ODD14 - C H A P T E R 14 Vector Analysis Section 14.1...

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Unformatted text preview: C H A P T E R 14 Vector Analysis Section 14.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 190 Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 193 Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 198 Section 14.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 202 Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 208 Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 211 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 CHAPTER Vector Analysis Section 14.1 14 Vector Fields Solutions to Odd-Numbered Exercises 1. All vectors are parallel to y-axis. Matches (c) 3. All vectors point outward. Matches (b) 5. Vectors are parallel to x-axis for y n. Matches (a) 9. F x, y F y 7. F x, y F i 2 j xi x2 yj y2 c 11. F x, y, z F 3y z 4 3y j c x2 y2 5 c2 y 1 −4 x x x 2 4 y −5 −4 −5 3 5 13. F x, y F x2 c 2 16 4x i 16x2 y c2 y yj y2 1 c 15. F x, y, z F 3 z i j k 17. 2 1 y 2 4 4 y −2 −1 −1 −2 x 1 2 2 −4 4 −2 −1 x 1 2 −4 x −2 19. 2 1 z 21. f x, y fx x, y fy x, y 1 2 y 5x2 10x 3x 10x 3xy 3y 20y 3y i 10y2 23. f x, y, z fx x, y, z fy x, y, z z ye x 2 2 x ye x ex 1 2 2 1 2 x F x, y 3x 20y j fz F x, y, z 2 x ye x i 2 ex j 2 k 178 S ection 14.1 25. g x, y, z gx x, y, z gy x, y, z gz x, y, z G x, y, z x y ln x y ln x x ln x 0 xy x 12xy i y 6 x2 6 x2 y ln x yi xy x y x ln x yj y y y xy x xy x y y Vector Fields 179 27. F x, y yj y have continuous first 29. F x, y sin y i x cos y j x cos y have continuous first partial M 12 xy and N partial derivatives. N x 12x M sin y and N derivatives. N x cos y M ⇒ F is conservative. y M ⇒ F is conservative. y 31. M N x 15y3, N 5y2 5xy2 M y 45y2 ⇒ Not conservative 33. M N x 2 2x y e ,N y 2x 2 x e y2 y y 2 y 2x 2x e y3 M ⇒ Conservative y x x2 x y x2 y2 y x x2 y2 x2 x2 y2 y x2 2 xy y2 2 xy y2 y2 2 35. F x, y y x 2 xy x2 2 xy i 2x 2x x2 j 37. F x, y y x 2 xye x x 2e x 2y xe x y 2y i 2y 2 xj 2y 39. F x, y 2 x 3 ye x 2y i j 2 xe x 2 xe x 2y 2 x 3 ye x 2y 2 Conservative fx x, y fy x, y f x, y 2 xy x2 x2y K Conservative fx x, y fy x, y f x, y 2 xye x x 2e x ex 2y 2y 2y Conservative fx x, y K fy x, y f x, y x x2 y x2 y2 y2 K 1 ln x2 2 y2 41. F x, y y x e x cos y i sin y j 43. F x, y, z x yz i i j yj k z z k z k, 1, 2, 1 e x cos y e x sin y e x sin y curl F e x sin y xy xyz y 2j xyj xz k Not conservative 45. F x, y, z e x sin y i i curl F x x sin y e e x cos y j, 0, 0, 3 j y x cos y e 2k k z 0 2e x cos y k curl F 1, 2, 1 curl F 0, 0, 3 180 Chapter 14 Vector Analysis 47. F x, y, z arctan i x i y ln x 2 j y y2 j k z y2 1 k curl F x x arctan y x x2 y2 1 x y2 xy 2 k 2x x2 y2 k 1 ln x 2 2 yi sin y j y 49. F x, y, z sin x i zj k z sin z xk curl F sin x x y sin y cos y x zi cos z xj cos x yk z sin z 51. F x, y, z sin y i i x cos y j j k k 53. F x, y, z ez y i i j xj k xyk curl F x sin y y z x cos y 1 2 cos yk 0 curl F x y z yez xe z xye z 0 Not conservative Conservative fx x, y, z fy x, y, z fz x, y, z f x, y, z ye z xe z x ye z x ye z K 55. F x, y, z i curl F 1 i y j x j y2 k 2z 1k 57. F x, y div F x, y 6x 2 i x 6x 2 xy2j y 2 xy x y2 x 1 y y x y2 z 2z 1 0 12 x Conservative fx x, y, z fy x, y, z fz x, y, z f x, y, z f x, y, z f x, y, z z2 f x, y, z x y 2z 1 y x y2 1 x y x y g y, z h x, z K1 K2 1 dx y x dy y2 2z z z2 1 dz p x, y z K K3 S ection 14.1 59. F x, y, z div F x, y, z sin x i x sin x cos y j y yj 1 1 z2 k cos y z z2 cos x sin y 2z Vector Fields 181 61. F x, y, z div F x, y, z div F 1, 2, 1 x yz i yz 4 zk yz 2 63. F x, y, z div F x, y, z div F 0, 0, 3 e x sin y i e x sin y 0 e x cos y j e x sin y 65. See the definition, page 1008. Examples include velocity fields, gravitational fields and magnetic fields. 