Test bank and solution of calculus 2nd (1)

INSTRUCTOR’S SOLUTIONS MANUAL MULTIVARIABLE MARK WOODARD Furman University C ALCULUS SECOND EDITION AND C ALCULUS E ARLY T RANSCENDENTALS SECOND EDITION William Briggs University of Colorado at Denver Lyle Cochran Whitworth University Bernard Gillett University of Colorado at Boulder with the assistance of Eric Schulz Walla Walla Community College Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo The author and publisher of this book have used their best efforts in preparing this book These efforts include the development, research, and testing of the theories and programs to determine their effectiveness The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs Reproduced by Pearson from electronic files supplied by the author Copyright © 2015, 2011 Pearson Education, Inc Publishing as Pearson, 75 Arlington Street, Boston, MA 02116 All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Printed in the United States of America ISBN-13: 978-0-321-95430-5 ISBN-10: 0-321-95430-0 CRK 17 16 15 14 13 www.pearsonhighered.com Contents Sequences and Infinite Series 8.1 An Overview 8.2 Sequences 8.3 Infinite Series 8.4 The Divergence and Integral Tests 8.5 The Ratio, Root, and Comparison Tests 8.6 Alternating Series Chapter Eight Review 3 10 23 34 43 49 55 Power Series 9.1 Approximating Functions With Polynomials 9.2 Properties of Power Series 9.3 Taylor Series 9.4 Working with Taylor Series Chapter Nine Review 63 63 82 88 100 111 10 Parametric and Polar Curves 10.1 Parametric Equations 10.2 Polar Coordinates 10.3 Calculus in Polar Coordinates 10.4 Conic Sections Chapter Ten Review 119 119 139 159 171 191 11 Vectors and Vector-Valued Functions 11.1 Vectors in the Plane 11.2 Vectors in Three Dimensions 11.3 Dot Products 11.4 Cross Products 11.5 Lines and Curves in Space 11.6 Calculus of Vector-Valued Functions 11.7 Motion in Space 11.8 Lengths of Curves 11.9 Curvature and Normal Vectors Chapter Eleven Review 209 209 217 227 235 243 251 257 273 279 289 12 Functions of Several Variables 12.1 Planes and Surfaces 12.2 Graphs and Level Curves 12.3 Limits and Continuity 12.4 Partial Derivatives 12.5 The Chain Rule 12.6 Directional Derivatives and the Gradient 12.7 Tangent Planes and Linear Approximation 303 303 324 335 340 348 355 366 Contents 12.8 Maximum/Minimum Problems 372 12.9 Lagrange Multipliers 381 Chapter Twelve Review 390 13 Multiple Integration 13.1 Double Integrals over Rectangular Regions 13.2 Double Integrals over General Regions 13.3 Double Integrals in Polar Coordinates 13.4 Triple Integrals 13.5 Triple Integrals in Cylindrical and Spherical Coordinates 13.6 Integrals for Mass Calculations 13.7 Change of Variables in Multiple Integrals Chapter Thirteen Review 407 407 413 432 446 455 463 471 482 14 Vector Calculus 14.1 Vector Fields 14.2 Line Integrals 14.3 Conservative Vector Fields 14.4 Green’s Theorem 14.5 Divergence and Curl 14.6 Surface Integrals 14.7 Stokes’ Theorem 14.8 The Divergence Theorem Chapter Fourteen Review 493 493 504 511 516 526 534 543 549 558 Copyright c 2015 Pearson Education, Inc Chapter Sequences and Infinite Series 8.1 An Overview 8.1.1 A sequence is an ordered list of numbers a1 , a2 , a3 , , often written {a1 , a2 , } or {an } For example, the natural numbers {1, 2, 3, } are a sequence where an = n for every n 8.1.2 a1 = 1 = 1; a2 = 12 ; a3 = 13 ; a4 = 14 ; a5 = 15 8.1.3 a1 = (given); a2 = · a1 = 1; a3 = · a2 = 2; a4 = · a3 = 6; a5 = · a4 = 24 8.1.4 A finite sum is the sum of a finite number of items, for example the sum of a finite number of terms of a sequence 8.1.5 An infinite series is an infinite sum of numbers Thus if {an } is a sequence, then a1 +a2 +· · · = ∞ ∞ is an infinite series For example, if ak = k1 , then k=1 ak = k=1 k1 is an infinite series k=1 k = + = 3; S3 = k=1 k = + = 5; S3 = 8.1.6 S1 = k=1 k = 1; S2 = + + + = 10 8.1.7 S1 = k=1 k = 1; S2 = + + + 16 = 30 8.1.8 S1 = 1 1 + + + 8.1.9 a1 = 1 k=1 k = 1 25 = 12 k=1 k = 1; S2 = = 1 + k=1 k=1 = 32 ; S3 = k = + + = 6; S4 = k = + + = 14; S4 = k=1 k = 1 + + = 11 ; S4 = ∞ k=1 k=1 k=1 ak k = k2 = k=1 k = 1 1 ; a2 = ; a3 = ; a4 = 10 100 1000 10000 8.1.10 a1 = 3(1) + = a2 = 3(2) + = 7, a3 = 3(3) + = 10, a4 = 3(4) + = 13 8.1.11 a1 = −1 , a2 = 22 = 14 a3 = −2 23 = −1 , a4 = 24 = 16 8.1.12 a1 = − = a2 = + = 3, a3 = − = 1, a4 = + = 8.1.13 a1 = 22 2+1 8.1.14 a1 = + = 43 a2 = 1 23 22 +1 = 2; a2 = + = 85 a3 = 24 23 +1 = 52 ; a3 = + = 16 = a4 = 10 ; 25 24 +1 = a4 = + 32 17 = 17 8.1.15 a1 = 1+sin(π/2) = 2; a2 = 1+sin(2π/2) = 1+sin π = 1; a3 = 1+sin(3π/2) = 0; a4 = 1+sin(4π/2) = + sin 2π = 8.1.16 a1 = · 12 − · + = 0; a2 = · 22 − · + = 3; a3 = · 32 − · + = 10; a4 = · 42 − · + = 21 Chapter Sequences and Infinite Series 8.1.17 a1 = 2, a2 = · = 4, a3 = 2(4) = 8, a4 = · = 16 8.1.18 a1 = 32, a2 = 32/2 = 16, a3 = 16/2 = 8, a4 = 8/2 = 8.1.19 a1 = 10 (given); a2 = · a1 − 12 = 30 − 12 = 18; a3 = · a2 − 12 = 54 − 12 = 42; a4 = · a3 − 12 = 126 − 12 = 114 8.1.20 a1 = (given); a2 = a21 − = 0; a3 = a22 − = −1; a4 = a23 − = 8.1.21 a1 = (given); a2 = · a21 + + = 2; a3 = · a22 + + = 15; a4 = · a23 + + = 679 8.1.22 a0 = (given); a1 = (given); a2 = a1 + a0 = 2; a3 = a2 + a1 = 3; a4 = a3 + a2 = 8.1.24 8.1.23 a 1 32 , 64 a −6, b a1 = 1; an+1 = c an = an b a1 = 1; an+1 = (−1)n (|an | + 1) 2n−1 c an = (−1)n+1 n 8.1.26 8.1.25 a −5, a 14, 17 b a1 = −5, an+1 = −an b a1 = 2; an+1 = an + c an = (−1) · n c an = −1 + 3n 8.1.27 8.1.28 a 32, 64 a 36, 49 b a1 = 1; an+1 = 2an √ b a1 = 1; an+1 = ( an + 1)2 c an = 2n−1 c an = n2 8.1.29 8.1.30 a 243, 729 a 2, b a1 = 1; an+1 = 3an b a1 = 64; an+1 = c an = 3n−1 c an = 64 2n−1 an = 27−n 8.1.31 a1 = 9, a2 = 99, a3 = 999, a4 = 9999 This sequence diverges, because the terms get larger without bound 8.1.32 a1 = 2, a2 = 17, a3 = 82, a4 = 257 This sequence diverges, because the terms get larger without bound 8.1.33 a1 = 10 , a2 = 100 , a3 = 1000 , a4 = 10,000 This sequence converges to zero 8.1.34 a1 = 10 , a2 = 100 , a3 = 1000 , a4 = 10,000 This sequence converges to zero 8.1.35 a1 = − 12 , a2 = 14 , a3 = − 18 , a4 = 16 This sequence converges to because each term is smaller in absolute value than the preceding term and they get arbitrarily close to zero 8.1.36 a1 = 0.9, a2 = 0.99, a3 = 0.999, a4 = 9999 This sequence converges to Copyright c 2015 Pearson Education, Inc 8.1 An Overview 8.1.37 a1 = + = 2, a2 = + = 2, a3 = 2, a4 = This constant sequence converges to 9.99 9.999 8.1.38 a1 = + 10 = 9.9, a2 = + 9.9 10 = 9.99, a3 = + 10 = 9.999, a4 = + 10 = 9.9999 This sequence converges to 10 8.1.39 a1 = 50 11 +50 ≈ 54.545, a2 = This sequence converges to 55 54.545 11 +50 ≈ 54.959, a3 = ≈ 54.996, a4 = 54.959 11 +50 54.996 11 +50 ≈ 55.000 8.1.40 a1 = − = −1 a2 = −10 − = −11, a3 = −110 − = −111, a4 = −1110 − = −1111 This sequence diverges 8.1.41 n 4 10 an 0.4636 0.2450 0.1244 0.0624 0.0312 0.0156 0.0078 0.0039 0.0020 0.0010 This sequence appears to converge to 8.1.42 n 10 an 3.1396 3.1406 3.1409 3.1411 3.1412 3.1413 3.1413 3.1413 3.1414 3.1414 This sequence appears to converge to π 8.1.43 n 10 an 12 20 30 42 56 72 90 This sequence appears to diverge 8.1.44 n 10 an 9.9 9.95 9.9667 9.975 9.98 9.9833 9.9857 9.9875 9.9889 9.99 This sequence appears to converge to 10 8.1.45 n 10 an 0.83333 0.96154 0.99206 0.99840 0.99968 0.99994 0.99999 1.0000 1.0000 1.0000 This sequence appears to converge to 8.1.46 n 10 11 an 0.9589 0.9896 0.9974 0.9993 0.9998 1.000 1.000 1.0000 1.000 1.000 1.000 This sequence converges to 8.1.47 a 2.5, 2.25, 2.125, 2.0625 b The limit is 8.1.48 a 1.33333, 1.125, 1.06667, 1.04167 b The limit is Copyright c 2015 Pearson Education, Inc Chapter Sequences and Infinite Series 8.1.49 n 10 an 3.500 3.750 3.875 3.938 3.969 3.984 3.992 3.996 3.998 3.999 This sequence converges to 8.1.50 n an −2.75 −3.688 −3.922 −3.981 −3.995 −3.999 −4.000 −4.000 −4.000 This sequence converges to −4 8.1.51 n 10 an 15 31 63 127 255 511 1023 This sequence diverges 8.1.52 n an 10 3.4 3.34 This sequence converges to 10 3.334 3.333 3.333 3.333 3.333 3.333 3.333 10 8.1.53 n an 1000 18.811 5.1686 4.1367 4.0169 4.0021 4.0003 4.0000 4.0000 4.0000 This sequence converges to 8.1.54 n 10 an 1.4212 1.5538 1.5981 1.6119 1.6161 1.6174 1.6179 1.6180 1.6180 1.6180 This sequence converges to √ 1+ ≈ 1.618 8.1.56 8.1.55 a 20, 10, 5, 2.5 a 10, 9, 8.1, 7.29 b hn = 20(0.5)n b hn = 10(0.9)n 8.1.58 8.1.57 a 30, 7.5, 1.875, 0.46875 n b hn = 30(0.25) a 20, 15, 11.25, 8.438 b hn = 20(0.75)n 8.1.59 S1 = 0.3, S2 = 0.33, S3 = 0.333, S4 = 0.3333 It appears that the infinite series has a value of 0.3333 = 13 8.1.60 S1 = 0.6, S2 = 0.66, S3 = 0.666, S4 = 0.6666 It appears that the infinite series has a value of 0.6666 = 23 Copyright c 2015 Pearson Education, Inc 8.1 An Overview 8.1.61 S1 = 4, S2 = 4.9, S3 = 4.99, S4 = 4.999 The infinite series has a value of 4.999 · · · = 8.1.62 S1 = 1, S2 = = 1.5, S3 = = 1.75, S4 = 15 = 1.875 