Calculus 2nd edition multivariable solutions

This page intentionally left blank Student’s Solutions Manual to accompany Jon Rogawski’s Multivariable CALCULUS SECOND EDITION GREGORY P DRESDEN Washington and Lee University JENNIFER BOWEN The College of Wooster RANDALL PAUL Oregon Institute of Technology W H FREEMAN AND COMPANY NEW YORK © 2012 by W H Freeman and Company ISBN-13: 978-1-4292-5508-0 ISBN-10: 1-4292-5508-0 All rights reserved Printed in the United States of America First Printing W H Freeman and Company, 41 Madison Avenue, New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England www.whfreeman.com CONTENTS Chapter 10 INFINITE SERIES (LT Chapter 11) 10.1 Sequences (LT Section 11.1) 10.2 Summing an Infinite Series (LT Section 11.2) 10.3 Convergence of Series with Positive Terms 10.4 10.5 10.6 10.7 (LT Section 11.3) Absolute and Conditional Convergence (LT Section 11.4) The Ratio and Root Tests (LT Section 11.5) Power Series (LT Section 11.6) Taylor Series (LT Section 11.7) Chapter Review Exercises Chapter 11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS (LT Chapter 12) 11.1 11.2 11.3 11.4 Parametric Equations (LT Section 12.1) Arc Length and Speed (LT Section 12.2) Polar Coordinates (LT Section 12.3) Area and Arc Length in Polar Coordinates (LT Section 12.4) 11.5 Conic Sections (LT Section 12.5) Chapter Review Exercises Chapter 12 VECTOR GEOMETRY (LT Chapter 13) 12.1 Vectors in the Plane (LT Section 13.1) 12.2 Vectors in Three Dimensions (LT Section 13.2) 12.3 Dot Product and the Angle between Two Vectors 12.4 12.5 12.6 12.7 (LT Section 13.3) The Cross Product (LT Section 13.4) Planes in Three-Space (LT Section 13.5) A Survey of Quadric Surfaces (LT Section 13.6) Cylindrical and Spherical Coordinates (LT Section 13.7) Chapter Review Exercises Chapter 13 CALCULUS OF VECTOR-VALUED FUNCTIONS (LT Chapter 14) 13.1 Vector-Valued Functions (LT Section 14.1) 13.2 Calculus of Vector-Valued Functions 13.3 13.4 13.5 13.6 (LT Section 14.2) Arc Length and Speed (LT Section 14.3) Curvature (LT Section 14.4) Motion in Three-Space (LT Section 14.5) Planetary Motion According to Kepler and Newton (LT Section 14.6) Chapter Review Exercises Chapter 14 DIFFERENTIATION IN SEVERAL VARIABLES (LT Chapter 15) 13 14.1 Functions of Two or More Variables 23 14.2 Limits and Continuity in Several Variables 38 44 52 64 81 14.3 Partial Derivatives (LT Section 15.3) 14.4 Differentiability and Tangent Planes (LT Section 15.1) (LT Section 15.2) (LT Section 15.4) 336 336 345 351 363 14.5 The Gradient and Directional Derivatives (LT Section 15.5) 14.6 The Chain Rule (LT Section 15.6) 14.7 Optimization in Several Variables (LT Section 15.7) 374 386 399 14.8 Lagrange Multipliers: Optimizing with a Constraint (LT Section 15.8) Chapter Review Exercises 96 96 112 120 133 143 154 Chapter 15 MULTIPLE INTEGRATION (LT Chapter 16) 15.1 Integration in Two Variables (LT Section 16.1) 15.2 Double Integrals over More General Regions (LT Section 16.2) 15.3 Triple Integrals (LT Section 16.3) 15.4 Integration in Polar, Cylindrical, and Spherical Coordinates 166 (LT Section 16.4) (LT Section 16.5) 15.6 Change of Variables (LT Section 16.6) 184 198 212 223 Chapter 16 LINE AND SURFACE INTEGRALS (LT Chapter 17) 250 458 458 471 489 502 15.5 Applications of Multiple Integrals 166 176 230 240 419 442 Chapter Review Exercises 16.1 16.2 16.3 16.4 Vector Fields (LT Section 17.1) Line Integrals (LT Section 17.2) Conservative Vector Fields (LT Section 17.3) Parametrized Surfaces and Surface Integrals (LT Section 17.4) 16.5 Surface Integrals of Vector Fields (LT Section 17.5) Chapter Review Exercises 521 537 554 576 576 581 599 606 622 636 250 261 273 282 303 319 326 Chapter 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT Chapter 18) 17.1 Green’s Theorem (LT Section 18.1) 17.2 Stokes’ Theorem (LT Section 18.2) 17.3 Divergence Theorem (LT Section 18.3) Chapter Review Exercises 649 649 665 678 693 1019763_FM_VOL-I.qxp 9/17/07 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 S 50 R 51 4:22 PM Page viii This page was intentionally left blank 1st Pass Pages 10 INFINITE SERIES 10.1 Sequences (LT Section 11.1) Preliminary Questions What is a4 for the sequence an = n2 − n? solution Substituting n = in the expression for an gives a4 = 42 − = 12 Which of the following sequences converge to zero? (a) n2 n2 + (b) 2n (c) −1 n solution (a) This sequence does not converge to zero: lim n2 n→∞ n2 + x2 = lim x→∞ x + = lim x→∞ + x2 = = 1+0 (b) This sequence does not converge to zero: this is a geometric sequence with r = > 1; hence, the sequence diverges to ∞ (c) Recall that if |an | converges to 0, then an must also converge to zero Here, − n = n , which is a geometric sequence with < r < 1; hence, ( 12 )n converges to zero It therefore follows that (− 12 )n converges to zero √ Let an be the nth decimal approximation to That is, a1 = 1, a2 = 1.4, a3 = 1.41, etc What is lim an ? n→∞ solution √ lim an = n→∞ Which of the following sequences is defined recursively? √ (b) bn = (a) an = + n + bn−1 solution (a) an can be computed directly, since it depends on n only and not on preceding terms Therefore an is defined explicitly and not recursively (b) bn is computed in terms of the preceding term bn−1 , hence the sequence {bn } is defined recursively Theorem says that every convergent sequence is bounded Determine if the following statements are true or false and if false, give a counterexample (a) If {an } is bounded, then it converges (b) If {an } is not bounded, then it diverges (c) If {an } diverges, then it is not bounded solution (a) This statement is false The sequence an = cos πn is bounded since −1 ≤ cos πn ≤ for all n, but it does not converge: since an = cos nπ = (−1)n , the terms assume the two values and −1 alternately, hence they not approach one value (b) By Theorem 5, a converging sequence must be bounded Therefore, if a sequence is not bounded, it certainly does not converge (c) The statement is false The sequence an = (−1)n is bounded, but it does not approach one limit May 18, 2011 C H A P T E R 10 INFINITE SERIES (LT CHAPTER 11) Exercises Match each sequence with its general term: a1 , a2 , a3 , a4 , General term (a) 12 , 23 , 34 , 45 , (i) cos πn (b) −1, 1, −1, 1, n! (ii) n (c) 1, −1, 1, −1, (iii) (−1)n+1 (d) 21 , 24 , 68 , 24 16 (iv) n n+1 solution (a) The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator; n , n = 1, 2, 3, hence an = n+1 (b) The terms of this sequence are alternating between −1 and so that the positive terms are in the even places Since cos πn = for even n and cos πn = −1 for odd n, we have an = cos πn, n = 1, 2, (c) The terms an are for odd n and −1 for even n Hence, an = (−1)n+1 , n = 1, 2, (d) The numerator of each term is n!, and the denominator is 2n ; hence, an = 2n!n , n = 1, 2, 3, In Exercises 3–12, calculate the first four terms of the sequence, starting with n = 1 Let ann = for n = 1, 2, 3, Write out the first three terms of the following sequences 2n − 3 c(a) n =b = a (b) cn = an+3 nn! n+1 solution n = 1, 2, 3, in the formula for cn gives (d) en = 2an − an+1 (c) dn =Setting an2 c1 = 31 = = 3, 1! c3 = 27 33 = = , 3! c2 = 32 = , 2! c4 = 81 27 34 = = 4! 24 a1 = 2, an+1 = 2an2 − (2n − 1)! bn = For n = 1, 2, we have: solution n! a2 = a1+1 = 2a12 − = · − = 5; a3 = a2+1 = 2a22 − = · 25 − = 47; a4 = a3+1 = 2a32 − = · 2209 − = 4415 The first four terms of {an } are 2, 5, 47, 4415 bn = + cos πn b1 = For 1, nb= n= solution 1, b2,n−1 3, 4+we have bn−1 b1 = + cos π = 4; b2 = + cos 2π = 6; b3 = + cos 3π = 4; b4 = + cos 4π = The first four terms of {bn } are 4, 6, 4, 1 cn = +2n+1 + ··· + cn 1=+(−1) n solution c1 = 1; = ; 2 11 1 ; =1+ + = + = 3 1 11 25 =1+ + + = + = 12 c2 = + c3 c4 May 18, 2011 Sequences S E C T I O N 10.1 (LT SECTION 11.1) 11 b1 = 2, b2 = 3, bn = 2bn−1 + bn−2 an = n + (n + 1) + (n + 2) + · · · + (2n) solution We need to find b3 and b4 Setting n = and n = and using the given values for b1 and b2 we obtain: b3 = 2b3−1 + b3−2 = 2b2 + b1 = · + = 8; b4 = 2b4−1 + b4−2 = 2b3 + b2 = · + = 19 The first four terms of the sequence {bn } are 2, 3, 8, 19 13 Find a formula for the nth term of each sequence cn = n-place decimal approximation to e −1 , , (b) , , , (a) , 27 solution (a) The denominators are the third powers of the positive integers starting with n = Also, the sign of the terms is alternating with the sign of the first term being positive Thus, (−1)1+1 ; a1 = = 13 (−1)2+1 a2 = − = ; 23 (−1)3+1 a3 = = 33 This rule leads to the following formula for the nth term: an = (−1)n+1 n3 (b) Assuming a starting index of n = 1, we see that each numerator is one more than the index and the denominator is four more than the numerator Thus, the general term an is an = n+1 n+5 In Exercises 15–26, Theorem to determine the7.limit of the sequence or state that the sequence diverges an = 41and lim bn = Determine: Suppose thatuse lim n→∞ n→∞ 12 (an + bn ) 15 a(a) n = lim (b) lim an3 n→∞ n→∞ solution We have an = f (n) where f (x) = 12; thus, (c) lim cos(πbn ) (d) lim (an2 − 2an bn ) n→∞ n→∞ lim an = lim f (x) = lim 12 = 12 n→∞ x→∞ x→∞ 5n − 17 bn = an 12n = 20+−9 n 5x − ; thus, solution We have bn = f (n) where f (x) = 12x + lim 5n − n→∞ 12n + 5x − = lim x→∞ 12x + = 12 19 cn = −2−n + n − 3n2 an = We have solution c = f (n) where f (x) = −2−x ; thus, 4n2 + 1n lim n→∞ −2−n = lim −2−x = lim − x = x→∞ x→∞ 21 cn = 9n n zn = We have cn = f (n) where f (x) = 9x ; thus, solution lim 9n = lim 9x = ∞ n→∞ x→∞ Thus, the sequence 9n diverges n 23 an = −1/n zn =n10 2+1 solution We have an = f (n) where f (x) = lim n→∞ n n2 + = lim x→∞ May 18, 2011 x x2 + x x2 + ; thus, x = lim √ x x→∞ x +1 x = lim x→∞ x +1 x2 = lim x→∞ = = √ 1+0 + 12 x C H A P T E R 10 INFINITE SERIES (LT CHAPTER 11) 12nn + 25 an = an ln = −9 + 4n n3 + solution We have an = f (n) where f (x) = ln lim ln n→∞ 12n + −9 + 4n 12x + ; thus, −9 + 4x 12x + −9 + 4x = lim ln x→∞ = ln