69. F x, y, z G x, y, z F i G1 x i xi 2x j yj 3y k zk 2 xz i curl F G 2 xz 71. F x, y, z x yz i i curl F x xyz j y y i curl curl F x 0 x yz i i curl F x xyz x Nj M j y y x yj k z z j y xy yj k z z 0 P k and G Qi i curl F G M y P y curl F x Q P N S N i z curl G N Qi Rj j y R z N P x P Rj P k z S Ri M j z x P S xyj xzk xyj xz k x 3y 2 67. See the definition on page 1014. j k 2x 3y yz 3y 2 i j y 3xy z z 3x y j k z y 2x 2 y 2x2 k 1 1i 4x 2x j 3y 6y k 6xj 3y k zk 73. F x, y, z G x, y, z F G i xi 2x j yj 3y k zk ij k 1 2x 3y x yz 2 xz 3y 2 i 2z 3x z 3x y j y 2x2 k k z xz zk zj yk div F G 75. F x, y, z div curl F 77. Let F F Mi G S k where M, N, P, Q, R, and S have continuous partial derivatives. Sk z M S y Qj R i z x N S x R Q j z y M R x Qk Q k y N x M k y 182 Chapter 14 Vector Analysis 79. Let F Mi G Nj P k and G x M R Ri y N Sj S T k. z P T M x M x div F R x N y div G N y P z S y R x P z S y T z T z div F 81. F Mi f Nj Pk F curl curl curl f f F F curl F F (Exercise 77) (Exercise 78) 83. Let F Mi Nj x P k, then f F fM y fN fMi z f Nj fP f f f Pk. M x M x M f x N y f F f N y N z N f y f M x f P z f N y P f z f P z div f F f div F In Exercises 85 and 87, F x, y, z 1 ln x 2 2 x x2 y2 xi yj z k and f x, y, z F x, y, z x2 y2 z 2. 85. ln f ln f y2 z2 i z2 x2 y y2 z2 j x2 z y2 z2 k xi x2 yj y2 zk z2 F f2 87. f n fn n x2 x2 y2 y2 n z2 n z2 x2 z2 n 1 x y2 n 2 2 n x y2 1 z x2 2 i z y2 n x2 k F y2 z2 n 1 x 2 y y2 z2 j z2 xi z2 2 n x2 y2 yj zk nfn 89. The winds are stronger over Phoenix. Although the winds over both cities are northeasterly, they are more towards the east over Atlanta. Section 14.2 Line Integrals 183 Section 14.2 Line Integrals t i, 0≤t≤ 3i t 3 j, 3 ≤ t ≤ 9 t i 3 j, 6 ≤ t ≤ 12 t j, 9≤t≤ 3 6 9 12 1. x2 x2 9 cos2 t y2 y 9 2 9 1 1 x2 9 y2 9 3 cos t 3 sin t 3 cos t i 3 sin t j 3. r t sin2 t cos2 t sin2 t x y rt 0≤t≤2 ti 2 4ti x C 5. r t t j, ti 2 0≤t≤1 t j, 1 ≤ t ≤ 2 4i 3 2 7. r t 3t j, 0 ≤ t ≤ 2; r t 2 3j 2 y ds 0 4t 3t 4 2 dt 0 5t dt 5t 2 2 2 10 0 9. r t x2 C sin t i y2 cos t j z2 ds 8 t k, 0 ≤ t ≤ 2 2 ;rt cos t i 64 t 2 sin t j 2 8k sin t 2 sin2 t 0 2 cos2 t 64t 2 d t cos t 64 dt 8 2 3 3 65 1 0 65 t 64 t 3 3 2 65 0 65 6 3 16 2 11. r t x2 C t i, 0 ≤ t ≤ 3 3 y y 2 ds 0 3 t2 t 2 dt 0 02 2 1 0 dt 1 x 1 2 3 13 t 3 3 −1 9 0 13. r t x2 C cos t i y2 ds sin t j, 0 ≤ t ≤ 2 y 2 1 cos 2 t 0 2 sin2 t sin t 2 cos t 2 d t x dt 0 2 1 184 15. r t Chapter 14 ti x C Vector Analysis y t j, 0 ≤ t ≤ 1 1 4 y ds 0 t t2 2 4t 83 t 3 1 1 2 0 1 dt 19 2 6 1 (1, 1) 2 x 1 17. r t t i, 2 3 x ti t j, t 1 0≤t≤1 1 j, 1 ≤ t ≤ 2 2≤t≤3 t dt 0 2 y (0, 1) C2 4 y ds 4 y ds 1 C1 1 2 t t2 2 t dt 8 3 4t 8 t 3 1 1 1 2 32 1 C3 x C2 2 2 2t 3 1 dt 19 2 6 3 32 2 C1 (1, 0) x x C3 4 y ds 2 43 1 2 19 2 6 8 3 3 19 t 8 3 2 6 x C 4 y ds 19 2 6 19 1 19. x, y, z rt rt rt Mass C 12 x 2 3 cos t i 3 sin t i y2 z2 2 t k, 0 ≤ t ≤ 4 2k 2 3 sin t j 3 cos t j 2 3 sin t 3 cos t 4 2 2 13 3 sin t 2 x, y, z d s 0 1 3 cos t 2 4 2 2t 2 13 d t 4t 3 3 4 0 13 2 2 13 3 21. F x, y C: r t Ft rt xyi 4t i 4t 2 i 4i F C 9 0 4t 2 dt 64 2 13 9t 2 4973.8 27 yj t j, 0 ≤ t ≤ 1 tj j 1 23. F x, y C: r t Ft rt 16 t 2 t dt C 3x i 4yi 2 sin t j, 0 ≤ t ≤ 8 sin t j 2 cos