The infinite series has a value of 8.1.63 a S1 = 23 , S2 = 45 , S3 = 67 , S4 = 89 b It appears that Sn = 2n 2n+1 c The series has a value of (the partial sums converge to 1) 8.1.64 a S1 = 12 , S2 = 34 , S3 = 78 , S4 = b Sn = − 15 16 2n c The partial sums converge to 1, so that is the value of the series 8.1.65 a S1 = 13 , S2 = 25 , S3 = 37 , S4 = 49 b Sn = n 2n+1 c The partial sums converge to 12 , which is the value of the series 8.1.66 a S1 = 23 , S2 = 89 , S3 = b Sn = − 26 27 , S4 = 80 81 3n c The partial sums converge to 1, which is the value of the series 8.1.67 a True For example, S2 = + = 3, and S4 = a1 + a2 + a3 + a4 = + + + = 10 b False For example, 12 , 34 , 78 , · · · where an = − previous one 2n converges to 1, but each term is greater than the c True In order for the partial sums to converge, they must get closer and closer together In order for this to happen, the difference between successive partial sums, which is just the value of an , must approach zero 8.1.68 The height at the nth bounce is given by the recurrence hn = r · hn−1 ; an explicit form for this sequence is hn = h0 · rn The distance traveled by the ball between the nth and the (n + 1)st bounce is thus n 2hn = 2h0 · rn , so that Sn+1 = i=0 2h0 · ri a Here h0 = 20, r = 0.5, so S1 = 40, S2 = 40 + 40 · 0.5 = 60, S3 = S2 + 40 · (0.5)2 = 70, S4 = S3 + 40 · (0.5)3 = 75, S5 = S4 + 40 · (0.5)4 = 77.5 b n an 40 60 70 75 77.5 78.75 n 10 11 12 an 79.375 79.688 79.844 79.922 79.961 79.980 n 13 14 15 16 17 18 an 79.990 79.995 79.998 79.999 79.999 80.000 n 19 20 21 22 23 24 an 80.000 80.000 80.000 80.000 80.000 80.000 The sequence converges to 80 Copyright c 2015 Pearson Education, Inc Chapter Sequences and Infinite Series 8.1.69 Using the work from the previous problem: a Here h0 = 20, r = 0.75, so S1 = 40, S2 = 40 + 40 · 0.75 = 70, S3 = S2 + 40 · (0.75)2 = 92.5, S4 = S3 + 40 · (0.75)3 = 109.375, S5 = S4 + 40 · (0.75)4 = 122.03125 b n an 40 70 92.5 109.375 122.031 131.523 n 10 11 12 an 138.643 143.982 147.986 150.990 153.242 154.932 n 13 14 15 16 17 18 an 156.199 157.149 157.862 158.396 158.797 159.098 n 19 20 21 22 23 24 an 159.323 159.493 159.619 159.715 159.786 159.839 The sequence converges to 160 8.1.71 8.1.70 a s1 = −1, s2 = 0, s3 = −1, s4 = a 0.9, 0.99, 0.999, 9999 b The limit does not exist b The limit is 8.1.72 8.1.73 a 1.5, 3.75, 7.125, 12.1875 a b The limit does not exist b The limit is 1/2 13 40 , , 27 , 81 8.1.75 8.1.74 a 1, 3, 6, 10 a −1, 0, −1, b The limit does not exist b The limit does not exist 8.1.76 a −1, 1, −2, b The limit does not exist 8.1.77 a 10 = 0.3, 33 100 = 0.33, 333 1000 = 0.333, 3333 10000 = 0.3333 b The limit is 1/3 8.1.78 a p0 = 250, p1 = 250 · 1.03 = 258, p2 = 250 · 1.032 = 265, p3 = 250 · 1.033 = 273, p4 = 250 · 1.034 = 281 b The initial population is 250, so that p0 = 250 Then pn = 250 · (1.03)n , because the population increases by percent each month c pn+1 = pn · 1.03 d The population increases without bound Copyright c 2015 Pearson Education, Inc 554 Chapter 14 Vector Calculus c The left-hand side is the flux across the boundary of D, while the right-hand side is the sum of the charge densities at each point of D The statement says that the flux across the boundary, up to multiplication by a constant, is the sum of the charge densities in the region d By the Divergence theorem, and using part (c), E · n dS = q (x, y, z) dV = D S ∇ · EdV D q(x,y,z) and because this holds for all regions D, we conclude that ∇ · E = e ∇2 ϕ = ∇ · ∇ϕ = ∇ · E = q(x,y,z) 0 14.8.40 a By Exercise 36, the flux of F across a sphere of radius a is 4πGM a3−p = 4πGM b Since the outward flux across a sphere, from part (a), is independent of the radius of the sphere, the outward flux across the spheres of radii a and b are equal, so their difference, which is the net flux across the spherical shell bounded by them, is zero c The left hand side is the flux across the boundary of D, while the right-hand side is the sum of the mass density inside D The statement says that the flux across the boundary is determined by (is a constant multiple of) the sum of the mass density inside D d By the Divergence theorem, and using part (c), 4πG F · n dS = ρ (x, y, z) dV = D S ∇ · FdV D and because this holds over all regions D, we have ∇ · F = 4πGρ (x, y, z) e ∇2 ϕ = ∇ · ∇ϕ = ∇ · F = 4πGρ (x, y, z) 14.8.41 F = −∇T = −1, −2, −1 , so that ∇ · F = and the heat flux is zero 14.8.42 F = −∇T = −2x, −2y, −2z , so that ∇ · F = −6, and the heat flux is −6 times the volume of the region, or −6 14.8.43 F = −∇T = 0, 0, e−z ; then ∇ · F = −e−z The heat flux is then 1 0 −e−z dx dy dz = e−1 − 14.8.44 From Exercise 42, ∇ · F = −6, so the heat flux is −6 times the volume of the sphere, or −8π 14.8.45 F= −∇T = 200xe−x −y −z , 200ye−x ∇ · F = 200e−x so that 2π π 0 −y −z −y −z a ∇ · F dV = 200 D e−r , 200ze−x −y −z Then − 2x2 − 2y − 2z 2 − 2r2 r2 sin u dr du dv = 800πa3 e−a 14.8.46 a By Exercise 36, the net flux across a sphere of radius a centered at the origin is 4πa3−p , which is independent of a only if p = Copyright c 2015 Pearson Education, Inc 14.8 The Divergence Theorem 555 3−p |r|p , b In the general case, we have ∇ · F = b ϕ2 ∇ · F dV = a D so 3−p r sin u du dv dr rp θ2 ϕ1 θ1 = a3−p − b3−p (ϕ1 cos θ1 − ϕ2 cos θ1 − ϕ1 cos θ2 + ϕ2 cos θ2 ) = a3−p − b3−p (ϕ1 − ϕ2 ) (cos θ1 − cos θ2 ) , and these are in general zero only if − p = 14.8.47 a ϕx (x, y, z) = G (ρ) ρx = G (ρ) · √ = G (ρ) xρ , so that ∇ϕ = F = G (ρ) ρr x x2 +y +z x y z, z,1 b The normal to the sphere of radius a is F · n = G (a) x2 z y2 z + , so on that sphere (where ρ = a) +z a = G (a) a2 −z z +z a = G (a) , z a and then the surface integral over the upper hemisphere is 2π a √ F · n dS = aG (a) S r dθ dr = 2πa2 G (a) , a2 − r so the total surface integral is twice that, or 4πa2 G (a) c By the Chain Rule, ∂ x x y2 + z2 G (ρ) = G (ρ) ρx + G (ρ) ∂x ρ ρ ρ3 so that (noting that ρx = xρ ) ∇ · F = G (ρ) x2 + y + z ρ2 + G (ρ) x2 + y + z 2 G (ρ) = G (ρ) + ρ3 ρ d By the Divergence theorem, the flux is also given by a 2π π ∇ · F dV = ρ2 sin u G (ρ) + D 0 a π G (ρ) ρ dv du dρ sin u ρ2 G (ρ) + 2ρ G (ρ) du dρ = 2π 0 a u=π u=0 (− cos u) ρ2 G (ρ) + 2ρ G (ρ) = 2π dρ a ρ2 G (ρ) + 2ρ G (ρ) dρ = 4π It remains to evaluate this integral Using integration by parts, we have a ρ2 G (ρ) + 2ρ G (ρ) dρ = 4π ρ2 G (ρ) 4π a ρ=a ρ=0 − 2ρG (ρ) dρ a = 4π a2 G (a) − 2ρ G (ρ) dρ and the remaining integrals cancel, giving 4πa2 G (a) as the final result Copyright c 2015 Pearson Education, Inc 556 Chapter 14 Vector Calculus 14.8.48 a Rearrange the given equation and integrate over D to get u ∇ · F dS = D ∇ · (u F) dS − D F · ∇u dS D u F · n dS where S is the By the Divergence theorem, the first term on the right side is equal to S boundary of D The result follows b If you set F = f (x) , 0, andu (x, y, z) = g (x), then ∇ · F = f (x) u∇ · F = f (x) g (x) uF = f (x) g (x) , 0, ∇u = g (x) , 0, ∇ · (uF) = (f g) (x) F · ∇u = f (x) g (x) so that (f g) (x) dV − f (x) g (x) dV = D f (x) g (x) dV = f (x) g (x) − D D f (x) g (x) dV, D which is the usual rule for integration by parts c One approach is to set u = and F = x2 z , x2 y , y z Gauss’ formula then gives x2 y + y z + z x = D x2 z , x2 y , y z dS S The integral on the right is computed by integrating over each face of the cube; on faces where x is constant, the normal is either 1, 0, or −1, 0, depending on whether x is or 0; similarly for y and z Considering x first, when x = 0, this surface integral becomes on that face (the integrand is x2 z at x = 0), while for x = it becomes z In that case, the integral is 13 This holds for each dimension, so the total integral on the right side is · 13 = 1, and thus the integral on the left is 12 14.8.49 Suppose F= f, g wheref = f (x, y), g = g (x, y), and supposeu = u (x, y) Then ∇ · F = fx + gy , and ∇u = ux , uy Then we have for this case u ∇ · F dS = D u (fx + gy ) dA R uF · n dS = uF · n ds C S F · ∇u dS = D (f ux + g uy ) dA, R and the result follows Setting u = then gives (fx + gy ) dA = R C F · n ds, which is the flux form of Green’s Theorem 14.8.50 Apply the Divergence Theorem to the vector field F = u∇u By the product rule (Thm 14.11), we have ∇ · (v∇v) = ∇u · ∇v + u∇2 v, so the Divergence theorem says that u∇2 v + ∇u · ∇v dV = D (∇ · (v∇v)) dV = D Copyright c 2015 Pearson Education, Inc u∇v · n dS S 14.8 The Divergence Theorem 557 14.8.51 From Exercise 50, we have both u∇2 v + ∇u · ∇v dV = D u∇v · n dS S v∇2 u + ∇v · ∇u dV = D v∇u · n dS, S where the second formula is obtained by switching u and v in the first formula Subtracting the second from the first, and using the fact that the dot product is commutative and integrals are linear, we have the desired result: u∇2 v − v∇2 u dV = (u∇v − v∇u) · n dS D S 14.8.52 A computation shows that ∇ϕ = −p r |r|p+2 Thus the potential field ∇ϕ = for p = 0, so certainly ∇ ϕ = as well Otherwise, for