We also deﬁne the positive orientation of the boundary of a solid (assuming that it is orientable) to be the outward orientation. Theorem 26.1 (Gauss’ divergence theorem). Let S be a bounded, simply-connected solid in R , whose boundary is the piecewise smooth, positively oriented surface Σ. Let F be a vector ﬁeld whose ﬁrst-order partials are continuous on an open set U containing S . Then
3 Σ

F · dS =
S

div F dV.

(26.1)

Example 26.1. Evaluate
Σ

(x3 + y sin z)e1 + (y3 + z sin x)e2 + 3ze3 · dS,

where Σ is the positively oriented boundary of the solid S bounded by the xy-plane and by the hemispheres z = 1 − x2 − y2 , z = 4 − x2 − y2 .

Figure 26.1. The solid 1 ≤ x2 + y2 + z2 ≤ 4, z ≥ 0
205

Solution. We write F = (x3 + y sin z)e1 + (y3 + z sin x)e2 + 3ze3 . Then, by Gauss’ theorem,
Σ

F · dS =
S

div F dV =
S

3x2 + 3y2 + 3 dV.

The solid S is displayed on Figure 26.1. It is described by the inequalities 1 ≤ x2 + y2 + z2 ≤ 4, 1 ≤ ρ ≤ 2,
2 Σ π/2 0 2 1 2 0 π/2 0 π/2

z ≥ 0. 0 ≤ φ ≤ π/2.

These inequalities are translated in spherical coordinates as 0 ≤ θ ≤ 2π,
2π

Hence, converting the triple integral over S to spherical coordinates, we obtain F · dS =
1

(3ρ2 sin2 φ + 3)ρ2 sin φ dθdφdρ (ρ4 sin2 φ + ρ2 ) sin φ dφdρ ρ4 (1 − cos2 φ) + ρ2 sin φ dφ dρ
1 2 0 1

= 6π = 6π = 6π
1 2 0 2 4 ρ 3

ρ4 (1 − u2 ) + ρ2 du dρ + ρ2 dρ =
194 π. 5

= 6π
1

Example 26.2. Evaluate where Σ is the positively oriented surface of the solid S shown on Figure 26.2 (a cube of side length 2 with a corner cube of side length 1 removed). (0, 2, 2) (2, 0, 2)
Σ

x2 e1 + y2 e2 + z2 e3 · dS,

(2, 2, 0) Figure 26.2. The solid E Solution. It is possible to evaluate the surface integral using double integrals. However, were we to do that, we would have to parametrize each face separately, and there are nine faces. Furthermore, we would have to split the three double integrals corresponding to the L-shaped faces in
206

two, so we would have to evaluate twelve double integrals altogether. On the other hand, S is the diﬀerence of two cubes, so we can calculate a triple integral over S by writing it as the diﬀerence of two triple integrals over those cubes. This suggests that we use the divergence theorem. By Gauss’ theorem,
Σ

x2 e1 + y2 e2 + z2 e3 · dS =
S

(2x + 2y + 2z) dV.

Since E is the diﬀerence of the cubes [0, 2]3 and [1, 2]3 , it is easier to calculate the triple integral over E as the diﬀerence of the triple integrals over those two cubes: (2x + 2y + 2z) dV =
S [0,2]3 2 2 0 2 2 0 2

(2x + 2y + 2z) dV −
[1,2]3

(2x + 2y + 2z) dV
2 2 1 2 1 2 1 2

=
0

(2x + 2y + 2z) dzdydx −
1 2 z2 0 2

(2x + 2y + 2z) dzdydx 2xz + 2yz + z2 dydx

=
0 2 0 2

2xz + 2yz +

dydx −
2 1 2 1 1

=
0 2 0

(4x + 4y + 4) dydx − 4xy + 2y2 + 4y 0 dx −
2 2 2 1 2 1

(2x + 2y + 3) dydx 2xy + y2 + 3y 1 dx
2 0 2

=
0

=
0

(8x + 16) dx −
1

(2x + 6) dx = 4x2 + 16x

− x2 + 6x

2 1

= 39.

With a little bit of extra eﬀort, we can also use the divergence theorem to evaluate surface integrals of scalar functions. Example 26.3. Evaluate where Σ is the sphere x2 + y2 + z2 = 4.
Σ

x2 z + y2 z dS ,

Solution. Note that we have to deal with the surface integral of a scalar function and not of a vector ﬁeld. Thus, Gauss’ theorem is not directly applicable. However, we know that the sphere Σ can be parametrized using spherical coordinates by the vector function r(θ, φ) = 2 cos θ sin φe1 + 2 sin θ sin φe2 + 2 cos φe3 , where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. We also know (we went through this calculation several times already) that the (outward) unit normal vector is then n = rφ × rθ = cos θ sin φe1 + sin θ sin φe2 + cos φe3 . Note that in Cartesian coordinates n = 1 xe1 + ye2 + ze3 . We can use this observation to represent 2 the integrand in the given integral in the form F · n, with F some nice vector ﬁeld; we can then appeal to Gauss’ theorem. For example, if we choose F = 2(x2 + y2 )e3 , we obtain F·n=0
1 x 2 1 + 0 1 y + 2(x2 + y2 ) 2 z = x2 z + y2 z, 2

207

whence Here, B is the ball x2 + y2 + z2 ≤ 4. Note that the above choice of F is not unique. We could have chosen also F = 2xze1 + 2yze2 , or any other vector ﬁeld F that satisﬁes F · n = x2 z + y2 z. However, in most cases, those choices would yield triple integrals that equal 0 for less obvious reasons than the one we encountered. For example, the choice F = 2xze1 + 2yze2 yields
Σ Σ

x2 z + y2 z dS =

Σ

(F · n) dS =

Σ

F · dS =
B

div F dV = 0.

F = xze1 + yze2 + (x2 + y2 )e3 ,

x2 z + y2 z dS =
B

div F dV =
B

4z dV.

The last integral is still 0, but this requires some thought (try to convince yourself by interpreting it in terms of the average value of f (x, y, z) = 4z). We can also use the divergence theorem to calculate volumes by means of surface integrals similarly to how we earlier used Green’s theorem to calculate areas by means of line integrals. For example, if S is a simply-connected solid and Σ is its boundary, then
Σ

xe1 · dS =
S

div(xe1 ) dV =
S

1 dV = vol.(S ).

Similarly, we obtain formulas for the volume of S that use the vector ﬁelds ye2 and ze3 . Here are some of the surface integrals representing vol.(S ): 1 (xe1 + ye2 + ze3 ) · dS. (26.2) vol.(E) = xe1 · dS = ye2 · dS = ze3 · dS = 3 Σ Σ Σ Σ Example 26.4. Find the volume of the spiral pipe shown on Figure 26.3. The outer surface of the pipe is parametrized by the vector function r(u, v) = (2 + sin v) cos 2ue1 + (2 + sin v) sin 2ue2 + (u + cos v)e3
2 2 2 2

(0 ≤ u, v ≤ 2π),

and its two openings are the disks (x − 2) + z ≤ 1 and (x − 2) + (z − 2π) ≤ 1 in the xz-plane.

Figure 26.3. The surface r(u, v) = (2 + sin v) cos 2ue1 + (2 + sin v) sin 2ue2 + (u + cos v)e3
208

Solution. Let Σ be the positively oriented, piecewise smooth surface consisting of the parametric surface Σ0 that is the outer surface of the pipe and the two disks representing its two openings. We denote the disks Σ1 and Σ2 . Then Σ is the boundary of the simply-connected solid S whose volume we want to ﬁnd. By (26.2), vol.(S ) =
Σ

ze3 · dS =

Σ0

+

Σ1

+

Σ2

ze3 · dS.

The unit normal vectors to Σ1 are e2 and −e2 , and of those two, it is −e2 that points out of the solid S . Thus, by (26.2),
Σ1

ze3 · dS =

Σ1

(ze3 ) · (−e2 ) dS = 0.

Similarly,
Σ2

ze3 · dS = vol.(S ) =

Σ2

(ze3 ) · (e2 ) dS = 0.

Hence,
Σ0

ze3 · dS.

We evaluate the last surface integral using the orientation ru × rv . n= ru × rv Thus, by (24.6),
Σ0

ze3 · dS =
0≤u,v≤2π

(z(u, v)e3 ) · (ru × rv ) dA.