t j 2 2 cos t i 6 cos t i 2 sin t i F dr 0 2 dr 0 12 sin t cos t 2 16 sin t cos t d t 16 3 t 3 12 t 2 1 0 35 6 2 sin2 t 0 2 S ection 14.2 25. F x, y, z C: r t Ft rt t4i i F C Line Integrals 185 x2yi ti t 2tj 1 x zj x yz k 27. F x, y, z rt Ft dr ti t2 i F x2z i t2j ln t i 2tj dr 1 6yj yz2k t2 j 2j 2 k, 0 ≤ t ≤ 1 2 t 3k ln t k, 1 ≤ t ≤ 3 6t 2 j t 2 ln 2 t k 1 k dt t 3 dr 0 t4 t 5 5 2t t 2t 3 3 2 dt 1 t 2 ln t 249.49 12 t 3 t ln t 2 dt 2t 2 0 17 15 C 29. F x, y C: y rt rt Ft F r x3 xi 2y j from 0, 0 to 2, 8 t 3 j, 0 ≤ t ≤ 2 3t 2 j ti t F C ti i 2t 3 j 6t 5 2 Work dr 0 t 6t5 dt 12 t 2 2 t6 0 66 31. F x, y 2xi yj C: counterclockwise around the triangle whose vertices are 0, 0 , 1, 0 , 1, 1 rt On C1: t i, i 3 t 1 j, ti 3 0≤t≤1 1≤t≤2 t j, 2 ≤ t ≤ 3 i 1 Ft Work 2 t i, r t F C1 dr 0 2t dt j t 1 1 On C2: Ft Work 2i t F C2 1 j, r t 2 dr 3 3 1 dt 1 2 i t 3 j t dt 3 2 On C3: Ft Work 23 ti F dr t j, r t 23 2 C3 Total work C F dr 1 1 2 3 2 0 33. F x, y, z C: r t rt Ft F Work C xi 2 cos t i yj 5z k 2 sin t j t k, 0 ≤ t ≤ 2 k 5t k 35. r t F dr F 5t dt 10 2 C 3 sin t i 150k 3 cos t i 2 3 cos t j 10 t k, 0 ≤ t ≤ 2 2 2 sin t i 2 cos t i 5t 2 2 cos t j 2 sin t j 3 sin t j 1500 dt 2 10 k dt 2 1500 t 2 2 r F dr 0 1500 ft 0 lb dr 0 186 Chapter 14 x2 i r1 t r1 t Ft F C1 Vector Analysis xyj 2t i 2i 4t i 3 2 37. F x, y (a) t j 2t t 1 j, 1 ≤ t ≤ 3 (b) r2 t r2 t 23 2i 43 2 ti j t 2i 83 t 2 t j, 0 ≤ t ≤ 2 1j 1 dt 236 3 F C2 Ft dr 23 2 t2 23 tj t2 t dt dr 1 8t2 2t t 0 Both paths join 2, 0 and 6, 2 . The integrals are negatives of each other because the orientations are different. 236 3 39. F x, y C: r t rt Ft F r yi ti i xj 2tj 2j 2ti 2t tj 2t 0. 0 41. F x, y C: r t rt Ft F r F C x3 ti i t3 t3 dr 2x 2 i t 2j 2tj 2t 2 i 2t 2 0. t x y j 2 t2 j 2 t2 2 0 Thus, C F dr 2t t Thus, 43. x 2t, y x C 10 t, 0 ≤ t ≤ 1 ⇒ y 10 5x or x y2 10 y , 0 ≤ y ≤ 10 5 10 3y 2 d y 0 y 5 3y 2 dy y3 0 1010 45. x 2t, y xy dx C 10 t, 0 ≤ t ≤ 1 ⇒ x 10 y , 0 ≤ y ≤ 10, dx 5 y3 75 y2 2 10 0 1 dy 5 190 OR 3 y dy 0 y2 25 y dy y 5x, d y xy dx C 5 d x, 0 ≤ x ≤ 2 2 y dy 0 5x 2 25x d x 5x 3 3 25x 2 2 2 0 190 3 y 3 47. r t xt dx t i, 0 ≤ t ≤ 5 t, y t d t, 2x C 0 0 5 2 1 x 1 2 3 4 5 dy y dx x 3y d y 0 2t dt 25 −1 −2 S ection 14.2 t i, 3i xt dx 2x C1 Line Integrals 187 49. r t C1: t t, y t d t, dy 0≤t≤3 3 j, 3 ≤ t ≤ 6 0, 0 3 y dx 3, y t 0, dy t dt x 3 3y d y 0 2t dt 9 C2: x t dx 2x C2 6 y dx y dx x x 3y d y 3 3 9 45 2 3t 63 2 3 dt 3t 2 2 6 6t 3 45 2 2x C y 3y d y (3, 3) 3 2 C2 1 C1 x 1 2 3 51. x t t, y t 2x C 1 y dx t2, 0 ≤ t ≤ 1, dx 1 dt, dy 2t 1 t2 t2 4t 2t dt t 1 dt 3 3t2 3t 4 2 2 t dt t3 3 1 x 3y dy 0 1 6t3 0 2t2 t 0 11 6 53. x t dx t, y t d t, dy 2x C 2 t 2, 0 ≤ t ≤ 2 4t dt 2 y dx x 3y d y 0 2 2t 24t3 0 2t2 dt 2t2 t 2t dt 6t 2 4 t d t 6t 4 23 t 3 2 t2 0 316 3 55. f x, y h 57. f x, y C: x 2 xy y2 rt rt 1 from 1, 0 to 0, 1 cos t i sin t i 1 sin t j, 0 ≤ t ≤ cos t j 2 C: line from 0, 0 to 3, 4 r rt rt 3t i 3i 5 4 t j, 0 ≤ t ≤ 1 4j Lateral surface area: 1 rt 5h d t 0 C Lateral surface area: 5h f x, y d s 0 2 f x, y d s C cos t sin t d t sin2 t 2 2 0 1 2 188 Chapter 14 h 1 rt rt rt Vector Analysis 59. f x, y C: y x 2 from 1, 0 to 0, 1 1 i 1 ti 21 41 1 tj t 2 1 t 2 j, 0 ≤ t ≤ 1 Lateral surface area: 1 f x, y ds C 0 h1 h 21 4 h 25 4 41 t ln 2 1 t 2 dt 1 41 5 t 2 ln 2 1 t 1 41 t 2 0 1.4789h 61. f x, y C: y 1 xy x 2 from 1, 0 to 0, 1 cos t, then: You could parameterize the curve C as in Exercises 59 and 60. Alternatively, let x y 1 rt rt rt cos2 t sin2 t