p = 1, we have ∇ϕ = −r ; |r|3 this is an inverse square field, and we have seen many times (e.g Exercise 36(b)) that the divergence of such a field is zero Thus ∇2 ϕ = for p = as well 14.8.53 The Divergence theorem applied to the field∇ϕ says that ∇2 ϕ dV = D ∇ϕ · n dS S and if ϕ is harmonic, the left side is zero 14.8.54 Apply Green’s First Identity (Exercise 49) to u and u to give u∇2 u + ∇u · ∇u dV = D u∇u · n dS S Because ∇2 u = and ∇u · ∇u = |∇u| , the result follows 14.8.55 If T is a vector field t, u, v , then by T, we mean t, u, v a Let F = f, g, h and suppose n = n1 , n2 , n3 Then n × F = n2 h − n3 g, n3 f − n1 h, n1 g − n2 f ∂h ∂g ∂f ∂h ∂g ∂f − , − , − ∇×F= ∂y ∂z ∂z ∂x ∂x ∂y Considering first the i component of these vectors, note that for the vector field F1 = 0, h, −g , the Divergence theorem says that (n2 h − n3 g) dS = S 0, h, −g · n1 , n2 , n3 dS = S S ∂h ∂g − ∂y ∂z = F1 · n dS = (∇ · F1 ) dR D dA D and similarly for the second and third components b Similarly to part (a), note that n × ∇ϕ = n × ϕx , ϕy , ϕz = n2 ϕz − n3 ϕy , n3 ϕx − n1 ϕz , n1 ϕy − n2 ϕx Copyright c 2015 Pearson Education, Inc 558 Chapter 14 Vector Calculus Looking first at the x component of this vector, we have n2 ϕz − n3 ϕy = 0, ϕz , −ϕy · n1 , n2 , n3 = (∇ × ϕ, 0, ) · n1 , n2 , n3 so that Stokes’ theorem says that, writing F = ϕ, 0, , (∇ × ϕ, 0, ) · n1 , n2 , n3 dS = S (∇ × F) · n dS = F · dr = C S ϕ, 0, · dr = C ϕ dr C and similarly for the second and third components Chapter Fourteen Review a False The curl is ∂ ∂x (x) − ∂ ∂y (−y) = b True The curl of a conservative vector field is zero c False For example, −y, x and 0, 2x both have curl d False For example, x, 0, and 0, y, both have divergence ∇ · F dV = e True By the Divergence theorem, the integral is equal to D dV D a Choice F, because this is a radial vector field whose magnitude increases with distance from the origin b Choice E, because this is a rotational field c Choice D, because this is also a radial field, but the magnitude is constant d Choice C, because this field is vertical along the line y = x e Choice B, because the magnitude decreases rapidly away from the origin f Choice A, because this field is periodic ∇ϕ = 2x, 8y y 6 2 x Copyright c 2015 Pearson Education, Inc Chapter Fourteen Review 559 ∇ϕ = x, −y y 2 1 x ∇ϕ = − |r|r ∇ϕ = −e−x −y −z x, y, z a x, y is an outward normal; for (x, y) on the circle, | x, y | = normal is 12 x, y b The normal component is y, −x · c The normal component is x2 +y 2 x, y · x2 + y = 2, so the unit outward x, y = x, y = x2 +y · · x2 + y = 12 We have | r (t)| = 5, so the line integral is π x2 − 2xy + y ds = C Here| r (t)| = π 25 cos2 t − 10 cos t sin t + 25 sin2 t dt = 25 (5 − sin t cos t) dt = 125π √ √ + + 36 = 46, so √ ds = 46 ln √ 46 54 (ln 2)2 e −1 C √ √ √ 10 Parameterize the line by −3t, + 6t, − 3t for ≤ t ≤ Then| r (t)| = + 36 + = 54 = 6, so ye −xz 3te √ xz − y ds = 6t2 dt = 3t (3t − 2) − (6t + 1) √ dt = −57 C 11 For the first parameterization we have| r (t)| = 2, so 2π (x − 2y − 3z) ds = (2 cos t − sin t) dt = 0 C For the second parameterization we have | r (t)| = √ (x − 2y − 3z) ds = C 16t2 sin2 (t2 ) + 16t2 cos2 (t2 ) = 4t, so 2π 8t cos t2 − 16t sin t2 dt = 0 12 a Parameterize the path from P to Q by − t, t, for ≤ t ≤ 1; then the work done is 1 dr = 1, 2t, · −1, 1, dt = (2t − 1) dt = Copyright c 2015 Pearson Education, Inc C 1, 2y, −4z · 560 Chapter 14 Vector Calculus b Parameterize the path from P to O by − t, 0, , and the path from O to Q by 0, t, , both for ≤ t ≤ Then the work done is C 1, 2y, −4z · dr = ( 1, 0, · −1, 0, + 1, 2t, · 0, 1, ) dt = (2t − 1) dt = 0 c Parameterize the quarter circle by cos t, sin t, for ≤ t ≤ dr = π/2 1, sin t, · − sin t, cos t, dt = π/2 π 2; then the work done is C 1, 2y, −4z · (2 cos t sin t − sin t) dt = d Yes, the work is independent of the path; this is a conservative vector field with potential function x + y − 2z 13 Parameterize the first path by r1 (t) = 0, t, , and the second by r1 (t) = 0, 1, 4t , both for ≤ t ≤ Then r1 (t) = 0, 1, and r2 (t) = 0, 0, −t, 0, · 0, 1, + −1, 4t, · 0, 0, F · dr = dt = 0 C 14 r (t) = 2t, 6t, −2t , so F·dr = C 3/2 (11t4 ) t2 , 3t2 , −t2 · 2t, 6t, −2t dt = 11−3/2 t−6 ·22t3 dt = 11−3/2 1 22t−3 dt = 3√ 11 44 15 The circulation is 2π 2π F · Tds = C −4 sin2 t + sin t cos t dt = −4π sin t − cos t, sin t · −2 sin t, cos t dt = The outward flux is 2π F · n ds = ((2 sin t − cos t) (2 cos t) − (2 