Note that the dot product depends only on the third component of ru × rv . We have ru (u, v) = −2(2 + sin v) sin 2ue1 + 2(2 + sin v) cos 2ue2 + e3 , rv (u, v) = cos v cos 2ue1 + cos v sin 2ue2 − sin ve3 . Thus, the cross product ru × rv is the sum of six nonzero terms: ru × rv = −2 cos v(2 + sin v) sin2 2u(e1 × e2 ) + 2 cos v(2 + sin v) cos2 2u(e2 × e1 ) + other terms = −2 cos v(2 + sin v) sin2 2u + cos2 2u e3 + other terms = −2 cos v(2 + sin v)e3 + other terms. The omitted terms are cross products involving e1 × e3 , e2 × e3 , etc., which do not contribute to the coeﬃcient of e3 in ru × rv . Thus, (z(u, v)e3 ) · (ru × rv ) = ((u + cos v)e3 ) · (ru × rv ) = −2 cos v(u + cos v)(2 + sin v). We conclude that
Σ0

ze3 · dS =
0≤u,v≤2π 2π

− 2 cos v(u + cos v)(2 + sin v) dA
2π

=
0 0

− 2u cos v(2 + sin v) − 2 cos2 v sin v − 4 cos2 v dv du.
209

We have
2π 2π 2π

(2 + sin v) cos v dv = 0,
0 0 2π Σ0

cos2 v sin v dv = 0,
0

cos2 v dv = π,

so ze3 · dS =
0

− 2u(0) − 2(0) − 4π du = −8π2 .

We obtained a negative volume, because we did not check whether the unit vector n gives the inward or the outward orientation of the surface. It turns out that it gives the inward (negative) orientation; hence, the negative volume. Consequently, the volume of the pipe is 8π2 . Exercises
Use Gauss’ theorem to evaluate
Σ

F · dS for the given ﬁeld F(x, y, z) and the given surface Σ oriented outward.

26.1. F(x, y, z) = y3 e1 + z3 e2 + x3 e3 ; Σ is the boundary of the solid S deﬁned by x2 + y2 + z2 ≤ 3, x ≥ 0, y ≥ 0 26.2. F(x, y, z) = x3 e1 + y3 e2 + z3 e3 ; Σ as in Exercise 26.1 26.3. F(x, y, z) = x2 z3 e1 + x2 ye2 + y2 ze3 ; Σ is the boundary of the solid S enclosed by the sphere x2 + y2 + z2 = 4 inside the cylinder x2 + y2 = 1 26.4. F(x, y, z) = xy2 e1 + yz2 e2 + zx2 e3 ; Σ is the boundary of the solid shown on Figure 26.4, that is, the part of the solid sphere x2 + y2 + z2 ≤ 4 outside the ﬁrst octant √ 26.5. F(x, y, z) = y3 ez e1 + y 4 − x2 e2 + ey cos xe3 ; Σ is the boundary of the solid S that lies below the surface z = 2 − x3 − y3 , −1 ≤ x, y ≤ 1, and above the xy-plane

Figure 26.4. Seven eights of the sphere x2 + y2 + z2 ≤ 4 26.6. Use Gauss’ theorem to evaluate
Σ

x2 + y2 + z3 dS , where Σ is the sphere x2 + y2 + z2 = 9.

Use the volume formulas (26.2) to calculate the volume of the given solid. 26.7. The solid enclosed by the ellipsoid x2 + 2y2 + 4z2 = 4, parametrized by √ (0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π) r(θ, φ) = 2 cos θ sin φe1 + 2 sin θ sin φe2 + cos φe3 26.8. The solid bounded by the torus r(u, v) = cos u(4 + 2 cos v)e1 + sin u(4 + 2 cos v)e2 + 2 sin ve3 (0 ≤ u, v ≤ 2π)

210

APPENDIX A

Determinants
An n × n matrix is an array of real numbers with n rows and n columns. For example, A below is a 2 × 2 matrix and B is a 3 × 3 matrix:   1 −3 1 1 2 A= , B = 0 2 −1 . 3 4 1 1 −2 The determinant of a 2 × 2 matrix a11 a12 a21 a22 is the following number (the product of its diagonal entries minus the product of its oﬀ-diagonal entries): a a det(A) = 11 12 = a11 a22 − a12 a21 . a21 a22 The determinant of a 3 × 3 matrix   a11 a12 a13 A = a21 a22 a23  a31 a32 a33 A= is given by expansion along the ﬁrst row, a process which requires the computation of three 2 × 2 determinants: a11 a12 a13 a a a a a a det(A) = a21 a22 a23 = a11 22 23 − a12 21 23 + a13 21 22 . a31 a32 a31 a33 a32 a33 a31 a32 a33 Example A.1. Compute the determinant of the matrix B above. Solution. We have det(B) = 1 2 −1 0 −1 0 2 − (−3) +1 1 −2 1 −2 1 1 = (2 · (−2) − 1 · (−1)) + 3(0 · (−2) − 1 · (−1)) + (0 · 1 − 2 · 1) = −2.

211

APPENDIX B

Deﬁnite Integrals of Single-Variable Functions
B.1. Deﬁnition of the deﬁnite integral Definition. Let f (x) be a bounded function on [a, b]. For each n = 1, 2, . . . , set ∆n = (b − a)/n and deﬁne the numbers x0 = a, x1 = a + ∆n , x2 = a + 2∆n ,
∗ x1 ≤ x2 ≤ x2 ,

...,

xn = a + n∆n = b.

∗ ∗ ∗ Also, choose sample points x1 , x2 , . . . , xn , with ∗ x0 ≤ x1 ≤ x1 ,

..., n ∗ xn−1 ≤ xn ≤ xn .

The deﬁnite integral of f from a to b is b f (x) dx = lim a n→∞

f (xi∗ )∆n , i=1 (B.1)

provided that the limit exists. The sums on the right side of (B.1) are called Riemann sums of f on [a, b]. Remarks. 1. In the above deﬁnition, the limit is supposed to exist and be independent of the choice of the sample points. This is not true for any old function, but it does hold for functions that are continuous on [a, b] or have only a ﬁnite number of jump discontinuities inside [a, b]. For such functions, we may evaluate the integral using any choice of the sample points (e.g., we can take each xi∗ to be the midpoint of the respective subinterval).

y = f (x)

R a b

Figure B.1. The deﬁnite integral as area 2. In single-variable calculus, the deﬁnite integral of a continuous function is usually motivated by the area problem: If f (x) is a continuous function on [a, b], with f (x) ≥ 0, calculate the area of the region (see Figure B.1) R = (x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f (x) .
213

The answer turns out to be the deﬁnite integral of f (x): b area(R) = a f (x) dx.

3. A diﬀerent interpretation of the deﬁnite integral arises, if we divide both sides of (B.1) by b − a. Since ∆n = (b − a)/n, we have 1 b−a n f (xi∗ )∆n = i=1 ∆n b−a

n

f (xi∗ ) = i=1 1 n

n

f (xi∗ ). i=1 Hence, dividing both sides of (B.1) by b − a, we derive 1 b−a b a

1 f (x) dx = lim n→∞ n

n

f (xi∗ ). i=1 (B.2)

∗ ∗ ∗ Note that the expression 1 i f (xi∗ ) is the average of the sample values f (x1 ), f (x2 ), . . . , f (xn ), n which are about equally spaced. Thus, the limit on the right side of (B.2) represents the average of “inﬁnitely many, equally spaced” sample values of f (x). Consequently, we may argue that the left side of (B.2) is the average value of f (x):

average f (x) = x∈[a,b] 1 b−a

b

f (x) dx. a (B.3)

This alternative interpretation of the deﬁnite integral is rarely emphasized in single-variable calculus, but is much more appropriate for the purpose of these notes, because we want to draw parallels between the properties of deﬁnite integrals and the properties of other types of integrals (double, triple, line, surface) whose geometric meaning may be less obvious. B.2. Properties of the deﬁnite integral The following is a list of some general properties of deﬁnite integrals: b b b

i) a b

[ f (x) + g(x)] dx = a b

f (x) dx + a g(x) dx

ii) a c

c f (x) dx = c a b

f (x) dx b iii) a f (x) dx + c f (x) dx = a b

f (x) dx b iv) If f (x) ≤ g(x), then a f (x) dx ≤ a b

g(x) dx f (x) dx ≤ M(b − a). a v) If m ≤ f (x) ≤ M, then m(b − a) ≤ a b a

vi) b f (x) dx = − a f (x) dx, a f (x) dx = 0.