sin2 t j, 0 ≤ t ≤ 2 sin t cos t j 4 sin2 t cos2 t sin t 1 4 cos2 t 2 cos t i sin t i sin2 t Lateral surface area: 2 2 f x, y ds C 0 cos t sin2 t sin t 1 1 4 cos2 t 12 4 cos2 t dt 0 sin2 t 1 4 cos2 t 12 sin t cos t dt 1 4 cos2 t 3 2. Let u sin2 t and dv f x, y ds C sin t cos t, then du 2 32 0 2 sin t cos t dt and v 2 1 12 1 sin2 t 1 12 1 sin2 t 1 12 1 12 1 120 4 cos2 t 4 cos2 t 1 5 120 1 6 1 0 4 cos2 t 2 52 0 32 sin t cos t dt 32 1 1 120 4 cos2 t 11 52 1 25 5 120 0.3742 63. (a) f x, y rt rt rt S C 1 y2 2 sin t j, 0 ≤ t ≤ 2 2 cos t j (c) 5 4 z 2 cos t i 2 sin t i 2 2 −3 f x, y ds 0 1 2 4 sin2 t 2 d t x 3 3 y 2t (b) 0.2 12 4t 12 5 sin t cos t 0 12 37.70 cm2 7.54 cm3 S ection 14.2 Line Integrals 189 65. S 25 60 50 40 30 20 10 3 x z Matches b 3 y 67. (a) Graph of: r t z 3 cos t i 3 sin t j 1 sin2 2 t k 0 ≤ t ≤ 2 3 2 1 3 4 x 3 4 y (b) Consider the portion of the surface in the first quadrant. The curve z 3 sin t j, 0 ≤ t ≤ 2. Hence, the total lateral surface area is 2 1 sin2 2t is over the curve r1 t 3 cos t i 4 C f x, y ds 4 0 1 sin2 2 t 3 d t 12 3 4 9 sq. cm 4y3 2 (c) The cross sections parallel to the xz-plane are rectangles of height 1 3 1 y 2 9 and base 2 9 y 2. Hence, Volume 2 0 29 y2 1 4 y2 1 9 y2 9 dy 42.412 cm3 69. See the definition of Line Integral, page 1020. See Theorem 14.4. 71. The greater the height of the surface over the curve, the greater the lateral surface area. Hence, z 3 < z1 < z 2 < z 4 . y 4 3 2 1 x 1 2 3 4 73. False 1 75. False, the orientations are different. xy ds C 2 0 t 2 dt 190 Chapter 14 Vector Analysis Section 14.3 1. F x, y (a) r1 t r1 t Ft F C Conservative Vector Fields and Independence of Path xy j t 2 j, 0 ≤ t ≤ 1 2t j t3j 1 x2i ti i t2i dr (b) r2 r2 Ft 11 15 C sin i cos i sin dr 0 2 sin2 j, 0 ≤ 2 sin cos j sin3 j 2 ≤ 2 i t2 0 2t 4 dt F sin2 cos sin3 3 2 sin5 5 2 sin4 cos 2 0 d 11 15 3. F x, y (a) r1 r1 F F C yi xj tan j, 0 ≤ sec2 j ≤ 3 sec i sec tan i dr 0 tan i sec j 3 3 sec 3 tan2 sec3 ln sec d 0 sec 3 sec2 ln 2 1 3 sec3 d 1.317 sec d 0 tan 0 (b) r2 t r2 t Ft F C t 1 2t ti 1i 1 i t j, 1 j 2t 1j t 2t 1 t 3 0 0≤t≤3 t 3 dr 0 t1 dt 2t 1 12 2 1 2 dt 3 0 1 tt 1 ln 2 t 1 dt 1 2 t2 1.317 3 0 t2 3 t 1 14 14 dt 1 2 14 ln 1 2 1 2 t 0 1 7 ln 2 2 23 1 ln 7 2 43 5. F x, y N x Since e x sin y i e x cos y N x ex cos y j M y e x cos y 7. F x, y N x Since 1 y2 N x 1 i y x j y2 M y 1 y2 M , F is conservative. y M , F is not conservative. y 9. F x, y, z curl F y2z i 2 xyz j xy 2k 0 ⇒ F is conservative. Section 14.3 11. F x, y (a) r1 t r1 t Ft F C Conservative Vector Fields and Independence of Path 191 2 xy i ti i 2t 3 i dr x2j t 2 j, 0 ≤ t ≤ 1 2t j t2j 1 (b) r2 t r2 t Ft ti i 2t 4 i dr t 3 j, 0 ≤ t ≤ 1 3t 2 j t2j 1 4t 3 dt 0 1 C F 5t 4 d t 0 1 13. F x, y (a) r1 t r1 t Ft F C yi ti i ti dr xj t j, j tj 0 C 0≤t≤1 (b) r2 t r2 t Ft F ti i t2i dr t 2 j, 2t j tj 1 0≤t≤1 (c) r3 t r3 t Ft ti i t3i dr t 3 j, 3t 2 j tj 1 0≤t≤1 t 2 dt 0 1 3 F C 2t 3 dt 0 1 2 15. C y2 dx 2 xy d y xy2 k. Therefore, we Since M y N x 2y, F x, y y 2 i 2 xy j is conservative. The potential function is f x, y can use the Fundamental Theorem of Line Integrals. 4, 4 1, 0 (a) C y2 dx 2 xy d y x2 y 0, 0 64 y2 dx C (b) C y2 dx 2 xy d y x2 y 1, 0 0 (c) and (d) Since C is a closed curve, 2 xy d y 0. 