sin t) (−2 sin t)) dt C 2π sin t cos t − cos2 t + sin2 t dt = =4 16 The circulation is 2π F · Tds = cos t, sin t · −2 sin t, cos t dt = 0 C The outward flux is 2π F · n ds = ((2 cos t) (2 cos t) − (2 sin t) (−2 sin t)) dt = 8π C 17 The circulation is F · Tds = C 2π cos t, sin t · −2 sin t, cos t dt = 0 The outward flux is F · n ds = C 2π ((2 cos t) (2 cos t) − (2 sin t) (−2 sin t)) dt = 2π 18 The circulation is 2π F · Tds = 2π cos t − sin t, cos t · −2 sin t, cos t dt = C (− sin t cos t + 1) dt = 4π The outward flux is 2π F · n ds = C 2π ((2 cos t − sin t) (2 cos t) − (2 cos t) (−2 sin t)) dt = cos2 t dt = 4π Copyright c 2015 Pearson Education, Inc Chapter Fourteen Review 561 19 The normal to the plane x = is 1, 0, , so the flux is 1/2 F · n ds = C −1/2 L −L v0 L2 − y dy dz = v0 L3 20 A potential function is xy 21 A potential function is xy + yz 22 A potential function is ex cos y 23 A potential function is xyez 24 a F = 2xy, x2 , so = − 5t2 C F · dr = 2 − t2 t, − t2 · −2t, dt = −4t2 − t2 + − t2 dt − t2 dt = b Because F = ∇ϕ, where ϕ (x, y) = x2 y, we have C F · dr = ϕ (3) − ϕ (0) = − = 25 a F = yz, xz, xy , so π F · dr = C π = t t sin t, cos t, sin t cos t · − sin t, cos t, dt π π π t cos2 t − sin2 t + sin t cos t dt = π π b F = ∇ (xyz) = ∇ϕ, so F · dr = ϕ (cos π sin π) − ϕ cos sin · C π = − = 26 a Parameterize C by the four paths r1 (t) = −1 + 2t, −1 , r2 (t) = 1, −1 + 2t , r3 (t) = − 2t, , r4 (t) = −1, − 2t , for ≤ t ≤ Then r1 (t) = 2, , r2 (t) = 0, , r3 (t) = −2, , r4 (t) = 0, −2 , and F · dr C −1 + 2t, · 2, + 1, − 2t · 0, + − 2t, −1 · −2, + −1, 2t − · 0, −2 = dt 1 (4t − + − 4t + 4t − + − 4t) dt = = 0 dt = 0 b F = ∇ϕ, where ϕ (x, y) = x2 − y , so the integral around any closed curve is zero 27 a C F · dr = 2π sin t, 4, − cos t · − sin t, cos t, dt = b The vector field is not conservative, since for example 28 For p = 2, F = ∇ϕ where ϕ = −1 , (p−2)|r|p−2 2π ∂ ∂y − sin2 t + cos t dt = −π (y) = while for p = 2, ϕ = ∂ ∂z (x) ln |r| gradient Thus F is conservative on all of R for p < Copyright c 2015 Pearson Education, Inc , as can be seen by taking the 562 Chapter 14 Vector Calculus 29 By the circulation form of Green’s Theorem, ∂ ∂ x2 y − xy ∂x ∂y xy dx + x2 y dy = C (2xy − 2xy) dA = dA = R R 30 By the circulation form of Green’s Theorem, ∂ ∂ x − y 2/3 − −3y + x3/2 ∂x ∂y −3y + x3/2 dx + x − y 2/3 dy = C dA = 4π dA = R R 31 By the circulation form of Green’s Theorem, ∂ ∂ x3 + xy − 2y − 2x2 y ∂x ∂y x3 + xy dy + 2y − 2x2 y dx = C dA R 3x2 + y − 4y + 2x2 dA = = R 32 By the flux form of Green’s Theorem, C 3x3 dy − 3y dx = 1 −1 −1 5x2 − 3y dy dx = 9x2 + 9y dA = R 2π 0 20 r3 dθ dr = 1152π Because the orientation is clockwise, the answer is −1152π 33 The ellipse is of the region is x2 16 + y2 = 1; parameterize it by r(t) = x, y = cos t, sin t , ≤ t ≤ 2π Then the area 2π ((4 cos t) (2 cos t) − (2 sin t) (−4 sin t)) dt = cos2 t + sin2 t dt = 8π C 34 dx = −3 cos2 t sin t dt, anddy = sin2 t cos t dt, so the area of the hypocycloid is x dy − y dx = C = 2π sin2 t cos t − sin3 t cos3 t −3 cos2 t sin t dt 2π cos4 t sin2 t + sin4 t cos2 t dt = 3π 35 a F = x2 + y −1/2 x, y , so the circulation is C y ∂ ∂x F · dr = x2 R −xy = x2 + y R + − y2 ∂ ∂y −xy − x2 + y x x2 + y2 dA dA = b The flux is C x ∂ ∂x F · n ds = x2 R + y2 = R (x2 + y ) 3/2 y2 + + y ∂ ∂y x2 x2 (x2 + y ) 3/2 dA + y2 dA = R x2 + y Copyright c 2015 Pearson Education, Inc π dr dθ = 2π dA = Chapter Fourteen Review 563 36 a The circulation is b The flux is C C F · dr = ∂ ∂x R F · n ds = R ∂ ∂x (x cos y) − (− sin y) + ∂ ∂y ∂ ∂y (− sin y) dA = (x cos y) dA = π/2 π/2 0 π/2 π/2 0 cos y dy dx = π (−x sin y) dy dx = − 18 π 37 ∂ ∂y a For F to be conservative, we must have ∂ ∂x b For F to be source-free, we must have (ax + by) = ∂ ∂x (cx + dy), or b = c ∂ (ax + by) = − ∂y (cx + dy), or a = −d c F is both conservative and source-free if b = c and a = −d, i.e if F = ax + by, bx − ay 38 The divergence is ∂ ∂x (yz) + ∂ ∂y (xz) + ∂ ∂z (xy) = The curl is ∂ ∂ ∂ ∂ ∂ ∂ (xy) − (xz) , (yz) − (xy) , (xz) − (yz) = ∂y ∂z ∂z ∂x ∂x ∂y The field is both source-free and irrotational 39 The divergence is |r| The curl is ∂ ∂ ∂ ∂ ∂ ∂ (z |r|) − (y |r|) , (x |r|) − (z |r|) , (y |r|) − (x |r|) = ∂y ∂z ∂z ∂x ∂x ∂y The field is irrotational but not source-free ∂ ∂ ∂ (sin xy) + ∂y (cos yz) + ∂z (sin xz) = y cos xy − z sin yz + x cos xz The curl is 40 The divergence is ∂x y sin yz, −z cos xz, −x cos xy The field is neither irrotational nor