Of course, the most important property of deﬁnite integrals is the Fundamental Theorem of Calculus, also known as the Newton–Leibnitz theorem.
214

Theorem B.1 (Fundamental theorem of calculus). Suppose that f (x) is a continuous function on [a, b]. i) We have x d f (t) dt = f (x). dx a x that is, a f (t) dt is an antiderivative of f (x). ii) If F(x) is any antiderivative of f (x), then b f (x) dx = F(b) − F(a). a 215

APPENDIX C

Polar Coordinates
C.1. Deﬁnition Consider a point P in the plane. Recall that its Cartesian coordinates (x, y) are the signed distances from P to two perpendicular axes (the x- and y-axes). In this appendix, we recall another set of coordinates: the polar coordinates of P. The polar coordinate system consists of a point O called the pole (or the origin) and a ray starting at O called the polar axis. The polar coordinates of a point P O are a pair of real numbers (r, θ), where r is the distance from O to P and θ ∈ [0, 2π) is the oriented angle between the polar axis and the line OP (see Figure C.1); the polar coordinates of the pole O are (0, θ), where θ is any number in [0, 2π). y P(x, y) P(r, θ) r θ θ O Figure C.1. Polar coordinates O Q(x, 0) x
2

x2 r=

+

y

Figure C.2. Polar and Cartesian coordinates

C.2. Relations between polar and Cartesian coordinates We want to be able to convert polar coordinates to Cartesian and vice versa. In order to do so, we need to set a Cartesian and a polar coordinate systems in the same copy of the plane. We choose the two coordinate systems so that they share the same origin and the polar axis coincides with the positive x-axis (see Figure C.2). Since the distance between the origin and P(x, y) is the same no matter what coordinate system we use, we have r = |OP| = x2 + y2 .

The relation between θ and (x, y) is more technical, but not much more complicated. First, consider a point P(x, y) in the ﬁrst quadrant. From OPQ on Figure C.2, x = r cos θ, y = r sin θ,
217

y tan θ = . x

The ﬁrst two of these equations hold for all points P(x, y), even if they do not lie in the ﬁrst quadrant, that is, the Cartesian coordinates of a point P(r, θ) are obtained from the formulas x = r cos θ, y = r sin θ. (C.1) On the other hand, the relation tan θ = y/x extends only to the points with x 0 and determines θ uniquely only on the half-planes x > 0 or x < 0. Thus, when we write the formulas for converting Cartesian into polar coordinates y (C.2) r = x2 + y2 , tan θ = , x we must remember that the latter formula must be applied cautiously.
2 Example C.1. The point P whose polar coordinates are (2, 3 π) has Cartesian coordinates √ 2 x = 2 cos 2 π = −1, y = 2 sin 3 π = 3. 3

Example C.2. The Cartesian point P(2, 2) has polar coordinates √ √ π 2 r = 22 + 22 = 2 2, θ = arctan = . 2 4 Here, we used that since P lies in the ﬁrst quadrant, the solution of the second equation in (C.2) is given by θ = arctan(y/x). The Cartesian point P(2, −2) has polar coordinates (r, θ), where √ r = 22 + (−2)2 = 2 2 and θ is such that √ that is, P(2 2, 7 π). 4 tan θ = −1, θ∈
3 π, 2π 2

,

Example C.3. The Cartesian point P(−3, 4) has polar coordinates(r, θ), where r= and θ is such that Thus, P(5, arctan(− 4 ) + π). 3 (−3)2 + 42 = 5 θ∈
1 π, π 2

4 tan θ = − , 3

.

C.3. Polar curves A polar curve is the set of points in the plane whose (polar) coordinates (r, θ) satisfy an equation r = f (θ), where f is some function. Note that using equations (C.1), we can write every polar curve in parametric form: x = r cos θ = f (θ) cos θ, y = r sin θ = f (θ) sin θ. Example C.4. The curve r = 3 is the set of points at distance 3 from the origin, that is, a circle of radius 3 centered at O (whose Cartesian equation is x2 + y2 = 9, of course). Notice that we reach the same conclusion if we use (C.2) to convert the given polar equation into a Cartesian equation: r=3 ⇐⇒ x2 + y2 = 3
218

⇐⇒

x2 + y2 = 9.

O

Figure C.3. The cardioid r = 1 + cos θ Example C.5. The curve θ = 3 π is the ray starting at the origin and intersecting the polar axis 4 at an angle of 135◦ . Example C.6. The curve r = −2 cos θ is a circle. To see this, we pass to Cartesian coordinates: r = −2 cos θ x2 + y2 = −2x ⇐⇒ r2 = −2r cos θ ⇐⇒ x2 + y2 = −2x. (x + 1)2 + y2 = 1.

The last equation is the Cartesian equation of a circle of radius 1 centered at (−1, 0): ⇐⇒ x2 + 2x + 1 + y2 = 1 ⇐⇒

Example C.7. The curve r = 1 + cos θ is known as a cardioid, because its graph is shaped like a heart (see Figure C.3). Unlike the curves in the previous examples, the cardioid does not have a simple Cartesian equation which we can use to sketch its graph.

C.4. Area and arc length in polar coordinates Theorem C.1. Let 0 ≤ a < b ≤ 2π and let f be a positive continuous function deﬁned on [a, b]. Then the area A of the region in the plane bounded by the curves θ = a, is given by A= 1 2 b θ = b,

r = f (θ),

[ f (θ)]2 dθ. a Example C.8. Find the area enclosed by the cardioid (see Example C.7). Solution. By Theorem C.1, the area is given by the integral 1 2
2π 0

(1 + cos θ)2 dθ =

1 2π 1 + 2 cos θ + cos2 θ dθ 2 0 1 2π 1 = 1 + 2 cos θ + 2 (1 + cos 2θ) dθ 2 0 1 3π 2π 1 = θ + 2 sin θ + 2 θ + 1 sin 2θ 0 = . 4 2 2
219

Theorem C.2. Let 0 ≤ a < b ≤ 2π and let f be a positive continuous function deﬁned on [a, b]. Then the length L of the curve r = f (θ) is given by b L= a f (θ)2 + f (θ)2 dθ.

Example C.9. Find the length L of the cardioid. Solution. By Theorem C.2,
2π 2π

L=
0 2π

(1 + cos θ)2 + (− sin θ)2 dθ =
0 2π

√

2 + 2 cos θ dθ π =
0 π/2

4 cos2 (θ/2) dθ =
0 π

2| cos(θ/2)| dθ = 4
0

| cos u| du

=4
0

cos u du − 4 π/2 cos u du = 8.

220

APPENDIX D

Introduction to Mathematica
D.1. Mathematica notebooks A standard Mathematica ﬁle is called a notebook and has extension .nb. To create a new notebook, select File→New→Notebook from the main menu or press Ctrl+N. That will result in a new window ﬁlled with white space and named Untitled-1, Untitled-2, or similar. Notebooks consist of one or more cells. Each cell represents a logical unit of the notebook and is shown by a blue square bracket on the right side of the screen. There are several diﬀerent types of cells, but the three main types are: input cells, output cells, and text cells. In addition to these, there are types of cells that provide special formatting: title cells, subtitle cells, section cells, etc. D.1.1. Selecting and editing cells. To select an existing cell, click on its cell bracket. A selected cell is indicated by a thick black bracket. To select more than one consecutive cell, click on one cell bracket and then drag over the others or click on the desired cells while pressing the Ctrl key down. To select cells that are not next to each other, click on the ﬁrst one then, while holding the Ctrl key down, point and click on the next one and so on. This is particularly useful in printing a selection of cells. One can also copy and paste selected cells. To edit the content of a cell, simply click on the desired location and start editing. One can also change an entire cell by selecting it and choosing a diﬀerent style type from the pull-down menu in the toolbar. D.1.2. Creating a new cell. If you position the mouse between two existing cells or after the last cell and then click, a horizontal line (called the cell insertion point or line) will be shown across the screen. This indicates where the next new cell will be inserted. Now, if you begin to type, Mathematica will create a new cell containing the text that you type. If you type several lines (using the Enter key to begin new lines), the cell bracket will grow to enclose everything you type. Note that, by default, every new cell you type in is created as an input cell. That means that Mathematica treats the symbols inside that cell as Mathematica code and may try to execute it. That is not always desirable, so you must always convert the style of cells you consider text from Input to Text. For example, a cell containing your name should have type Text, Subtitle or Subsubtitle, not Input. D.1.3. Executing cells. To execute a cell (to get an answer to an arithmetic operation, for example), placing the cursor anywhere within the cell and then press Shift+Enter simultaneously or press the Enter key on the numeric keypad. As soon as a cell is executed, it will be labeled with In[1] for the ﬁrst cell. The result will be shown in a new cell labeled Out[1]. The next cell executed will be labeled In[2] regardless of its actual location relative to the ﬁrst cell. Mathematica pays attention only to the order in which the cell is executed rather than to its relative position with respect to other cells. Note that Mathematica will execute only input cells.
221