17. C 2x y dx Since M F x, y y x2 N 2 xy i y2 dy x x2 2x, y 2 j is conservative. x2y x2 y y3 3 y3 3 y3 3 k. 19. F x, y, z yz i xz j xyk Since curl F 0, F x, y, z is conservative. The potential function is f x, y, z x yz k. (a) r1 t ti F 2j dr tj t k, 0 ≤ t ≤ 4 4, 2, 4 The potential function is f x, y (a) C x yz 0, 2, 0 32 0, 4 5, 0 0, 4 2, 0 2x y dx x2 y2 dy 64 3 64 3 C (b) r 2 t t2i F C t 2 k, 0 ≤ t ≤ 2 4, 2, 4 (b) C 2x y dx x2 y2 dy x2 y dr x yz (0, 0, 0 32 21. F x, y, z 2y xi x2 zj 2y 4z k F x, y, z is not conservative. (a) r1 t r1 t Ft F C ti i 2t dr 2 t 2j 2t j ti 1 k, 0 ≤ t ≤ 1 t2 2t 3 2t 2 1j 2t 2 t dt 2 3 4k 0 —CONTINUED— 192 Chapter 14 Vector Analysis 21. —CONTINUED— (b) r2 t r2 t Ft ti i 3t i F C tj j t2 2t 4 2t 2t 1 1 2 k, 0 ≤ t ≤ 1 1k 1 t2 2 j 2t 2t 1 2 4 2t 1 2 k 1 16 2t 3 dr 0 1 3t 17t 2 0 8t 2t 2 1 3 dt 5t 2 2 2t 6 1 3 1 4 0 5t 2t 1 16 2t 1 dt 17t 3 3 2 2t 1 17 6 3, 8 23. F x, y, z ez y i xj xyk 25. C yi xj dr xy 0, 0 24 F x, y, z is conservative. The potential function is f x, y, z x ye z k . (a) r1 t 4 cos t i F C 4 sin t j x ye z 4, 0, 3 3k , 0 ≤ t ≤ 0 4, 0, 3 dr 4 8t i dr (b) r2 t F C 3k, 0 ≤ t ≤ 1 4, 0, 3 x ye z 4, 0, 3 0 3 2, 2 27. C cos x sin y d x sin x cos y d y sin x sin y 0, 1 2 ,0 29. C e x sin y d x e x cos y d y e x sin y 0, 0 0 31. C y 2z d x x 3z d y 2x 3y d z xy 3yz 2 xz . F x, y, z is conservative and the potential function is f x, y, z 1, 1, 1 (a) xy 3 yz 2 xz 0, 0, 0 0, 0, 1 0 0 0 1, 1, 1 (b) xy 3yz 2 xz 0, 0, 0 1, 0, 0 xy 3 yz 2 xz 0, 0, 1 1, 1, 0 0 0 0 1, 1, 1 (c) xy 3yz 2 xz 0, 0, 0 xy 3yz 2xz 1, 0, 0 xy 3yz 2 xz 1, 1, 0 0 1 1 0 2, 3, 4 33. C sin x d x z dy y dz cos x yz 0, 0, 0 12 1 11 35. F x, y Work 9x 2 y 2 i 3x 3 y 2 y 6x3y 5, 9 1 j is conservative. 30,366 0, 0 S ection 14.4 37. r t rt at Ft W C Green’s Theorem 193 2 cos 2 t i 2 sin 2 t j 4 cos 2 t j 8 2 4 sin 2 t i 8 m 2 cos 2 t i 1 at 32 sin 2 t j 2 at F 4 2 cos 2 t i sin 2 t j sin 2 t j 4 sin 2 t i cos 2 t j d t 3 C dr C 4 cos 2 t i 0 dt 0 39. Since the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing at a rate of 10 units per minute, then the potential energy is increasing at a rate of 10 units per minute. 41. No. The force field is conservative. 45. (a) The direct path along the line segment joining 4, 0 to 4, 4 and then to 3, 4 . 43. See Theorem 14.5, page 1033. 4, 0 to 3, 4 requires less work than the path going from (b) The closed curve given by the line segments joining 4, 0 , 4, 4 , 3, 4 , and 4, 0 satisfies C F dr 0. 47. False, it would be true if F were conservative. 51. Let F Then Mi M y 2f x2 49. True Nj f y f i y 2f f j. x y2 and M y N x N . x x f x 2f y 2f y2 x2 . Since 0 we have Thus, F is conservative. Therefore, by Theorem 14.7, we have f dx y f dy x M dx C N dy C F dr 0 C for every closed curve in the plane. Section 14.4 t i, 4i 12 16 y dx C 2 2 Green’s Theorem 0≤t≤4 4≤t≤8 8 ≤ t ≤ 12 12 ≤ t ≤ 16 8 y 1. r t t 4 j, t i 4 j, t j, 4 (4, 4) 4 3 2 x dy 0 0 dt 12 t0 4 2 t dt 0 0 4 0 4 t 2 0 0 16 d t 16 1 x 1 2 3 4 16 8 12 2 16 12 t 2 0 0 dt 0 By Green’s Theorem, R 64 64 N x M dA y 4 4 2x 0 2y dy dx 0 8x 16 d x 0. 194 Chapter 14 ti 8 y 2 dx C Vector Analysis 0≤t≤4 4≤t≤8 t4 dt 16 t4 16 t2 t3 dt 2 t dt 2 8 8 y 4 3 3. r t t 2 4 j, ti 8 x2 dy t j, 4 0 4 0 (4, 4) 8 4 t 2 dt 224 5 8 128 3 t 2 dt 2 1 C2 C1 x 1 2 3 4 28 4 t 2 dt 32 15 By Green’s Theorem, N x y2 4 2 sin t, 0 ≤ t ≤ 2 . 2 R M dA y 4 0 x 4 2x x2 4 2y dy dx 0 x2 x3 2 x4 dx 16 32 . 15 5. C: x 2 Let x 2 cos t and y xey d x C ex dy 0 2 cos te2 sin t 2 2 4 4 x2 2 sin t e2 cos t 2 cos t d t 2 19.99 R N x M dA y N x M y ex x2 xey d y d x 2 24 x2 ex xe 4 x2 xe 4 x2 dx 19.99 In Exercises 7 and 9, 1. 