lim x→∞ 12x + −9 + 4x = ln In Exercises 27–30, use Theorem to determine the limit of the sequence rn = ln n − ln(n2 + 1) 27 an = + n solution We have lim + n→∞ Since 1 = lim + = n x→∞ x √ x is a continuous function for x > 0, Theorem tells us that lim n→∞ 4+ = n lim + n→∞ √ = 4=2 n n3 = −1 e4n/(3n+9) an cos 29 an = 2n + solution We have n3 lim n→∞ 2n3 + = Since cos−1 (x) is continuous for all x, Theorem tells us that lim cos−1 n→∞ n3 2n3 + = cos−1 lim n3 n→∞ 2n3 + = cos−1 (1/2) = π n 31 Let an = −1 −n Find a number M such that: an = tan n + 1(e ) (a) |an − 1| ≤ 0.001 for n ≥ M (b) |an − 1| ≤ 0.00001 for n ≥ M Then use the limit definition to prove that lim an = n→∞ solution (a) We have |an − 1| = n − (n + 1) −1 n −1 = = = n+1 n+1 n+1 n+1 ≤ 0.001, that is, n ≥ 999 It follows that we can take M = 999 Therefore |an − 1| ≤ 0.001 provided n+1 ≤ 0.00001, that is, n ≥ 99999 It follows that we can take M = 99999 (b) By part (a), |an − 1| ≤ 0.00001 provided n+1 We now prove formally that lim an = Using part (a), we know that n→∞ |an − 1| = provided n > − Thus, Let < , n+1 > and take M = − Then, for n > M, we have |an − 1| = 1 < = n+1 M +1 33 Use the limit definition to prove that lim n−2 = n n→∞ Let bn = 13 solution We see that (a) Find a value of M such that |bn | ≤ 10−5 for n ≥ M (b) Use the limit definition to prove that lim −2 −bn0|==0 = < n→∞ |n n2 n2 provided n> √ May 18, 2011 Divergence Theorem (LT SECTION 18.3) S E C T I O N 17.3 691 ϕ = We compute the partial derivatives: We show that ϕ is harmonic, that is, ∂ϕ = xz ∂x ∂ 2ϕ =z ∂x ⇒ ∂ϕ = −yz ∂y ⇒ x2 − y2 ∂ϕ = ∂z ∂ 2ϕ = −z ∂y ∂ 2ϕ =0 ∂z2 ⇒ Therefore, ϕ= ∂ 2ϕ ∂ 2ϕ ∂ 2ϕ + + =z−z+0=0 ∂x ∂y ∂z2 Since F is the gradient of a harmonic function, we know by part (c) that div(F) = Therefore, by the Divergence Theorem, the flux of F through a closed surface is zero: S F · dS = W n whereChallenges n is any number, r = (x LetInsights F = r er , and Further div(F) dV = W dV = + y + z2 )1/2 , and e = r −1 x, y, z is the unit radial vector r (a) SCalculate div(F) surface of a region W in R and let D ϕ denote the directional derivative of ϕ, where e is 39 Let be the boundary en n (b) Calculate the flux of F through ofoperator a spheredefined of radius R centered at the origin For which values of n the outward unit normal vector Let bethe thesurface Laplace earlier is this independent of R? to prove that (a) Use theflux Divergence Theorem (c) Prove that ∇(r n ) = n r n−1 er (d) Use (c) to show that F is conservative forDnen=ϕ −1 Then show that F = r −1 er is also conservative by computing dS = ϕ dV S W the gradient of ln r (b) Show that is if the ϕ isvalue a harmonic (defined Exercise 37),that then (e) What of Ffunction · ds, where C is ainclosed curve does not pass through the origin? C is harmonic (f) Find the values of n for which the function ϕ =Dr n ϕ en dS = S solution (a) By the theorem on evaluating directional derivatives, Den ϕ = ∇ϕ · en , hence, S Den ϕ dS = S ∇ϕ · en dS (1) By the definition of the vector surface integral, we have S ∇ϕ · dS = S (∇ϕ · en ) dS Combining with (1) gives S Den ϕ dS = S ∇ϕ · dS We now apply the Divergence Theorem and the identity div(∇ϕ) = S (b) If ϕ is harmonic, then Den ϕ dS = S ∇ϕ · dS = W ϕ shown in part (a) of Exercise 27, to write div(∇ϕ) dV = W ϕ dV ϕ = 0; therefore, by the equality of part (a) we have S Den ϕ dS = W ϕ · dV = W dV = 41 Let F = P , Q, R be a vector field defined on R such that div(F) = Use the following steps to show that F has that ϕ is harmonic Show that div(ϕ∇ϕ) = ∇ϕ and conclude that a vectorAssume potential (a) Let A = f, 0, g Show that ϕDen ϕ dS = ∇ϕ dV S W ∂g ∂f ∂g ∂f curl(A) = , − ,− ∂y ∂z ∂x ∂y May 20, 2011 692 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) (b) Fix any value y0 and show that if we define y f (x, y, z) = − g(x, y, z) = y0 y y0 R(x, t, z) dt + α(x, z) P (x, t, z) dt + β(x, z) where α and β are any functions of x and z, then ∂g/∂y = P and −∂f/∂y = R (c) It remains for us to show that α and β can be chosen so Q = ∂f/∂z − ∂g/∂x Verify that the following choice works (for any choice of z0 ): α(x, z) = z z0 Q(x, y0 , t) dt, β(x, z) = Hint: You will need to use the relation div(F) = solution (a) If A = f, 0, g , then the curl of A is the following vector field: curl(A) = i j ∂ ∂x f ∂ ∂y k ∂ ∂z g ∂g −0 i− ∂y = ∂g ∂f − ∂x ∂z j+ 0− ∂f ∂y k= ∂g ∂f ∂g ∂f , − ,− ∂y ∂z ∂x ∂y (b) Using the Fundamental Theorem of Calculus, we have y ∂g ∂ ∂ (x, y, z) = β(x, z) = P (x, y, z) + = P (x, y, z) P (x, t, z) dt + ∂y ∂y y0 ∂y − y ∂ ∂ ∂f (x, y, z) = α(x, z) = R(x, y, z) + = R(x, y, z) R(x, t, z) dt + ∂y ∂y y0 ∂y (c) We verify that the functions α(x, z) = z z0 Q(x, y0 , t) dt, β(x, z) = satisfy the equality Q= ∂g ∂f − ∂z ∂x We differentiate to obtain y y ∂f ∂g − Rz (x, t, z) dt + αz (x, z) − Px (x, t, z) dz − βx (x, z) =− ∂z ∂x y0 y0 =− y y0 (Px (x, t, z) + Rz (x, t, z)) dt + αz (x, z) (1) By the Fundamental Theorem of Calculus, αz (x, z) = z ∂ Q(x, y0 , t) dt = Q(x, y0 , z) ∂z z0 (2) Also, since div(F) = 0, we have div(F) = Px + Qy + Rz = ⇒ Px + Rz = −Qy Substituting (2) and (3) in (1) gives y ∂f ∂g − = Qy (x, t, z) dt + Q(x, y0 , z) = Q(x, y, z) − Q(x, y0 , z) + Q(x, y0 , z) = Q(x, y, z) ∂z ∂x y0 Parts (a)–(c) prove that F = curl(A), or A is a vector potential for F 43 Show that Show that −z F = 2yez − xy, y, yz F = 2y − 1, 3z2 , 2xy has a vector potential and find one has a vector potential and find one May 20, 2011 (3) Chapter Review Exercises 693 solution As shown in Exercise 41, if F is divergence free, then F has a vector potential We show that div(F) = 0: div(F) = ∂ ∂ ∂ (2yez − xy) + (y) + (yz − z) = −y + + y − = ∂x ∂y ∂z We find a vector potential A, using the result in Exercise 41: A = f, 0, g (1) Using z0 = 0, we have y f (x, y, z) = − g(x, y, z) = y0 R(x, t, z) dt + z Q(x, y0 , t) dt y P (x, t, z) dt y0 Hence, P (x, y, z) = 2yez − xy, Q(x, y, z) = y, and R(x, y, z) = yz − z We choose y0 = and compute the functions f and g: y f (x, y, z) = − g(x, y, z) = y (tz − z) dt + z (2tez − xt) dt = t ez − y t 2z − zt dt = − t=0 = zy − y2 y2z =z y− 2 xt y xy x = y ez − = y ez − t=0 2 Substituting in (1) we obtain A= z y− CHAPTER REVIEW y2 , 0, y ez − x er In the text, we observed that although the inverse-square radial vector field F = satisfies div(F) = 0, F cannot r EXERCISES have a vector potential on its domain {(x, y, z) = (0, 0, 0)} because the flux of F through a sphere containing the origin is nonzero (a) F(x, Showy)that a vector potential A such that FEvaluate = curl(A) F on· ds thedirectly restricted Let = xthe + method y , x −ofy Exercise and let C41 beproduces the unit circle, oriented counterclockwise as domain D consisting of R with the y-axis removed C a line(b) integral Theorem Showand thatusing F alsoGreen’s has a vector potential on the domains obtained by removing either the x-axis or the z-axis from R solution We parametrize the unit circle by c(t) = (cos t, sin t), ≤ t ≤ 2π Then, c (t) = − sin t, cos t and (c) = Does of2 at vector potential on these domains contradict the fact that the flux of F through F(c(t)) (costhe t +existence sin2 t, cos − sin t) We compute therestricted dot product: a sphere containing the origin is nonzero? F(c(t)) · c (t) = cos t + sin2 t, cos2 t − sin t · − sin t, cos t = (− sin t)(cos t + sin2 t) + cos t (cos2 t − sin t) = cos3 t − sin3 t − sin t cos t The line integral is thus C F (c(t)) · c (t) dt = = = 2π 2π cos3 t − sin3 t − sin t cos t dt cos3 t dt − 2π sin3 t dt − sin t 2π cos2 t sin t + + 3 2π sin 2t dt cos t sin2 t cos t + 3 2π + cos 2t 2π =0 We now compute the integral using Green’s Theorem We compute the curl of F Since P = x + y and Q = x − y, we have ∂Q ∂P − = 2x − 2y ∂x ∂y Thus, C May 20, 2011 F · ds = D (2x − 2y) dx dy 694 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) y C D x We compute the double integral by converting to polar coordinates We get C F · ds = = 2π 1 (2r cos θ − 2r sin θ)r dr dθ = 2π 2r dr 0 (cos θ − sin θ) dθ 2π = 2r (cos θ − sin θ) dr dθ 31 r sin θ + cos θ 2π = (1 − 1) = In Exercises 3–6, use Green’s Theorem to evaluate the line integral around the given closed curve Let ∂R be the boundary of the rectangle in Figure and let ∂R1 and ∂R2 be the boundaries of the two triangles, all oriented counterclockwise xy dx + x y dy, where C is the rectangle −1 ≤ x ≤ 2, −2 ≤ y ≤ 3, oriented counterclockwise C Determine (a) F · ds if F · ds = and F · ds = −2 ∂ R1 ∂R ∂ R2 solution (b) What is the value of ∂R y F ds if ∂R is oriented clockwise? C D −1 x −2 Since P = xy , Q = x y the curl of F is ∂Q ∂P − = 3x y − 3xy ∂x ∂y By Green’s Theorem we obtain C xy dx + x y dy = = D −2 (3x y − 3xy ) dx dy = x3y − −2 −1 (3x y − 3xy ) dx dy 3x y 2 3y dy = (8y − 6y ) − −y − x=−1 −2 dy 3y 9y 81 81 + − (12 + 18) = −30 = − + 2 2 −2 −2 √ y dx − x dy, where C consists of the arcs y = x and y = x, ≤ x ≤ 1, oriented clockwise (3x + 5y − cos y) dx + x sin y dy, where C is any closed curve enclosing a region with area 4, oriented counC C We compute the curl of F solution terclockwise = − 9y + 9y dy = − y C y= x D y = x2 We have P = y and Q = −x , hence ∂Q ∂P = −2x − 2y − ∂y ∂x May 20, 2011 x Chapter Review Exercises 695 We now compute the line integral using Green’s Theorem Since the curve is oriented clockwise, we consider the negative of the double integrals We get C y dx − x dy = − = = D 1 (−2x − 2y) dA = − √ 2xy + y x y=x dx = √ x2 x (−2x − 2y) dy dx √ 2x x + x − (2x · x + x ) dx (−x − 2x + 2x 3/2 + x) dx = − x4 4x 5/2 x2 x5 − + + 5 1 =− − + + = 5 Let c(t) = t (1 − t), t(t − 1)2 x ye xey c(t) dy, where (a) Plotdx the+path for ≤C tis≤the triangle with vertices (−1, 0), (0, 4), and (0, 1), oriented counterclockwise C (b) Calculate the area A of the region enclosed by c(t) for ≤ t ≤ using the formula A = (x dy − y dx) C solution (a) The path c(t) for ≤ t ≤ is shown in the figure: y 0.1 x 0.1 Note that the path is traced out clockwise as t goes from to (b) We