Sometimes, one wants to execute more than one cell. That can be done by selecting the desired cells and then pressing Shift+Enter. One can also execute the entire notebook by selecting Evaluation→Evaluate Notebook from the main menu. This will make Mathematica run through the entire notebook and execute all input cells in the order of appearance. Note that if there is text contained in an input cell, Mathematica will try to execute that text. The results of the latter can be comical—for example, executing an input cell containing the text “Today is 10/10/2010.” results in the following: In[1]:= Today is 10/10/2010. Out[1]:= 0.000497512 is Today D.2. Basic syntax D.2.1. Arithmetic operations and numerical values. Mathematica supports the basic arithmetic operations. The following symbols are used: + indicates addition - indicates subtraction * indicates multiplication / indicates division ˆ indicates exponentiation Here are some examples of those rules in action: In[1]:= 2+2*3 Out[1]:= 8 In[2]:= 2+2 3 Out[2]:= 8 In[3]:= 5ˆ3 Out[3]:= 125 In[4]:= 2/3-1/6 Out[4]:= 1/2 Note the second example in this list: the space between the numbers 2 and 3 is interpreted as a multiplication. A space between two expressions is another way to indicate multiplication in Mathematica. Sometimes, even the space is unnecessary: e.g., “2*x”, “2 x” and “2x” are all interpreted by Mathematica as “2x”. Notice that Mathematica returns the sum of two rational numbers as a rational number. That is because by default, Mathematica uses exact arithmetic. This means that a rational number, such as 1 , will normally be displayed as a fraction and not as the decimal 0.5. In fact, the latter 2 is considered an approximation having a limited precision. If we want the numerical value of a fraction, we can use the built-in command N[]. In general, the command N[..., n] returns the numerical value of the expression in place of the ..., with n correct decimal digits. When the value of n is missing, the default precision is used. For example, to get the numeric value of 7 , we can try one of the following: 94
222

In[1]:= N[7/94] Out[1]:= 0.0744681 In[2]:= N[7/94, 30] Out[2]:= 0.0744680851063829787234042553191 An alternative way to get the numeric value of an expression is to add a left apostrophe ‘ after it. For example, In[1]:= 7/94‘ Out[1]:= 0.0744681 It is important to remember that Mathematica performs arithmetic computations in the usual mathematical order: exponentiations are performed ﬁrst; then multiplications and divisions, from left to right; ﬁnally, additions and subtractions, also from left to right. When this natural order is to be changed, we use parentheses. For example: 1 4 2ˆ3 - 1/3*4 + 13 represents 23 − 3 ∗ 4 + 13 = 8 − 3 + 13 = 19 2 ; 3 23 − 1 7 = . (2ˆ3 - 1)/(3*4 + 13) represents 34 + 13 94 D.2.2. Elementary built-in functions and constants. Some of Mathematica’s built-in constants are: Pi π = 3.14159265358979 . . . E e = 2.71828182845904 . . . √ I i = −1 Note that Mathematica is case-sensitive, so pi is not the number π! All built-in constants and functions in Mathematica begin with an upper-case letter. Furthermore, function arguments in Mathematica are enclosed in square brackets [], as opposed to the parentheses used in usual mathematical notation. Hence, the input Sqrt[16] returns the value 4, whereas sqrt[16], Sqrt(16) and sqrt(16) do not work. Table D.1 below lists some of the elementary built-in functions that you will use often. Mathematica function Sin[x],Cos[x],Tan[x],Sec[x] ArcSin[x],ArcCos[x],ArcTan[x] Sqrt[x] Abs[x] Log[x] Log10[x] Log[b,x] N[x,n] Exp[x], Eˆx n! Usual notation/Description sin(x), cos(x), tan(x), sec(x) arcsin(x), arccos(x), arctan(x) √ x |x| ln(x) log(x), log10 (x) logb (x) Approximate value of x with n digits e x , exponential function base e Factorial of n: n! = 1 · 2 · 3 · · · n

Table D.1. Mathematica syntax of some basic functions
223

Mathematica gives exact results whenever possible. So, typing Pi as input will return π. If it is a numerical approximation to π that we want, we should ask for it by using the N[]-command, as shown in the following examples: In[1]:= N[Pi] Out[1]:= 3.14159 In[1]:= N[Pi,50] Out[2]:= 3.1415926535897932384626433832795028841971693993751 Here are some more examples of how Mathematica tries to use exact calculations: In[1]:= Sin[Pi/2] Out[1]:= 1 In[2]:= Log10[1000] Out[2]:= 3 In[3]:= Log10[200] Out[3]:= Log[200]/Log[10] In[4]:= N[Log10[1000]] Out[4]:= 2.30103 In[5]:= 5! Out[5]:= 120 As we mentioned earlier, Mathematica considers decimals, such as 0.5, approximate numbers. Therefore, most numerical calculations using such decimals are limited in precision. For example, compare the outputs of the following two calculations: N[Eˆ(1/2),50] 1.6487212707001281468486507878141635716537761007101 N[Eˆ0.5,50] 1.64872 √ In the ﬁrst calculation, we ask Mathematica to evaluate e1/2 = e, which is an exact number having an inﬁnite precision. The second calculation, on the other hand, is limited in precision by the presence of the exponent 0.5, which Mathematica treats diﬀerently from 1 . For this reason, it 2 is preferred to represent decimals by fractions in numerical calculations. D.2.3. Deﬁning functions of one variable. To deﬁne a new function f (x) in Mathematica, we typically use syntax of the form f[x_] := **** where the stars should be replaced by the appropriate formula. The symbols := (the delayed assignment operator) separate the deﬁnition in two. The left-hand side of the deﬁnition (f[x_] above) speciﬁes the name and the variable of the function; the right-hand side describes the function, usually by a formula involving the variable. For example, the command f[x_] := xˆ2
224

deﬁnes the function f (x) = x2 . Note that we enclose the argument between square brackets and that we use an underscore character after the argument’s name on the left-hand side of the deﬁnition. Mathematica becomes aware of the deﬁnition of a user-deﬁned function not when one has typed in the deﬁnition of the function, but only after the cell containing that deﬁnition has been executed. A common mistake is to type the deﬁnition of a function one wants to use and then to move on to using that function without executing the deﬁnition ﬁrst. If a function has not been previously deﬁned, Mathematica is likely to simply repeat any input that uses the undeﬁned function. For example, In[1]:= someNewFunction[5] Out[1]:= someNewFunction[5] Note that when Mathematica does not recognize the name of a function or a variable, it displays that name in blue: e.g., someNewFunction[]. Once the deﬁnition of the function has been loaded (so that Mathematica knows what that name means), the highlighting changes from blue to black and calls to the function result in its evaluation. For example, In[1]:= someNewFunction[x_] := Sin[x] In[2]:= someNewFunction[5] Out[2]:= Sin[5] In[3]:= N[someNewFunction[5],10] Out[3]:= -0.9589242747 Remark. There are two conventions that you should follow when deﬁning your own functions: • To avoid conﬂicts with Mathematica’s built-in commands, always start the name of your functions with a small letter. Since all built-in commands start with a capital letter, this convention avoids the accidental overwriting of a built-in command. • To avoid conﬂicts with the deﬁnitions of namesake functions, clear the name of the function before you deﬁne it. Mathematica remembers old deﬁnitions even if we delete the cells containing them (or even close the notebook in which they were introduced), and this can occasionally cause problems. To force Mathematica to forget such old deﬁnitions and start afresh, we use of the built-in function Clear. It is best to precede each function’s deﬁnition with a call to Clear to erase potential prior deﬁnitions. For example, to deﬁne the function g(x) = x3 + 2x, we should use Clear[g] g[x_] := xˆ3+2x D.2.4. Lists. Many of Mathematica’s more sophisticated commands have inputs and outputs in the form of lists. A list is a collection of expressions (numbers, variables, functions, other lists, etc.) separated by commas and enclosed in a pair of curly braces. For example, {x,0,2} is a list of three elements: the variable x and the numbers 0 and 2. Here is another example: L = {1,3,5,7,9,11} This deﬁnes a list L whose elements are the odd integers 1, 3, . . . , 11. When a list has been given a name, we can use that name followed by a number in double brackets ([[ ]]) to access a speciﬁc element of the list. For example, the fourth element of the list L above can be obtained by L[[4]].
225