source-free 41 The divergence is ∂ ∂x 2xy + z + ∂ ∂y x2 + ∂ ∂z 4xz = 2y + 12xz The curl is ∂ ∂ ∂ ∂ ∂ ∂ 4xz − x2 , 2xy + z − 4xz , x2 − 2xy + z ∂y ∂z ∂z ∂x ∂x ∂y = 0, so the field is irrotational but not source-free 42 |r| = x2 + y + z , so ∇ x2 + y + z Now −2 = −4x x2 + y + z ∂ −4x x2 + y + z ∂x −3 −3 , −4y x2 + y + z = −4 y + z − 5x2 −3 , −4z x2 + y + z x2 + y + z −4 −3 =− 4r |r| , so that −4 ∇ · ∇ |r| = −4 x2 + y + z −4 y + z − 5x2 + x2 + z − 5y + x2 + y − 5z = 12 |r| 43 a The curl is ∂ ∂ ∂ ∂ ∂ ∂ (−y) − (x) , (z) − (−y) , (x) − (z) = −1, 1, ∂y ∂z ∂z ∂x ∂x ∂y So the scalar component in the direction of 1, 0, is −1, 1, · 1, 0, = −1, and the scalar component in the direction of 0, − √12 , √12 is 0, − √12 , √12 · 1, 0, = Copyright c 2015 Pearson Education, Inc 564 Chapter 14 Vector Calculus b The scalar component of the curl is a maximum in the direction of the curl, i.e in the direction −1, 1, , whose unit vector is √13 −1, 1, 44 The curl of the vector field is ∇ × F = 0, 0, , and the component of the curl along a unit vector n is thus 0, 0, · n a It does not spin, because 0, 0, · 1, 0, = b The scalar component of the curl in the direction 0, 0, is c It spins the fastest when the paddle wheel is aligned with 0, 0, 45 Parameterize the sphere by sin u cos v, sin u sin v, cos u , ≤ u ≤ so 2π dS = ≤ v ≤ 2π Then |n| = sin u, π/2 sin u dA = S π 2; sin u du dv = 18π R 46 Parameterize the surface by v cos u, v sin u, v for ≤ v ≤ 4, ≤ u ≤ 2π Then √ dS = S 47 The volume element is v dA = √ 2π R √ v du dv = 12π √ √ 12 + 12 + 12 = 3, so the area is dS = √ dA = S R zx2 + zy2 + = 48 The volume element is √ −1 −1 (x2 + y ) + 1, so the integral is S r R 49 The volume element for z = − x − y is (1 + yz) dS = 2π (x2 + y ) + dA = dS = √ S √ dx dy = 2r2 + dθ dr = 26 π √ 3, so the integral is (1 + yz) dA = √ R 2−x √ (1 + y (2 − x − y)) dy dx = 50 The normal to the curved surface of the cylinder at (x, y, z) is 0, y, z , so 0, y, z · n dS = a S y + z dA = a2 · area of R = 32πa3 R 51 Parameterize the curved surface using spherical coordinates, so that |n| = sin u; then for the curved surface we have 2π π/2 (x − y + z) dS = (sin u cos v − sin u sin v + cos u) sin u du dv = 8π S For the planar surface, n = 0, −1, so that |n| = and (x − y + z) dS = S 2π (x − y + z) dA = R r (cos θ − sin θ) dθ dr = 0 and the total integral is thus 8π Copyright c 2015 Pearson Education, Inc Chapter Fourteen Review 565 52 The normal to the cylinder is x, y, with magnitude 1, so F · n dS = S R 53 F = x2 + y + z −1/2 x2 + y dA = R dA = 32π R x, y, z ; using spherical coordinates to parameterize the sphere gives a sin u cos v, a sin u sin v, a cos u · a2 sin2 u cos v, a2 sin2 u sin v, a2 sin u cos u dA a F · n dS = S x, y, z · x, y, dA = R 2π π a2 sin u dA = = a2 sin u du dv = 4πa2 R 54 a Using the explicit description, we have zz2 + zy2 + = 2π (x2 + y ) + dA = dS = S (x2 + y ) + 1, so 4r2 + dθ dr = r R √ π 17 17 − √ b Using the given parametric description, we have |n| = v + 4v , so 2π dS = S v + 4v dA = R + 4v dv dθ = v √ π 17 17 − c Using the given parametric description, we have √ √ 1 − v sin u, v cos u, × − v −1/2 cos u, v −1/2 sin u, 2 |tu × tv | = so that dS = S √ 4v + dA = R 2π √ 4v + dv dθ = = 1√ 4v + √ π 17 17 − 55 a The base of S is the surface where z = 0, or the circle x2 + y = a2 Similarly, the base of the paraboloid is found by setting z = 0; simplifying gives again x2 + y = a2 The high point of the hemisphere (maximum z-coordinate) occurs when x = y = 0; then z = a Similarly, the high point on the paraboloid also occurs when x = y = and again this gives z = a b The graph of the paraboloid is inside that of the hemisphere everywhere, so we would expect it to have smaller surface area We know that the surface area of the hemisphere is 4πa2 · 12 = 2πa2 For the 4(x2 +y ) a2 2y paraboloid, we have zx = − 2x a , zy = − a , so that |n| = 1 dS = a S (x2 R + y2 ) + a2 dA = a 2π a r + 1, so 4r2 + which is in fact smaller than the area of the hemisphere Copyright c 2015 Pearson Education, Inc a2 √ 5−1 πa2 , dθ dr = 566 Chapter 14 Vector Calculus c n = a2 sin2 u cos v, a2 sin2 u sin v, a2 cos u sin u , so F · n dS = S a sin u cos v, a sin u sin v, a cos u · a2 sin2 u cos v, a2 sin2 u sin v, a2 cos u sin u dA R 2π π/2 sin3 u cos2 v + sin3 u sin2 v + cos2 u sin u du dv = a3 0 2π 2π π/2 π/2 sin3 u + cos2 u sin u du dv = a3 = a3 0 sin u du dv = 2πa3 d For the paraboloid, the parameterization is v cos u, v