use the formula for the area enclosed by a closed curve, A= (x dy − y dx) C We compute the line integral Since x = t (1 − t) and y = t(t − 1)2 , we have dx = 2t (1 − t) − t dt = 2t − 3t dt dy = (t − 1)2 + t · 2(t − 1) = (t − 1)(3t − 1) dt Therefore, x dy − y dx = t (1 − t) · (t − 1)(3t − 1) dt − t(t − 1)2 · (2t − 3t ) dt = t (t − 1)2 dt We obtain the following integral (note that the path must be counterclockwise): A= 1 1 −t (t − 1)2 dt = (t − 2t + t ) dt = 2 t5 t4 t3 − + = 60 In Exercises 9–12,state calculate the the curlequation and divergence of the(valid vectorfor field In (a)–(d), whether is an identity all F or V ) If it is not, provide an example in which the F =equation yi − zk does not hold (a) curl(∇V ) = solution We compute the curl of the vector field, (c) div(curl(F)) = i j k ∂ ∂ ∂ curl(F) = ∂x ∂y ∂z y −z = ∂ ∂ (−z) − (0) i − ∂y ∂z = 0i + 0j − 1k = −k May 20, 2011 (b) div(∇V ) = (d) ∇(div(F)) = ∂ ∂ (−z) − (y) j + ∂x ∂z ∂(0) ∂(y) − k ∂x ∂y 696 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) The divergence of F is div(F) = ∂ ∂ ∂ (y) + (0) + (−z) = + − = −1 ∂x ∂y ∂z −x −y −z 11 F =F∇(e = ex+y , ey+z), xyz solution 2 2 In Exercise we proved the identity curl(∇ϕ) = Here, ϕ = e−x −y −z , and we have 2 curl ∇ e−x −y −z = To compute div F, we first write F explicitly: 2 2 2 2 2 2 F = ∇ e−x −y −z = −2xe−x −y −z , −2ye−x −y −z , −2ze−x −y −z = P , Q, R div(F) = ∂Q ∂R ∂P + + ∂x ∂y ∂z 2 2 2 2 2 2 = −2e−x −y −z + 4x e−x −y −z + −2e−x −y −z + 4y e−x −y −z 2 2 2 + −2e−x −y −z + 4z2 e−x −y −z 2 = 2e−x −y −z 2(x + y + z2 ) − 13 Recall that if F1 , F2 , and F3 are differentiable functions of one variable, then er = r −1 x, y, z r = x + y + z2 curl ( F1 (x), F2 (y), F3 (z) ) = Use this to calculate the curl of F = x + y , ln y + z2 , z3 sin(z2 )ez solution We use the linearity of the curl and the property mentioned in the exercise to compute the curl of F: curl F = curl x + y , ln y + z2 , z3 sin z2 ez = + curl y , z2 , = = curl x , ln y, z3 sin(z2 )ez + curl y , z2 , ∂ ∂ ∂ ∂ ∂ ∂ (0) − z2 , y − (0), z − y = −2z, 0, −2y ∂y ∂z ∂z ∂x ∂x ∂y 15 Verify the identities of Exercises and 34 in Section 17.3 for the vector fields F = xz, yex , yz and G = Give an example of a nonzero vector field F such that curl(F) = and div(F) = z2 , xy , x y solution We first show div(curl(F)) = Let F = P , Q, R = xz, yex , yz We compute the curl of F: curl(F) = i ∂ ∂x P j ∂ ∂y Q k ∂ ∂z R = ∂Q ∂P ∂R ∂Q ∂P ∂R − , − , − ∂y ∂z ∂z ∂x ∂x ∂y Substituting in the appropriate values for P , Q, R and taking derivatives, we get curl(F) = z − 0, x − 0, yex − Thus, div (curl(F)) = (z)x + (x)y + (yex )z = + + = Likewise, for G = P , Q, R = z2 , xy x y , we compute the curl of G: curl(G) = i ∂ ∂x P j ∂ ∂y Q k ∂ ∂z R = ∂Q ∂P ∂R ∂Q ∂P ∂R − , − , − ∂y ∂z ∂z ∂x ∂x ∂y Substituting in the appropriate values for P , Q, R and taking derivatives, we get curl(G) = x − 0, 2z − 2xy, y − May 20, 2011 Chapter Review Exercises 697 Thus, div (curl(G)) = (x )x + (2z − 2xy)y + (y )z = 2x − 2x = We now work on the second identity For F = xz, yex , yz and G = z2 , xy , x y , it is easy to calculate F × G = x y ex − xy z, yz3 − x yz, x y z − yz2 ex Thus, div(F × G) = (2xy ex + x y ex − y z) + (z3 − x z) + (x y − 2yzex ) On the other hand, from our work above, curl(F) = z, x, yex curl(G) = x , 2z − 2xy, y So, we calculate G · curl(F − F) · curl(G) = z2 · z + xy · x + x y · yex − xz · x − yex · (2z − 2xy) − yz · y = z3 + x y + x y ex + 2xy ex − x z − 2yzex − y z = (2xy ex + x y ex − y z) + (z3 − x z) + (x y − 2yzex ) = div(F × G) 17 Prove that if F is a gradient vector field, then the flux of curl(F) through a smooth surface S (whether closed or not) Suppose that S1 and S2 are surfaces with the same oriented boundary curve C Which of the following conditions is equal to zero guarantees that the flux of F through S1 is equal to the flux of F through S2 ? solution F is gradient vectorVfield, then F is conservative; therefore the line integral of F over any closed curve (i) F = If ∇V fora some function is zero Combining with Stokes’ Theorem (ii) F = curl(G) for some vector fieldyields G S curl(F) · dS = ∂S F · ds = 19 Let F = z2 , x + z, y and let S be the upper half of the ellipsoid Verify Stokes’ Theorem for F = y, z − x, and the surface z = − x − y , z ≥ 0, oriented by outwardpointing normals x2 + y + z2 = oriented by outward-pointing normals Use Stokes’ Theorem to compute S curl(F) · dS solution We compute the curl of F = z2 , x + z, y : i ∂ ∂x z2 curl(F) = j ∂ ∂y x+z k ∂ ∂z y2 = (2y − 1)i − (0 − 2z)j + (1 − 0)k = 2y − 1, 2z, Let C denote the boundary of S, that is, the ellipse x4 + y = in the xy-plane, oriented counterclockwise Then by Stoke’s Theorem we have S curl(F) · dS = C F · ds We parametrize C by C : r(t) = (2 cos t, sin t, 0), ≤ t ≤ 2π Then F (r(t)) · r (t) = 0, cos t, sin2 t · −2 sin t, cos t, = cos2 t Combining with (1) gives S curl(F) · ds = Use Stokes’ Theorem to evaluate May 20, 2011 C 2π cos2 t dt = t + sin 2t 2π = 2π y, z, x · ds, where C is the curve in Figure (1) 698 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) 21 Let S be the side of the cylinder x + y = 4, ≤ z ≤ (not including the top and bottom of the cylinder) Use Stokes’ Theorem to compute the flux of F = 0, y, −z through S (with outward pointing normal) by finding a vector potential A such that curl(A) = F solution We can write F = curl(A) where A = yz, 0, The flux of F through S is equal to the line integral of A around the oriented boundary which consists of two circles of radius with center on the z-axis (one at height z = and one at height z = 2) However, the line integrals of A about both circles are zero This is clear for the circle at z = because then A = 0, but it is also true at z = because the vector field A = 2y, 0, integrates to zero around the circle 2+ Verify23–26, the Divergence Theorem for F = 0, z and the region x 2for + ythe z2 =vector field and surface In Exercises use the Divergence Theorem to 0, calculate F · dS given S 23 F = xy, yz, x z + z2 , S is the boundary of the box [0, 1] × [2, 4] × [1, 5] solution z x y We compute the divergence of F = xy, yz, x z + z2 : div(F) = ∂ ∂ ∂ xy + yz + (x z + z2 ) = y + z + x + 2z = x + y + 3z ∂x ∂y ∂z The Divergence Theorem gives S xy, yz, x z + z2 · dS = = = = 1 1 + y + 3z (x + y + 3z) dx dy dz = dy dz = 16 + + 12z − 100 20z 3z2 + = 75 + 3 x3 + (y + 3z)x dy dz x=0 y + y + 3zy dz y=2 + + 6z − 3+ 20 dz = = 20 + 6z dz 296 z) + ex , ex +y , S is the boundary of the solid bounded by the cylinder x + y = 16 25 F = xyz + xy, 122y (1 − F = xy, yz, x z + z2 , S is the boundary of the unit sphere and the planes z = and z = y − 2 solution We compute the divergence of F: div(F) = ∂ ∂ (xyz + xy) + ∂x ∂y y2 (1 − z) + ex + ∂ x +y (e ) = yz + y + y(1 − z) = 2y ∂z Let S denote the surface of the solid W The Divergence Theorem gives S F · dS = = May 20, 2011 W div(F) dV = D 2yz z=y−4 dx dy = W 2y dV = D D y−4 2y dz dx dy 2y (0 − (y − 4)) dx dy = D (8y − 2y ) dx dy Chapter Review Exercises 699 We convert the integral to polar coordinates: S F · dS = 2π =8 =0− 4 (8r cos θ − 2r cos2 θ)r dr dθ 2π r dr cos θ dθ 0 r4 4 θ+ sin 2θ 2π − =− 2π r dr cos2 θ dθ 44 · 2π = −128π 27 Find the volume of a region W if F = sin(yz), x + z4 , x cos(x − y) , S is any smooth closed surface that is the boundary of a region in R x + xy + z, x + 3y − y , 4z · dS = 16 ∂W solution Let F = x + xy + z, x + 3y − 12 y , 4z We compute the divergence of F: div(F) = ∂ ∂ (x + xy + z) + ∂x ∂y ∂ x + 3y − y + (4z) = + y + − y + = ∂z Using the Divergence Theorem and the given information, we obtain 16 = S F · dS = W div(F) dV = W dV = W dV = Volume (W) That is, 16 = Volume (W) or Volume (W) = In Exercises 29–32, let F be a vector field whose curl and divergence at the origin are Show that the circulation of F = x , y , z(x + y ) around any curve C on the surface of the cone z2 = x + y is equal to zero (Figure 3) curl(F)(0, 0, 0) = 2, −1, , div(F)(0, 0, 0) = −2 29 Estimate C F · ds, where C is the circle of radius 0.03 in the xy-plane centered at the origin solution We use the estimation C F · ds ≈ (curl(F)(0) · en ) Area(R) z en R C y x The unit normal vector to the disk R is en = k = 0, 0, The area of the disk is Area (R) = π · 0.032 = 0.0009π Using the given curl at the origin, we have C F · ds ≈ 2, −1, · 0, 0, · 0.0009π = · 0.0009π ≈ 0.0113 31 Suppose that v is the velocity field of a fluid and imagine placing a small paddle wheel at the origin Find the equation of the plane in whichFthe should be placed make of it rotate as quickly as possible · ds,paddle wherewheel C is the boundary of thetosquare side 0.03 in the yz-plane centered at the origin Does Estimate C solution Thedepend paddle on wheel maximum spinwithin when the circulationMight of thethe velocity v around the wheel is the estimate howhas thethe square is oriented the yz-plane? actualfield circulation depend on how maximum The maximum circulation occurs when e , and the curl of v at the origin (i.e., the vector 2, −1, ) point in n it is oriented? the same direction Therefore, the plane in which the paddle wheel should be placed is the plane through the origin with the normal 2, −1, This plane has the equation, 2x − y + 4z = May 20, 2011 700 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) 33 The velocity vector field of a fluid (in meters per second) is Estimate the flux of F through the box of side 0.5 in Figure Does the result depend on how the box is oriented relative to the coordinate axes? F = x + y , 0, z2 Let W be the region between the hemisphere S = (x, y, z) : x + y + z2 = 1, x, y, z ≥ and the disk D = (x, y, 0) : x + y ≤ in the xy-plane Recall that the flow rate of a fluid across a surface is equal to the flux of F through the surface (a) Show that the flow rate across D is zero (b) Use the Divergence Theorem to show that the flow rate across S, oriented with outward-pointing normal, is equal to W