Thus, we have now encountered four diﬀerent types of brackets in Mathematica, each with its own dedicated use. These types are: () Round parentheses indicate the order of operations. Do not use braces or brackets instead! [] Square brackets enclose the argument of a function. Do not use round parentheses instead! {} Curly brackets (braces) enclose a list. [[]] Double square brackets refer to a particular element of a list. D.3. Plotting commands D.3.1. Plotting functions of one variable. D.3.1.1. Basic plots. To graph a function (or an expression) of one variable, we use the Plot command. The command takes at least two arguments, a function (or an expression), and a range. The range is a list of three elements: the variable in the expression, for example x, a minimum value for x, and a maximum value for x. For example, the command Plot[Sin[x], {x,0,2Pi}] will plot the sine function over the range 0 ≤ x ≤ 2π. One can also graph a previously deﬁned function by using its name as the ﬁrst argument of the Plot command as shown in the following example. Clear[f] f[x_] := Cos[Eˆx] Plot[f[x], {x,0,3}]
1.0

0.5

0.5

1.0

1.5

2.0

2.5

3.0

0.5

1.0

D.3.1.2. Graphics options. An option is a rule of the form OptionName->OptionValue, where OptionName is the name of the option and OptionValue is the value that it is assigned. If an option is not supplied explicitly, the default value is used. The following command will display all the possible plotting options in Mathematica together with their default settings. Options[Plot] {AlignmentPoint->Center, AspectRatio->1/GoldenRatio, Axes->True, AxesLabel->None, AxesOrigin->Automatic, ... Ticks->Automatic, TicksStyle->{}, WorkingPrecision->MachinePrecision} However, a much better way to learn about the various options of the Plot command is to read the documentation entry on it in Mathematica’s Documentation Center (under the Help menu). Here,
226

we present only a few of the more commonly used options. GridLines. We can use this option to simulate graphing paper: Plot[Sin[x], {x,-4,4}, GridLines->Automatic]
1.0

0.5

4

2

2

4

0.5

1.0

The two basic values for this option are None (the default) and Automatic, which makes Mathematica generate the most reasonable rectangular grid for the given graph. There are also more sophisticated alternatives that can be used to control the size and formatting of the grid—see the documentation entry on GridLines. PlotStyle. We can use this option to change the appearance of the plot. The value of PlotStyle can be a single directive (e.g., Blue or Thick) or a list of such directives. Often used styles are: Thin or Thick, Dashed or Dotted, and various colors. In particular, there are several ways to specify the color of the plot. The most common is by the function RGBColor, which takes three arguments: the amount of red, the amount of green, and the amount of blue. These arguments must be numbers between 0 and 1, and they need not add up to 1. There are also several predeﬁned colors that can be assigned to PlotStyle; those include: Red, Green, Blue, Yellow, Cyan, Magenta, LightOrange, Lighter[Purple], Darker[Brown], and similar modiﬁcations. Here are a couple of examples using diﬀerent styling options: Plot[Sin[x], {x,-4,4}, PlotStyle->{RGBColor[0.8,0.2,0.2], Dashed}] Plot[Sin[x], {x,-4,4}, PlotStyle->{Thick, Darker[Green], Dotted}]
1.0 1.0

0.5

0.5

4

2

2

4

4

2

2

4

0.5

0.5

1.0

1.0

227

PlotRange. We can use this option to specify the range of y-coordinates to include in the plot. The most often used variants of this option are PlotRange->All PlotRange->{a,b} When PlotRange is set to All, Mathematica plots all the points on the graph. When PlotRange is set to an actual range {a,b}, Mathematica plots only those points on the graph that have ycoordinates in the range a ≤ y ≤ b. When this option is not set, Mathematica displays only points where it thinks the function is “interesting”. Sometimes, this does not work very well and Mathematica needs to be told to plot all points. For example, compare the outputs of the following two attempts to plot (sin x)/x: Plot[Sin[x]/x, {x,-20,20}] Plot[Sin[x]/x, {x,-20,20}, PlotRange->All]
0.4 1.0 0.3 0.8

0.2

0.6

0.1

0.4

0.2 20 10 0.1 10 20

20

10 0.2

10

20

0.2

D.3.1.3. Superimposing several graphs. One can superimpose the graphs of several functions by supplying the functions as a list in the ﬁrst argument of a Plot command. The following command will plot sin x and x − x3 /6 + x5 /120. Plot[{Sin[x], x-xˆ3/6+xˆ5/120}, {x,-5,5}] One can use the PlotStyle option to get more easily distinguishable curves as seen in the following example. Notice that the argument of PlotStyle is a list of two lists: the ﬁrst list is the list of plotting options applied to the graph of sin x and the second list is the list of plotting options applied to the graph of x − x3 /6 + x5 /120. Plot[{Sin[x], x-xˆ3/6+xˆ5/120}, {x,-5,5}, PlotStyle->{{Thick, Red}, {Blue}}]
3 2 1

4

2 1 2 3

2

4

228

D.4. Dynamic output D.4.1. Manipulate. A major strength of Mathematica is its ability to handle dynamically updated variables. The basic command to access Mathematica’s dynamic features is Manipulate. It takes two or more inputs: the ﬁrst input is the expression that we want to update dynamically and the subsequent inputs are lists indicating ranges of parameters that appear in the expression. Here are two examples of dynamic plots: Manipulate[ Plot[Cos[x], {x,-a,a}, PlotRange->{-1,1}], {a,5,0.001}] Manipulate[ Plot[Sin[x+c]+d, {x,-7,7}, PlotRange->{-3.5,3.5}], {c,0,3}, {d,-2,2}] c a
1.0 3

d

0.5

2

1

4

2

2

4

6

4

2 1

2

4

6

0.5 2

3 1.0

Note the little + buttons next to the sliders in the above dynamic outputs. Clicking on one of those opens a navigation menu for that particular slider. The navigation menu allows us to see animated versions of the output, to ﬁnd the current value of the parameter, or to set the parameter equal to a particular value. For example, we can see that setting the two parameters to c = 1.57 and d = −1 in the second dynamic plot results in a sine curve that looks exactly as the graph y = cos x − 1. c 1.57

d
1

3

2

1

6

4

2 1

2

4

6

2

3

229

Although dynamic evaluations are a wonderful tool, they also require some care to avoid falling in certain traps. Consider the following dynamic plot. Clear[f] f[x_] := Exp[-xˆ2] Manipulate[Plot[f[x], {x,-a,a}, PlotRange->{0,1}], {a,2,4}] a 1.0

0.8

0.6

0.4

0.2

2

1

0

1

2

We see a nice bell-shaped curve. Now, let us take a look at another dynamic plot. Clear[f] f[x_] := x*Exp[-x] Manipulate[Plot[f[x], {x,-a,a}, PlotRange->{-1,1}], {a,2,4}] You may expect that after executing these commands, the pair of plots will look something like this: a 1.0 1.0

a

0.8 0.5 0.6

0.4

2

1

1

2

0.2

0.5

2

1

0

1

2

1.0

230

However, the reality will look more like this: a 1.0

a
1.0

0.5

0.5

2

1

1

2

2

1

1

2

0.5

0.5

1.0

1.0

This is not a bug! Since both plots are supposed to present the graph of the function f (x) and are to be updated dynamically, changing the deﬁnition of f (x) will result in both plots changing to match the new deﬁnition of f (x). For example, if we execute the commands Clear[f] f[x_] := xˆ2 both plots above will change to dynamic plots of the parabola y = x2 . The safest way to avoid such confusion is not to reuse function names when using Manipulate. For example, the dynamic plots 2 of the functions e−x and xe−x above should have been entered as Clear[f1, f2] f1[x_] := Exp[-xˆ2] Manipulate[Plot[f1[x], {x,-a,a}, PlotRange->{0,1}], {a,2,4}] f2[x_] := x*Exp[-x] Manipulate[Plot[f2[x], {x,-a,a}, PlotRange->{-1,1}], {a,2,4}] D.5. Calculus-related Mathematica commands Mathematica has built-in commands for many of the operations that you will learn in calculus. In this section, we list some of those commands. D.5.1. Limits. To evaluate a limit in Mathematica, we use the built-in command Limit is easy. For example, to evaluate e x − 1 − x − x2 /2 lim , x→0 x3 we just execute the following the command. In[1]:= Limit[(Eˆx-1-x-xˆ2/2)/xˆ3, x->0] Out[1]:= 1/6 (Recall that Mathematica uses E for e = 2.71828 . . . ) One-sided limits can be evaluated using the option Direction. The value assigned to this option can be any nonzero number, but only the sign of the number really matters: if a > 0, then
231