2 x y 7. C y x dx 2x y dy 0 2 x2 x dy dx 2 (2, 2) y=x 2x 0 x2 dx 1 4 3 y = x2 − x x 1 2 9. From the accompanying figure, we see that R is the shaded region. Thus, Green’s Theorem yields y C y (− 5, 3) (− 1, 1) (− 1, − 1) 4 2 (5, 3) (1, 1) x 2 4 x dx 2x y dy R 1 dA Area of R 6 10 56. 22 −2 (1, − 1) (− 5, − 3) − 4 (5, − 3) 11. Since the curves y 2 xy d x C 0 and y x y dy 4 x 2 intersect at 1 2x d A 2, 0 and 2, 0 , Green’s Theorem yields 2 4 x2 1 20 2 2x d y d x 4 x2 R y 2 2 2 xy 8x 4x2 8 3 0 dx 2 x3 dx x4 2 32 . 3 2 2 4 2 x2 x3 3 4x 8 3 16 S ection 14.4 Green ’s Theorem 195 13. Since R is the interior of the circle x 2 x2 C y2 2y a 2, Green’s Theorem yields 2y d A x2 a y2 dx 2 xy d y R a a a2 a2 4y dy dx x2 4 a 0 dx 0. 15. Since M y 2x x2 y2 N , x we have path independence and N x M dA y 0. R 17. By Green’s Theorem, sin x cos y d x C xy cos x sin y d y R 1 0 x y x sin x sin y y dy dx 1 2 1 sin x sin y d A x 0 x2 dx 1 x2 22 x3 3 1 0 1 . 12 19. By Green’s Theorem, xy dx C x y dy R 2 0 1 3 x dA 2 1 1 r cos r dr d 0 4 26 cos 3 d 8. 21. F x, y C: x 2 xy i y2 4 x yj 2 2 2 Work C xy dx x y dy R 1 x dA 0 0 1 r cos r dr d 0 2 8 cos 3 d 4 23. F x, y x3 2 3y i 6x 5 yj C: boundary of the triangle with vertices 0, 0 , 5, 0 , 0, 5 Work C x3 2 3y d x 6x 5 y dy R 9 dA 9 1 2 55 225 2 25. C: let x A 1 2 a cos t, y x dy C a sin t, 0 ≤ t ≤ 2 . By Theorem 14.9, we have y dx 1 2 2 a cos t a cos t 0 a sin t a sin t dt 1 2 2 a 2 dt 0 a2 t 2 2 a 2. 0 196 Chapter 14 Vector Analysis y 27. From the accompanying figure we see that C1: y C2: y 2x 4 1, dy x 2, dy 2 dx 2 (1, 3) x 4 2 x d x. −6 −4 −4 −6 Thus, by Theorem 14.9, we have A 1 2 1 2 1 2 1 x2 3 1 2x 1 dx 1 2 1 2 1 dx 3 1 2 x2 3 x 1 2x 4 x2 dx (− 3, − 5) 4 dx 4 dx 1 2 1 3 1 1 1 1 dx 3 x2 3 3 3 x2 dx 1 3x 2 x3 3 1 3 32 . 3 29. See Theorem 14.8, page 1042. 31. Answers will vary. F1 x, y F2 x, y F3 x, y yi x2 i 2xy i xj y2 j x2 j 2 33. A 2 4 1 2A x2 dx 1 2A 4x x3 3 2 2 32 3 3 y y = 4 − x2 x x2 dy C1 x2 dy C2 2 C1 For C1, dy x 1 2 32 3 2x dx and for C2, dy 2 0. Thus, 3 64 x4 2 2 2 1 x x2 2 2 x dx 0. −2 −1 C2 1 2 To calculate y, note that y y x, y 1 2 32 3 0, 8 5 2 0 along C2. Thus, x2 2 dx 3 64 2 4 2 16 2 8x 2 x 4 dx 3 16x 64 8x 3 3 x5 5 2 2 8 . 5 1 35. Since A 0 x x3 dx x2 2 x4 4 1 0 1 1 , we have 4 2A 2. On C1 we have y x 3, dy 3x 2 d x and on C2 we have y x x, dy 2 C 1 d x. Thus, 2 C1 0 x2 dy x4 dx 0 x 2 3x 2 d x x2 dx 1 2 C2 x2 d x 8 15 1 y 6 y 2 2 6 5 2 3 (1, 1) C2 C1 x 1 y2 dx C 1 0 2 0 x6 dx 88 , 15 21 2 1 x2 dx 2 7 2 3 8 . 21 x, y S ection 14.4 Green ’s Theorem 197 37. A 1 2 a2 2 2 a2 1 0 2 cos 2 cos 2 d 1 2 cos 2 2 d a2 3 22 2 sin 1 sin 2 4 2 0 1 0 a2 3 2 3 a2 2 39. In this case the inner loop has domain A 1 2 1 2 4 2 4 2 3 2 ≤ 3 d d ≤ 4 . Thus, 3 1 3 3 4 cos 4 cos 4 cos2 2 cos 2 3 3 1 3 2 4 3 3 4 sin sin 2 2 33 . 2 41. I y dx x2 C x dy y2 y x2 y2 i x x2 N x y2 j. M y x2 x2 y2 . y2 2 (a) Let F F is conservative since F is defined and has continuous first partials everywhere except at the origin. If C is a circle (a closed path) that does not contain the origin, then F C dr C M dx N dy R N x M dA y 0. (b) Let r a cos t i a sin t j, 0 ≤ t ≤ 2 be a circle C1 oriented clockwise inside C (see figure). Introduce line segments C2 and C3 as illustrated in Example 6 of this section in the text. For the region inside C and outside C1, Green’s Theorem applies. Note that since C2 and C3 have opposite orientations, the line integrals over them cancel. Thus, C4 C1 C2 C C3 and F C4 dr F C1 dr C F dr 0. But, 2 F C1 dr 0 2 a sin t a 2 cos 2 t sin2 t a sin t a 2 sin2 t t a cos t a 2 cos2 t 2 a cos t dt a 2 sin2 t cos2 t d t dr 2. 