div(F) dV Then compute this triple integral solution (a) To show that no fluid flows across D, we show that the normal component of F at each point on D is zero At each point P = (x, y, 0) on the xy-plane, F(P ) = x + y , 0, 02 = x + y , 0, Moreover, the unit normal vector to the xy-plane is en = (0, 0, 1) Therefore, F(P ) · en = x + y , 0, · 0, 0, = Since D is contained in the xy-plane, we conclude that the normal component of F at each point on D is zero Therefore, no fluid flows across D (b) By the Divergence Theorem and the linearity of the flux we have S F · dS + D F · dS = W div(F) dV Since the flux through the disk D is zero, we have F · dS = S W div(F) dV (1) To compute the triple integral, we first compute div(F): div(F) = ∂ ∂ ∂ (x + y ) + (0) + (z2 ) = 2x + 2z = 2(x + z) ∂x ∂y ∂z z W y x Using spherical coordinates we get W div(F) dV = =2 = π/2 2π 1 (ρ sin φ cos θ + ρ cos φ)ρ sin φ dρ dφ π/2 ρ dρ 2π sin2 φ dφ cos θ dθ 0 π/2 sin 2φ dφ 0+π = π − Combining with (1) we obtain the flux: S May 20, 2011 F · dS = π cos 2φ + 2π π/2 π/2 cos φ sin φ dρ π π = − (−1 − 1) = Chapter Review Exercises 701 x 35 LetThe V (x, y) = xfield + of The vector field F = ∇V (Figure 5) provides a model in the plane of the velocity velocity 2 (in meters per second) is x +a yfluid field of an incompressible, irrotational fluid flowing past a cylindrical obstacle (in this case, the obstacle is the unit circle F = (3y − 4)i + e−y(z+1) j + (x + y )k x + y = 1) (a) Verify that F istheirrotational [bycubic definition, is irrotational if curl(F) = 0] (a) Estimate flow rate (in metersF per second) through a small surface S around the origin if S encloses a region of volume 0.01 m3 y (b) Estimate the circulation of F about a circle in the3 xy-plane of radius r = 0.1 m centered at the origin (oriented counterclockwise when viewed from above) (c) Estimate the circulation of F about a circle in the1 yz-plane of radius r = 0.1 m centered at the origin (oriented counterclockwise when viewed from the positive x-axis) x −3 −2 −1 −1 −2 −3 FIGURE The vector field ∇V for V (x, y) = x + x x2 + y2 (b) Verify that F is tangent to the unit circle at each point along the unit circle except (1, 0) and (−1, 0) (where F = 0) (c) What is the circulation of F around the unit circle? (d) Calculate the line integral of F along the upper and lower halves of the unit circle separately solution (a) In Exercise 8, we proved the identity curl(∇ϕ) = Since F is a gradient vector field, it is irrotational; that is, curl(F) = for (x, y) = (0, 0), where F is defined (b) We compute F explicitly: ∂ϕ ∂ϕ y2 − x2 2xy , = 1+ ,− 2 ∂x ∂y (x + y ) (x + y ) F = ∇ϕ = Now, using x = cos t and y = sin t as a parametrization of the circle, we see that F = + sin2 t − cos2 t, −2 cos t sin t = sin2 t, −2 cos t sin t , and so F = sin t sin t, − cos t = sin t y, −x , which is clearly perpendicular to the radial vector x, y for the circle (c) We use our expression of F from Part (b): F = ∇ϕ = + y2 − x2 (x + y ) ,− 2xy (x + y ) Now, F is not defined at the origin and therefore we cannot use Green’s Theorem to compute the line integral along the unit circle We thus compute the integral directly, using the parametrization c(t) = (cos t, sin t), ≤ t ≤ 2π y C x Then, F (c(t)) · c (t) = + sin2 t − cos2 t (cos2 t + sin2 t) ,− cos t sin t (cos2 t + sin2 t) · − sin t, cos t = + sin2 t − cos2 t, −2 cos t, sin t · − sin t, cos t = sin2 t, −2 cos t sin t · − sin t, cos t = −2 sin3 t − cos2 t sin t = −2 sin t (sin2 t + cos2 t) = −2 sin t Hence, C May 20, 2011 F · ds = 2π −2 sin t dt = 702 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) (d) We denote by C1 and C2 the upper and lower halves of the unit circle Using part (c) we have C1 F · ds + C2 F · ds = ⇒ F · ds = − C2 C1 F · ds (1) y C1 x C2 To compute the circulation along C1 , we compute the integral as in part (c), only that the limits of integration are now t = and t = π Using the computations in part (c) we obtain C1 F · ds = π −2 sin2 t dt = −4 Therefore, by (1), C2 F · ds = 37 In Section 17.1, we showed that if C is a simple closed curve, oriented counterclockwise, then the line integral is Figure shows the vector field F = ∇V , where dy x−2 y+dx Area enclosed by(yC − = 1)2 +x ln (y + 1)2 V (x, y) = ln x + C which is the velocity field for the flow of a fluid with sources of equal strength at (0, ±1) (note that V is undefined Suppose that C is a path from P to Q that is not closed but has the property that every line through the origin intersects C at these two points) Show that F is both irrotational and incompressible—that is, curlz (F) = and div(F) = [in in at most one point, as in Figure Let R be the region enclosed by C and the two radial segments joining P and Q to computing div(F), treat F as a vector field in R with a zero z-component] Is it necessary to compute curlz (F) to the origin Show that the line integral in Eq (1) is equal to the area of R Hint: Show that the line