Direction->a speciﬁes a left limit; if a < 0, then Direction->a speciﬁes a right limit. Here are several examples: In[1]:= Limit[Abs[x]/x, x->0, Direction->1] Out[1]:= -1 In[2]:= Limit[Abs[x]/x, x->0, Direction->-1] Out[2]:= 1 In[3]:= Limit[Exp[1/(x-1)], x->1, Direction->1] Out[3]:= 0 In[4]:= Limit[Exp[1/(x-1)], x->1, Direction->-1] Out[4]:= Infinity In particular, the last input/output pair demonstrates what happens when a limit is inﬁnite. It should be noted that if no direction is speciﬁed and the two-sided limit does not exist, then Mathematica automatically reverts to evaluation of the right limit (Direction->-1). This explains, for example, the following evaluation. In[1]:= Limit[Abs[x]/x, x->0] Out[1]:= 1 Mathematica can also handle limits at ±∞. To evaluate a limit as x → ∞, we simply replace the number in the second argument of Limit by the word Infinitiy (or by -Infinity, for limits as x → −∞). For example, In[1]:= Limit[Log[x]/x, x->Infinity] Out[1]:= 0 In[2]:= Limit[\Sqrt[xˆ2+4x]/(3x - 7), x->-Infinity] Out[2]:= -1/3 Finally, a few words about limits which do not exist. We already know what happens when a limit is inﬁnite. What if the limit fails to exist for a diﬀerent reason? For example, it is known from calculus that the limit lim sin(1/x) does not exist. Mathematica reports that fact as follows: x→0 In[1]:= Limit[Sin[1/x], x->0] Out[1]:= Interval[{-1,1}] So, in this case Mathematica returns a whole range as the value of the limit instead of a single number. That means that the limit does not exist. D.5.2. Diﬀerentiation. Mathematica has a built-in command to calculate the derivative of a function. To ﬁnd f (x), given f (x), we simply execute one of the following two commands In[1]:= f’[x] Out[1]:= D[f[x],x] For example, if f (x) = x3 + cos(x2 ), then In[1]:= f’[x] Out[1]:= 2x - 2x Sin[xˆ2]
232

In[2]:= D[f[x],x] Out[2]:= 2x - 2x Sin[xˆ2] Note that the latter version of the diﬀerentiation command works also for partial derivatives of multivariable functions. For example, if f (x, y) = sin(x + 2y) + (x2 − y2 ) exp(x2 + 3y2 ), then D[f[x,y],x] and D[f[x,y],y] calculate, respectively, the x- and y-partials of f (x, y). Clear[f] In[1]:= f[x_,y_] := Sin[x+2y] + (xˆ2-yˆ2)Exp[xˆ2+3yˆ2] D[f[x,y], x] Out[2]:= (2xˆ3-2xyˆ2+2x)Eˆ(xˆ2+3yˆ2) + Cos[x+2y] In[3]:= D[Sin[x+2y] + (xˆ2-yˆ2)Exp[xˆ2+3yˆ2], y] Out[3]:= (6yxˆ2-6yˆ3-2y)Eˆ(xˆ2+3yˆ2) + 2Cos[x+2y] D.5.3. Integration. Mathematica’s basic command for evaluation of integrals is Integrate. It can evaluate both deﬁnite and indeﬁnite integrals. In its two basic forms, this command looks like this Integrate[f[x], x] evaluates f (x) dx b Integrate[f[x], {x,a,b}] evaluates a f (x) dx The second version of Intergate tries to evaluate the given integral exactly in “closed form,” but that is not always possible. In such cases, it may be preferable to settle for a very close numeric approximation to the value of the deﬁnite integral. Mathematica provides another command for this purpose. The command NIntegrate[f[x], {x,a,b}] b uses sophisticated algorithms to approximate the deﬁnite integral a f (x) dx and the result is usually fairly close to the exact answer. By default, the answer that you obtain by executing this command is accurate to at least 6 digits; for higher precision, the options PrecisionGoal and WorkingPrecision can be used (see the documentation entry on NIntegrate). Here are some examples of integral evaluations: In[1]:= Integrate[ArcTan[2x+3], x] Out[1]:= (3/2)ArcTan[3+2x] + x*ArcTan[3+2x] - (1/4)Log[5+6x+2xˆ2] In[2]:= Integrate[Exp[-xˆ2], {x,0,2}] Out[2]:= (1/2) Sqrt[Pi] Erf[2] In[3]:= NIntegrate[Exp[-xˆ2], {x,0,2}] Out[3]:= 0.882081 In[4]:= NIntegrate[Exp[-xˆ2], {x,0,2}, PrecisionGoal->10] Out[4]:= 0.882081 In[5]:= NIntegrate[Exp[-xˆ2], {x,0,2}, WorkingPrecision->10] Out[5]:= 0.8820813907
233

In[6]:= NIntegrate[Exp[-xˆ2], {x,0,2}, WorkingPrecision->10, PrecisionGoal->10] Out[6]:= 0.8820813908 The last three examples in this series are meant as a warning that use of the high-precision options of NIntegrate requires some care. The safest solution is to set both PrecisionGoal and WorkingPrecision to the same number (as in the last example). What exactly goes wrong in the two earlier examples is beyond the reach of this introduction to Mathematica; suﬃce it to say that if you need to use these options, you should ﬁrst study carefully the documentation. D.6. User-deﬁned commands Mathematica’s includes some programming features which allow the sophisticated user to deﬁne his/her own commands. User-deﬁned commands share features with both built-in commands and user-deﬁned functions. Similar to built-in commands, user-deﬁned commands may have multiple inputs (of varying types) and may produce all kinds of output, including numeric answers, text, plots, etc. On the other hand, similar to user-deﬁned functions, we must type and execute the deﬁnition of a user-deﬁned command before we can use it. In the laboratories, we will make extensive use of such custom commands, so it is important to learn how to apply them properly, even if the actual “nuts and bolts” of the command seem mysterious. The deﬁnition of the command newCommand always takes one of the following two forms: newCommand[*list of arguments*] := Module[ *some Mathematica code* ] newCommand[*list of arguments*] := DynamicModule[ *some Mathematica code* ] Here, *list of arguments* and *some Mathematica code* are placeholders for the actual list of arguments and the deﬁnition of the command. In order to be able to use such a command, we must do two things: • We must familiarize ourselves with the arguments the command accepts and the output that it returns. These are usually described in the manual, shortly before or after the deﬁnition of the command. • We must execute the cell containing the deﬁnition of the command. Until then all occurrences of the name of the command will appear in blue (e.g., newCommand) and Mathematica will simply repeat any command containing a call to the new command. The next three pages demonstrate these points. They present printouts of the deﬁnition of a module that appears in an actual TU calculus laboratory assignment and several calls to it. On the ﬁrst printout, the deﬁnition of newton has not been executed and the “output” merely repeats the input. On the second printout, the deﬁnition has been executed but the arguments do not match the expected ones—in one case two inputs are missing, and in the other n is not an integer. On the third printout, all the rules have been followed and the call to newton has resulted in an actual plot.