2. 0 0 Finally, C F dr F C1 Note: If C were orientated clockwise, then the answer would have been 2 . y 3 2 C C2 x 4 C1 C3 −2 −3 43. Pentagon: 0, 0 , 2, 0 , 3, 2 , 1, 4 , A 1 2 1, 1 1 4 0 0 19 2 0 0 4 0 12 2 198 Chapter 14 Vector Analysis 45. C yn dx xn dy R N x M dA y y 2a For the line integral, use the two paths C1: r1 x C2: r2 x yn dx C1 a y = a2 − x2 C2 x i, xi xn dy xn dy a≤x≤a a2 0 a2 a a a0 a2 x 2 j, x a to x a −a C1 x a yn dx C2 x2 x2 n2 xn 1 x a2 nyn 1 x2 dx R N x M dA y nxn dy dx (a) For n 1, 3, 5, 7, both integrals give 0. (b) For n even, you obtain n 2: 43 3a n 4: 16 5 15 a n 6: 32 7 35 a n 8: 256 9 315 a (c) If n is odd and 0 < a < 1, then the integral equals 0. 47. C f DN g gD N f d s C f DNg ds C gDN f d s f g dA R f R 2g g 2f g f dA R f 2g g 2f dA 49. F N x F C Mi M y dr Nj 0⇒ M dx C N x N dy M y 0. N x M dA y 0 dA R 0 R Section 14.5 1. r u, v z xy ui vj Parametric Surfaces uvk 3. r u, v x2 y2 2 cos v cos u i z2 4 2 cos v sin u j 2 sin v k Matches c. v k 2 Matches b. 5. r u, v y 2z ui 0 vj 7. r u, v x2 z2 2 cos u i 4 vj 2 sin u k Plane z 3 2 −4 345 5 x y Cylinder z 3 x 5 −3 5 y Section 14.5 For Exercises 9 and 11, r u, v u cos v i u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 . 5 z Parametric Surfaces 199 Eliminating the parameter yields z x2 y 2, 0 ≤ z ≤ 4. 2 x 2 y 9. s u, v z x2 u cos v i y2 u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 The paraboloid is reflected (inverted) through the xy-plane. 11. s u, v u 2 k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 The height of the paraboloid is increased from 4 to 9. u cos v i u sin v j 2u cos v i 2u sin v j u 4 k, z 3 2 1 9 x 2 x 2 y 13. r u, v 15. r u, v 2 sinh u cos v i sinh u sin v j z 9 6 cosh u k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 z x2 16 y2 2 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 z2 1 x2 4 y2 1 1 6 3 6 9 y 17. r u, v 0≤u≤ u sin u cos v i 1 cos u sin v j z 5 4 3 −3 u k, ,0≤v≤2 −2 −3 3 x −2 2 −1 1 2 3 y 19. z y ui vj vk 21. x 2 r u, v y2 16 4 cos u i y2 4 sin u j vk r u, v 23. z x2 25. z ui vj uk 2 4 inside x 2 v cos u i 9. v sin u j 4 k, 0 ≤ v ≤ 3 r u, v r u, v 27. Function: y x ,0≤x≤6 2 u cos v, z 2 u sin v 2 29. Function: x x sin z, 0 ≤ z ≤ sin u sin v, z u Axis of revolution: x-axis x u, y Axis of revolution: z-axis sin u cos v, y 0≤u≤ ,0≤v≤2 0 ≤ u ≤ 6, 0 ≤ v ≤ 2 200 Chapter 14 u i vi Vector Analysis u vj i 1. i i 1 1 y 2z 0 j j 1 1 1 k k 0 1 2z i 1 j 0 2k j v k, 1, k 1, 1 33. r u, v ru u, v rv u, v 2u cos v i 2 cos v i 2 u sin v i 2 and v 4 k , rv 2, rv 2, j 3 0 k 4 0 2 3u sin v j 3 sin v j u 2 k, 0, 6, 4 2u k 31. r u, v ru u, v At 1, ru 0, 1 N j, rv u, v 0 and v j, rv 0, 1 rv 0, 1 1 y 1, 1 , u i 3u cos v j 2. 4i At 0, 6, 4 , u ru 2, N 2 ru 2, i 0 4 3j ru 0, 1 Tangent plane: x x 2 2 16 j 3 3z 12 4 0 12 k (The original plane!) Direction numbers: 0, 4, Tangent plane: 4 y 4y v j 2 v k , 0 ≤ u ≤ 2, 0 ≤ v ≤ 1 2 1 j 2 k 0 1 2 6 3z 35. r u, v ru u, v 2ui 2 i, rv u, v i 2 0 2 2 1 k 2 j k ru ru A rv rv 1 0 j 0 1 2 2 du dv 0 22 37. r u, v ru u, v rv u, v ru ru A 0 0 a cos u i a sin u i k a sin u j a cos u j v k, 0 ≤ u ≤ 2 , 0 ≤ v ≤ b rv rv b 2 i j k a sin u a cos u 0 0 0 1 a a du dv 2 ab a cos u i a sin u j 39. r u, v ru u, v rv u, v ru ru A 0 a u cos v i a cos v i a u sin v i a u sin v j a sin v j k u k, 0 ≤ u ≤ b, 0 ≤ v ≤ 2 a u cos v j a u cos v i a u sin v j a 2u k rv rv 2 b i j k a cos v a sin v 1 au sin v au cos v 0 au 1 a1 0 a2 a2 u du dv a b2 1 a2 S ection 14.5 41. r u, v ru u, v rv u, v u cos v i cos v i 2u u sin v j sin v j 2u k u cos v j j sin v 2u u cos v k 1 0 u cos v i u sin v j 1 k 2 u k, 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 Parametric Surfaces 201 u sin v i i cos v 2u u sin v ru rv ru A rv 2 0 4 u u 0 1 4 1 du dv 4 6 17 17 1 36.177 43. See the definition, page 1051. 