sin u, a − n = −v sin u, v cos u, × cos u, sin u, − 2v a so that the outward-pointing normal is S , ≤ v ≤ a, ≤u≤ 2π, and 1 2v = − 2v cos u, − 2v sin u, −v , a a a cos u, 2va sin u, v and v cos u, v sin u, a − F · n dS = v2 a v2 2v 2v · cos u, sin u, v dA a a a R 2π a v3 v + av − a a = 0 du dv = 3 πa 56 (a cos u sin v)2 a2 a For the given r, we have cos2 v = (b sin u sin v)2 b2 + = cos2 u sin2 v + sin2 u sin2 v + cos2 v = sin2 v + b The normal vector is determined by n = tu × tv = −bc cos u sin2 v, −ac sin u sin 2v, −ab sin v cos v , so that the outward pointing normal is the negative of this vector Then 2π π bc cos u sin2 v, ac sin u sin2 v, ab sin v cos v dS = S dv du 2π π b2 c2 cos2 u sin4 v + a2 c2 sin2 u sin4 v + a2 b2 sin2 v cos2 v dv du = 0 57 Parameterize x2 + y = for z = using r(t) = cos t, sin t, for0 ≤ t ≤ 2π 2π F · dr = C 2π F · r (t) dt = 0, 0, cos t sin t · −2 sin t, cos t, dt = 0 58 F = u2 − v , u, 2v (6 − 2u − v) , r (u, v) = u, v, − 2u − v , tu × tv = 2, 1, √ 6−2u 39 2 2u − 4v + 12v − 4uv dv du = F · dr = C 0 59 The boundary of this region is in the xy-plane, found by setting z = 0, so it is x2 + y = 99, the circle √ of radius 99 about the origin Parameterize the circle in the usual way; then (∇ × F) · n dS = F · dr = C S 0, √ 99 cos t, √ √ √ 99 sin t · − 99 sin t, 99 cos t, dt C 2π 99 cos2 t dt = 99π = Copyright c 2015 Pearson Education, Inc Chapter Fourteen Review 567 60 The boundary of this region is the circle x2 + z = for y = 0; parameterizing it in the usual way as cos t, 0, sin t gives 2π (∇ × F) · n dS = F · dr = cos2 t − sin2 t, 0, sin t cos t · −2 sin t, 0, cos t dt C S 2π 2π sin2 t − cos2 t sin t + sin t cos2 t dt = =8 sin3 t dt = 0 61 By Stokes’ theorem, the circulation around a closed curve C can be found by choosing a surface S of which C is the boundary; then F · dr = (∇ × F) · n dS C S But for F = ∇ 10 − x + y + z , ∇ × F = 0, so the right-hand side is zero 2 62 We have ∇ · F = −3 so that F · n dS = S ∇ · F dV = −3 · volume of cube = −3 D 63 ∇ · F = x + y + z , so 2 F · n dS = 2π π r2 · r2 sin u du dv dr = x2 + y + z dV = S D 0 972 π 64 ∇ · F = (x + y + z), so F · n dS = S 2π (x + y + z) dV = 2r (r cos θ + r sin θ + t) dθ dr dt = 256π D 0 65 ∇ · F = x2 + y , so the outward flux across the boundary S of a hemisphere D of radius a is F · n dS = S x2 + y dV = D a 2π 0 π/2 3r2 sin2 u · r2 sin u du dv dr = across the region bounded by the hemispheres of radii and is 45 π (32 − 1) = 5 πa , so that the net flux 124 π 66 ∇·F = 0, so the flux is zero across any surface that bounds a region where F is defined and differentiable; the given region does not include zero, so is one of these Thus the net outward flux is zero 67 Using the Divergence theorem, ∇ · F = 2x + sin y + 2y − sin y + 2z + sin y = (x + y + z), so that F · n dS = S 1−x (x + y + z) dV = (x + y + z) dy dx dz = D 0 32 68 a The normal vectors point outwards everywhere on S; that is, on the curved surface, they point upwards, and on the flat surface they point in the direction of negative x b Parameterize C by two paths: r1 (t) = a cos t, a sin t, for − π2 ≤ t ≤ ≤ t ≤ a Then r1 (t) = −a sin t, a cos t, and r2 (t) = 0, −2, So π and r2 (t) = 0, a − 2t, for π/2 F · dr = C −π/2 −a sin t, a cos t, a sin t − 2a cos t · −a sin t, a cos t, dt a π/2 2t − a, 0, a − 2t · 0, −2, dt = + a2 dt = πa2 −π/2 Copyright c 2015 Pearson Education, Inc 568 Chapter 14 Vector Calculus c ∇ × F = 2, 4, For the curved portion of S, using spherical coordinates, the normal vector is a2 sin2 u cos v, a2 sin2 u sin v, a2 cos u sin u , and for the flat portion, the normal vector is −1, 0, 2a2 sin2 u cos v + 4a2 sin2 u sin v + 2a2 cos u sin u dS + (∇ × F) · n dS = (−2) dS = Then 2a2 S π/2 π/2 −π/2 S1 S2 sin2 u cos v + sin2 u sin v + sin u cos u du dv − πa2 = 2πa2 − πa2 = πa2 Copyright c 2015 Pearson Education, Inc ... 1/n and bn = en , then lim an bn = ∞ n→∞ c True The definition of the limit of a sequence involves only the behavior of the nth term of a sequence as n gets large (see the Definition of Limit of. .. an+a+1 , and thus the limit of the sequence a(a+1) 1 −√ Then the second term of an cancels with the first term of an+2 , so the n+1 n+3 1 − √n+3 and thus the sum of the series is the limit of Sn... } as n → ∞ 8.2.8 The definition of the limit of a sequence involves only the behavior of the nth term of a sequence as n gets large (see the Definition of Limit of a Sequence) Thus suppose an ,