234

12.3. Geometric Interpretation
Newton's method for solving f x 0 is equivalent to the following geometric process. Pick a starting number x0 near the root r. Draw the tangent line to the graph of f at the point x0 , f x0 . The x-intercept, x1 , of the tangent line is generally closer to r than x0 was. The process is then repeated until some prescribed accuracy is attained. The module below, called newton, repeats the above construction n times, where n is a given positive integer. It stores the values of x0 , x1 , ..., xn in a list named data and produces a plot of the iteration. The calling parameters of the module are: f = the function f ; a = the left end of some interval that contains r; b = the right end of the interval; x0 = the initial value to start the iteration; n = the number of iterations desired. newton f_, a_, b_, x0_, n_ : Module data g , g x_ x f x f' x ; ;

NestList g, x0, n ; data2 Flatten Table data i , 0, data i , f data i , i, Length data data3 Partition data2, 2 ; data4 Drop data3, 1 ; p1 Plot f x , x, a, b , PlotStyle Red, Thick , AxesLabel "x", "y" , PlotRange All ; p2 Graphics Hue .6 , Line data4 ; Show p1, p2

We now apply this module to the function f x x2 2 with x0 obtain a geometric illustration of the first iteration (n 1). Clear f ; f x_ : x^2
In[1]:= Out[1]=

0.5 and the interval 0, 2.5 in order to

2

newton f, 0, 2.5, .5, 1 newton f, 0, 2.5, 0.5, 1

12.3. Geometric Interpretation
Newton's method for solving f x 0 is equivalent to the following geometric process. Pick a starting number x0 near the root r. Draw the tangent line to the graph of f at the point x0 , f x0 . The x-intercept, x1 , of the tangent line is generally closer to r than x0 was. The process is then repeated until some prescribed accuracy is attained. The module below, called newton, repeats the above construction n times, where n is a given positive integer. It stores the values of x0 , x1 , ..., xn in a list named data and produces a plot of the iteration. The calling parameters of the module are: f = the function f ; a = the left end of some interval that contains r; b = the right end of the interval; x0 = the initial value to start the iteration; n = the number of iterations desired.
In[1]:=

newton f_, a_, b_, x0_, n_ : Module data

g , g x_

x

f x

f' x ; ;

NestList g, x0, n ; data2 Flatten Table data i , 0, data i , f data i , i, Length data data3 Partition data2, 2 ; data4 Drop data3, 1 ; p1 Plot f x , x, a, b , PlotStyle Red, Thick , AxesLabel "x", "y" , PlotRange All ; p2 Graphics Hue .6 , Line data4 ; Show p1, p2

We now apply this module to the function f x x2 2 with x0 obtain a geometric illustration of the first iteration (n 1).
In[2]:=

0.5 and the interval 0, 2.5 in order to

Clear f ; f x_ : x^2

2

In[4]:= Out[4]=

newton f, .5, 2 newton f, 0.5, 2 newton f, 0, 2.25, .5, 1.5
NestList::intnm : Non negative machine sized integer expected at position 3 in NestList g$116, 0.5, 1.5 .

In[5]:=

12.3. Geometric Interpretation
Newton's method for solving f x 0 is equivalent to the following geometric process. Pick a starting number x0 near the root r. Draw the tangent line to the graph of f at the point x0 , f x0 . The x-intercept, x1 , of the tangent line is generally closer to r than x0 was. The process is then repeated until some prescribed accuracy is attained. The module below, called newton, repeats the above construction n times, where n is a given positive integer. It stores the values of x0 , x1 , ..., xn in a list named data and produces a plot of the iteration. The calling parameters of the module are: f = the function f ; a = the left end of some interval that contains r; b = the right end of the interval; x0 = the initial value to start the iteration; n = the number of iterations desired.
In[2]:=

newton f_, a_, b_, x0_, n_ : Module data

g , g x_

x

f x

f' x ; ;

NestList g, x0, n ; data2 Flatten Table data i , 0, data i , f data i , i, Length data data3 Partition data2, 2 ; data4 Drop data3, 1 ; p1 Plot f x , x, a, b , PlotStyle Red, Thick , AxesLabel "x", "y" , PlotRange All ; p2 Graphics Hue .6 , Line data4 ; Show p1, p2

We now apply this module to the function f x x2 2 with x0 obtain a geometric illustration of the first iteration (n 1).
In[3]:=

0.5 and the interval 0, 2.5 in order to

Clear f ; f x_ : x^2

2

In[5]:=

newton f, 0, 2.5, .5, 1 y 4 3

Out[5]=

2 1 0.5 1 2 1.0 1.5 2.0 2.5 x

APPENDIX E

Answers to Selected Exercises
√ √ √ §1. 1. Q; none; Q; R. 3. 14, 14, 14; equilateral. 4. (x −1)2 +(y−1)2 +(z−1)2 = 10. 6. A plane parallel to the yz-plane and 3 units behind it. 8. All points on or between the planes y = −1 and √ = 2. 10. All points between y √ the spheres centered at O that have radii 2 and 3. 13. x2 + y2 + z2 < 4, z > 0. 15. ( 3 2, 3 2, −3). 17. (0, −3, −1). 2 √ √ √ √2 √ √ 18. ( 2, 7π/4, 4). 20. (2, π/6, −2). 24. (− 3 3, − 3 , 0). 25. (0, 2, 2). 26. ( 6, π/4, arccos(2/ 6)). 2 2 27. (2, 3π/4, 3π/4). 30. r2 = 2z, ρ2 sin2 φ = 2ρ cos φ. 32. z = r, φ = π/4. 33. z = −2, a horizontal plane. 2 2 2 35. x + y + z = 4, a sphere with radius 2 and center O. 37. x2 + y2 − 2z = 0. 40. The sphere with radius 2 and center O. 42. All points above the xy-plane that lie on or between the spheres with center O and radii 2 and 4. 44. All points below the xy-plane that lie on or inside the inﬁnite cylinder with radius 1 and axis the z-axis. √ √ √ 1 8. 0. 10. 3. 12. π/3. §2. 2. 2, −5, −2 , 2 2, 4, −7 . 4. 3, 5, −3, −4 , 2 11, 3, −2, −1 . √ 14. arccos(−1/ 28) ≈ 101◦ . 15. π/2, π/3, π/6. 18. Neither. 19. −1, −2, 2 . 21. −7, −3, 5 . 23. 2, −1, 3 . √ √ 26. 2 1329. 28. 80. 30. Undeﬁned. 32. 1 110. 35. abe1 − b cos φe2 − b sin φe3 . 2 §3. 2. x = t, y = −2 + 3t, z = 3 + 2t (t ∈ R); x = y+2 3

=

z−3 2 .

3. Skew.

5. arccos −

7. 3x − 2z + 3 = 0. 9. −x + 2y + z + 5 = 0. 10. There is no such plane. 11. 4x + 5y − 3z + 1 = 0. 16. π/2. 17. 8 . 9 §4. 1. (a) ln 2, sin 2, e ; (b) 1, 2, 1 . 2. §5. 1. (a) Yes, t = 0; ; (c) no. z = 2. 8. (0, ±5, −2). 9. Smooth. (e) f(s) = 1+
3s 5 2t t √ 2 t2 +1 , 1−4t2 , e

14 √ 3 38 13. x−2.4 2

≈ 139◦ . = y+0.4 3 z = 5.

. 4. 3t2 , 0, −6e2t .
√3 , 0, √1 10 10

1 7. 0, 2 e2 − e−2 , 0 .

2. Yes, r1 (0) = r2 (2) = 0, 0, 0 . 4. 11. Piecewise smooth. , 1+
3s 5

. 5. x = −1, y = 1. 7. x = −5,

14.
3s 5)

sin 4 ln(1 +

3s 5)

cos 4 ln(1 +

, 5 , 0 ≤ t ≤ 5.