integral of F = −y, x conclude that it is zero? along the two radial segments is zero and apply Green’s Theorem y C Q R P x FIGURE solution y C Q R P x Q ∂P Let F = −y, x Then P = −y and Q = x, and ∂Q ∂x − ∂y = By Green’s Theorem, we have C −y dx + x dy + QO −y dx + x dy + OP −y dx + x dy = R dA = R dA Denoting by A the area of the region R, we obtain A= 1 −y dx + x dy + −y dx + x dy + −y dx + x dy C QO OP May 20, 2011 (1) Chapter Review Exercises 703 We parametrize the two segments by QO : c(t) = (t, t tan β) c (t) = 1, tan β ⇒ OP : d(t) = (t, t tan α) d (t) = 1, tan α Then, F (c(t)) · c (t) = −t tan β, t · 1, tan β = −t tan β + t tan β = F (d(t)) · d (t) = −t tan α, t · 1, tan α = −t tan α + t tan α = Therefore, QO F · ds = OP F · ds = Combining with (1) gives A= −y dx + x dy C 39 Prove the following generalization of Eq (1) Let C be a simple closed curve in the plane (Figure 8) Suppose that the curve C in Figure has the polar equation r = f (θ) (a) Show that c(θ) = (f (θ) cos θ, f (θ) sin counterclockwise S :θ) is axa + by + cz + d = 0parametrization of C (b) In Section 11.4, we showed that the area of the region R is given by the formula Then the area of the region R enclosed by C is equal to β of + R (cx = − az)fdy (θ)+2 (ay dθ − bx) dz (bz −Area cy) dx α n C Use result 37 to a new of this formula Hint: thetoline Eq (1) where n the = a, b, cofisExercise the normal to give S, and C is proof oriented as the boundary of Evaluate R (relative theintegral normal in vector n) using Hint: c(θ) Apply Stokes’ Theorem to F = bz − cy, cx − az, ay − bx z Plane S n = 〈a, b, c〉 R C y x FIGURE solution By Stokes’ Theorem, S curl(F) · dS = S (curl(F) · en ) dS = C F · ds We compute the curl of F: curl(F) = i ∂ ∂x bz − cy j ∂ ∂y cx − az k ∂ ∂z ay − bx = 2ai + 2bj + 2ck = a, b, c The unit normal to the plane ax + by + cz + d = is en = a, b, c a + b2 + c2 Therefore, curl(F) · en = a, b, c · = May 20, 2011 a + b2 + c2 a + b2 + c2 a, b, c (a + b2 + c2 ) = a + b2 + c2 (1) 704 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LT CHAPTER 18) Hence, S The sign of curl(F) · dS = S curl(F) · en dS = S a + b2 + c2 dS = a + b2 + c2 S dS (2) S dS is determined by the orientation of S Since the area is a positive value, we have S ds = Area (S) Therefore, (2) gives S curl(F) · dS = a + b2 + c2 Area(S) Combining with (1) we obtain a + b2 + c2 Area(S) = C F · ds or Area(S) = a + b2 + c2 = · n C (bz − cy) dx + (cx − az) dy + (ay − bx) dz 41 Show that G(θ, φ) = (a cos θ sin φ, b sin θ sin φ, c cos φ) is a parametrization of the ellipsoid Use the result of Exercise 39 to calculate the area of the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) as a line integral Verify your result using geometry y z x + + =1 a b c Then calculate the volume of the ellipsoid as the surface integral of F = 13 x, y, z (this surface integral is equal to the volume by the Divergence Theorem) solution For the given parametrization, x = a cos θ sin φ, y = b sin θ sin φ, z = c cos φ (1) We show that it satisfies the equation of the ellipsoid y z x + + = a b c a cos θ sin φ + a b sin θ sin φ + b c cos φ c = cos2 θ sin2 φ + sin2 θ sin2 φ + cos2 φ = sin2 φ(cos2 θ + sin2 θ) + cos2 φ = sin2 φ + cos2 φ = Conversely, for each (x, y, z) on the ellipsoid, there exists θ and φ so that (1) holds Therefore whole ellipsoid Let W be the interior of the ellipsoid S Then by Eq (10): Volume(W) = S F · dS, (θ, φ) parametrizes the F = x, y, z We compute the surface integral, using the given parametrization We first compute the normal vector: ∂ = −a sin θ sin φ, b cos θ sin φ, ∂θ ∂ = a cos θ cos φ, b sin θ cos φ, −c sin φ ∂φ ∂ ∂ × = −ab sin2 θ sin φ cos φk − ac sin θ sin2 φj − ab cos2 θ sin φ cos φk − bc cos θ sin2 φi ∂θ ∂φ = −bc cos θ sin2 φ, −ac sin θ sin2 φ, −ab sin φ cos φ May 20, 2011 Chapter Review Exercises Hence, the outward pointing normal is n = bc cos θ sin2 φ, ac sin θ sin2 φ, ab sin φ cos φ F ( (θ, φ)) · n = a cos θ sin φ, b sin θ sin φ, c cos φ · bc cos θ sin2 φ, ac sin θ sin2 φ, ab sin φ cos φ = abc cos2 θ sin3 φ + abc sin2 θ sin3 φ + abc sin φ cos2 φ = abc sin3 φ(cos2 θ + sin2 θ) + abc sin φ cos2 φ = abc sin3 φ + abc sin φ cos2 φ = abc sin3 φ + abc sin φ(1 − sin2 φ) = abc sin φ We obtain the following integral: Volume(W) = S F · dS = 2π π abc sin φ dφ dθ 0 π 2πabc 2πabc π − cos φ sin φ dϕ = = 3 0 May 20, 2011 = 4πabc 705 ... Technology W H FREEMAN AND COMPANY NEW YORK © 2012 by W H Freeman and Company ISBN-13: 97 8-1 -4 29 2-5 50 8-0 ISBN-10: 1-4 29 2-5 50 8-0 All rights reserved Printed in the United States of America First Printing...This page intentionally left blank Student’s Solutions Manual to accompany Jon Rogawski’s Multivariable CALCULUS SECOND EDITION GREGORY P DRESDEN Washington and Lee University JENNIFER... Three-Space (LT Section 13.5) A Survey of Quadric Surfaces (LT Section 13.6) Cylindrical and Spherical Coordinates (LT Section 13.7) Chapter Review Exercises Chapter 13 CALCULUS OF VECTOR-VALUED