16 √ . 3 53

16. 1 + ln 2. 17. (a) s(a) = 0, s(b) = L;

§6. 1. Domain: all points (x, y); range: [0, ∞). 3. Domain: the points (x, y, z) for which (x, y) is not on the hyperbola x2 − y2 = 1; range: all real numbers. 5. The points on and above the line y = −x/2, except for those that lie on the circle x2 + y2 = 1. 6. Levels −1 and 0: no level curves; level 1: the line y = 0; level 2: the lines y = 1 and √ √ y = −1; level 3: the lines y = 2 and y = − 2. 8. Level −1: no level curve; level 0: the level curve consists of a single point, (0, 0); for k = 1, 2, 3, the k-level curve is the circle x2 + y2 = k. 9. −29. 11. 0. 12. D.N.E. 14. 2. 15. All (x, y). 16. All (x, y) (0, 0). 17. The points with x2 + y2 + 4z2 ≤ 4. §7. 1. Parabolas, parabolas, hyperbolas, hyperbolic paraboloid. 3. Ellipses, ellipses, ellipses, ellipsoid. 6. (x − 2)2 + 4y2 + (z + 1)2 = 9, an ellipsoid centered at (2, 0, −1). 7. r(t) = 3 cos t, 3 sin t, 3 cos t + 3 sin t + 2 , 0 ≤ t ≤ 2π. 8. The elliptic paraboloid z = x2 + y2 . 11. r(y, z) = − 2 y2 + z2 + 4, y, z , −∞ < y, z < ∞. 2 2 2 2 13. (a) r(x, y) = x, y, 4 − x − y , x + y ≤ 4; (b) r(u, v) = u cos v, u sin v, 4 − u2 , 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. 14. r(u, v) = (b + a cos v) cos u, (b + a cos v) sin u, a sin v , 0 ≤ u, v ≤ 2π. §8. 1. 4x3 − 8xy3/2 , −6x2 y1/2 . 3. 3(3x + 7yz)−1 , 7z(3x + 7yz)−1 , 7y(3x + 7yz)−1 . x−1 x x−1 4. 2x2 + xy x2 + xy + y2 + x2 + xy + y2 ln x2 + xy + y2 , 2x2 + xy x2 + xy + y2 . 7. 12x2 − 8y3/2 , −12xy1/2 , 2 −1/2 2 2 2 2 2 2 −3x y . 8. −(2x + y) cos x + xy + y − 2 sin x + xy + y , −(2x + y)(x + 2y) cos x + xy + y2 − sin x2 + xy + y2 , −(x + 2y)2 cos x2 + xy + y2 − 2 sin x2 + xy + y2 . 10. −12y1/2 . 13. 6z2 + 4yz3 e2yz , 12xz + 24xyz2 + 8xy2 z3 e2yz . 2(y − 3z)et − 3(x + 2z)e−t + 2(2y − 3y)e2t 14. 2y(2t + 3) + (2x + 3y2 )(2t). 17. . 18. (4xy − y2 )6s + (2x2 − 2xy)3t, xy + 2yz − 3xz (4xy − y2 )2t + (2x2 − 2xy)3s. 21. (−y sin xy + y3 x ln 3)2s + (−x sin xy + 3 x ) + 2z(t + 1),
239

(−y sin xy + y3 x ln 3) + (−x sin xy + 3 x )(−2t) + 2z(sg + 1). 23. −6e−24 . 24. 9 . 27. ln 7 + 3 (x − 2) + 5 (y − 1). 7 7 7 29. 1 (x, y) = (y − 2), 2 (x, y) = 0. 30. (a) 0, 0. 31. (a) a, d. √ 1 2 §9. 1. √5 e2xy (x + 2y + x2 y + 2xy2 + y3 + 2x3 ). 4. 60 10. 6. 7, in the direction of −3, 2, −6 . x−2 y+1 z+2 = = z + 2. 3. x + y − 2z − 5 = 0 and x − 2 = y + 1 = . 3 2 −2 3 1 1 5. (a) x = x0 + tx0 , y = y0 + ty0 , z = z0 + tz0 . 9. 16 , − 2 , 8 . 10. The tangent plane to both is x − 2y + z = 2. 2 11. −x + π y − z = 0. 14. −2y + z + 1 = 0. 17. (− 3 , 10 , − 26 ). 18. Smooth. 20. Piecewise smooth. 21. Not even 2 9 27 piecewise smooth: it is non-smooth on each curve xy = − π + 2kπ. 2 §10. 1. 3x + 2y + z − 2 = 0 and
1 §11. 1. A saddle point at ( 2 , −2). 3. Local minima at (±1, ±1) and a saddle point at (0, 0). 6. Local maxima ±1 √1 at local minima at ( √2 , 2 ), and a saddle point at (0, 0). 8. A local minimum at (1, 1 ) and a saddle point 2 √ √ √ 4 at (0, 0). 10. 8, 20. 12. − 3 √2 , 4 √2 . 14. 2/3. 16. The cube of side length 5. 2 3 2 √ √ ( ±1 , ±1 ), 2 2 4 §12. 1. ± 5 5−1/4 . 8 §13. 1. (a) − 27 .

4. The minimum is 7; there is no maximum. 6. S /3, S /3, S /3. 3. 2.5. 4. 20 2 . 6. 8(ln 3)−2 . 8. 3
4 4−y 3 4

ln 3 − ln 2. 10.

25 8 π.

12. −1. 13. 4 ln 6. 14.
1 y2 0 0 f (x, y) dxdy. 21. 1 ln 17. 23. 4.9. 4

1 §14. 1. −21 3 . 3. sin 1−cos 1. 4. 257.2. 6. 1 ln 2. 8. 1 (1−sin 1). 10. 7. 2 2 2 √ 2y ln 2 2

20. 1 1 − e−1 . 16. 0 0 f (x, y) dxdy + 2 0 f (x, y) dxdy. 18. 0 e x f (x, y) dydx. 4 3 56 16 2 24. 6. 26. 2 π + 15 . 27. 243.2. 29. 3 . 30. 3 π. √ √ √ 8 §15. 2. (e − 1) π . 3. 15 15 − 7 7 π . 5. 1 1 − 5e−4 . 6. 8π. 8. 36 − 20 5 π. 10. 3 π − 2 6 8 3 1 − e−16 . 3. 0.7. 5. 64 . 7. 1 (1 − sin 1). 9. 16 . 15 2 3 √ 2π 34 2π §17. 1. 15 π. 3. 3 2 2 − 1 . 4. 2π2 . 6. 8π. 8. 35 . 11. 81 π. 13. 2 √ √ √ 9 §18. 1. 4 6. 3. 8π 2 − 2 . 5. π 2 6 − 8 . 8. 64 , ( 6 , 0). 9. 32, 3 9 5 §16. 2.
1 2

32 9 .

11. π . 2

14 3 π. 7 7 7 6, 6, 6

.

12. In =

13 22

· · · 2n−1 2

√ π 2 .

§19. 1. R : x2 + 4y2 ≤ 4. 3. R : x2 − 2xy + 2y2 ≤ 2. 5. R is a parallelogram with vertices (0, 0), (1, 1), (1, −3), 5 and (2, −2). 6. (c) u. 7. −6. 9. 1 ln 2 − 16 . 11. (a) D : [1, 4] × [1, 6]; (c) −4x2 − 2y2 ; (d) 7.5. 13. (a) T is not 2 one-to-one on the planes u = 0 and v = 0; (b) u2 v; (c) p!q!r!s!/(p + q + r + s + 3)!. z y 2x , , . 5. Open and simply-connected. x2 + yz x2 + yz x2 + yz 6. Simply-connected, but not open. 7. Open and connected, but not simply-connected. 9. Open, but not connected. 11. No. 13. Yes, f (x, y) = xey + c. 14. Yes, f (x, y) = y2 arctan x + c. 16. curl F = −4xz2 , x2 y + 4yz2 , −x2 z , div F = −6xyz. 18. curl F = 2xze2xy , −2yze2xy − x sin xz, 2y cos xy , div F = e2xy + 2x cos xy − z sin xz. 20. Yes, f (x, y, z) = xy2 cos z + c. 21. No. 23. No, because div(xy2 e1 + yz2 e2 + zx2 e3 ) 0. √ √ 7 2 1 §21. 1. 840 125 5 − 1 . 3. 24 . 5. 3π3 13. 7. 14. 9. 15π. 10. 3 − cos 1 + sin 2. §20. 2. ye2y cos xy, xe2y cos xy + 2e2y sin xy . 4. §22. 1. −1 + 2e−1 . 3. −4. 5. 256 cos 8. 6. 4π.
3 §23. 2. −24π. 4. 162. 6. 0. 7. 4 − 1 π2 . 9. 91 7 . 10. 3π. 12. 2π. 3 2 √ §24. 1. 171 14. 3. 16π. 5. 713/180. 7. − 2 π. 3 13 §25. 1. 8π. 3. − 12 . 5. −20π. 7. − 13 . 12 √ √ 27 3 8 §26. 1. 0. 2. 5 π. 5. 3 π + 4 3. 6. 216π. 7.

√ 8 2 3 π.

8. 32π2 .

240

Calc Iii Lecture Notes

Similar Documents

Technology

Finance

Kkswmddkmdkemwekdemkedmkdemkmdkmdkew

Facility Planning

Herbert Hoover

Exsm

Managing Information Technology (7th Edition)

It Notes for Business

Linux as a Personal Desktop Operating System

Gsl Scientific Library

Basic Mba

Library Management

Industrial Engineering

Researching and Writing

Computer Vision

Popular Essays