45. (a) From 10, 10, 0 (b) From 10, 10, 10 (c) From 0, 10, 0 (d) From 10, 0, 0 47. (a) r u, v 4 4 cos v cos u i cos v sin u j sin v k, (b) r u, v 4 4 2 cos v cos u i 2 cos v sin u j 2 sin v k, 0≤u≤2 ,0≤v≤2 z 4 −6 0 ≤ u ≤ 2 ,0 ≤ v ≤ 2 z −6 4 6 x −4 6 y x 6 6 y (c) r u, v 8 8 cos v cos u i cos v sin u j sin v k, (d) r u, v 8 8 3 cos v cos u i 3 cos v sin u j 3 sin v k, 0≤u≤2 ,0≤v≤2 z 9 0 ≤ u ≤ 2 ,0 ≤ v ≤ 2 z 12 3 3 x −9 y 12 x 12 − 12 y The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution. 202 Chapter 14 Vector Analysis 20 sin u sin v j 20 cos u k 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 49. r u, v ru rv ru rv 20 sin u cos v i 20 cos u cos v i 20 sin u sin v i 20 cos u sin v j 20 sin u cos v j 20 sin u k i j 20 cos u cos v 20 cos u sin v 20 sin u sin v 20 sin u cos v 400 sin2 u cos v i 400 sin2 u cos v i k 20 sin u 0 400 cos u sin u cos2 v cos u sin u k cos2 u sin2 u cos u sin u sin2 v k 400 sin2 u sin v j sin2 u sin v j sin u sin v 4 2 ru rv 400 sin u cos v 400 sin4 u 400 sin2 u 2 3 4 2 cos2 u sin2 u 400 sin u 2 3 S S dS 0 2 0 400 sin u d u d v 0 400 cos u 0 dv 200 d v 0 400 m2 51. r u, v ru u, v rv u, v ru ru A 0 u cos v i cos v i u sin v j sin v j u cos v j 2v k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 u sin v i 2k 2 sin v i 2 cos v j uk rv rv 2 3 i j cos v sin v u sin v u cos v 4 4 0 z 4π k 0 2 u2 u2 d u d v 3 13 4 ln 3 2 13 2π −4 −2 4 x 2 4 y 53. Essay Section 14.6 1. S: z 4 x S Surface Integrals z x 4 1, x z y 1 0 0 1 2 x, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 4 4 2y z dS 0 0 x 4 4 2y 4 0 2 dy dx 2 0 0 2y dy dx Section 14.6 z x 1 Surface Integrals 203 3. S: z 10, x 2 y 2 ≤ 1, z y 1 1 1 0 x2 x S 2y z dS 1 2 0 2 0 x2 x 2y 10 1 0 2 0 2 dy dx r cos 0 2r sin 2 sin 3 10 r dr d 1 cos 3 5d 2 1 sin 3 2 cos 3 5 0 10 5. S: z 6 x 2y, (first octant) 6 3 0 6 x2 z x 1 1, 2 z y 2 2 5 y xy dS S 0 xy 1 xy 2 2 x9 0 3 0 x2 2 dy dx 4 3 2 1 y=3− 2x 6 0 dx 12 x dx 4 x4 16 6 0 1 x −1 1 2 3 4 5 6 6 2 6 3x 6 9x 2 2 2 x3 27 6 2 7. S: z z x 9 x 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ x, 2x, z y 2 0 2 xy dS S 0 y xy 1 4x 2 dx dy 391 17 240 1 9. S: z 10 x2 S x2 y 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 2 2 2xy dS 0 0 x2 2xy 1 4x 2 4y 2 dy dx 11.47 11. S: 2x x, y, z m 3y x2 6z y2 x2 12 (first octant) ⇒ z 2 1 x 3 2 1 y 2 5 4 3 2 1 y y2 2x 3 1 x 2 R 1 3 y 2 2 1 2 y=4− 2 x 3 dA R x 1 2 3 4 5 6 7 6 7 6 6 0 6 4 0 dy dx −1 x2 4 0 2 x 3 1 4 3 2 x 3 3 dx 743 x 63 14 x 6 1 4 8 2 x 3 46 0 364 3 204 Chapter 14 Vector Analysis v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 2 k 2 1 13. S: r u, v ru rv y S ui vj 1 j 2 5 2 v 5 5 du dv 2 65 5 dS 0 0 15. S: r u, v ru rv 2 cos u i 2 cos u i 2 2 sin u j 2 sin u j 2 v k, 0 ≤ u ≤ 2 8 2 ,0≤v≤2 xy dS S 0 0 8 cos u sin u du dv 17. f x, y, z S: z x x2 2, x 2 y2 z2 y2 ≤ 1 1 1 1 2 1 x2 f x, y, z dS S 1 x2 x2 r2 0 2 0 1 y2 x 2 2 2 1 1 2 0 2 dy dx 2 r cos r 2 cos2 r4 cos2 4 11 4 2 r dr d 2 0 2 0 r2 r4 4 9 4 1 8 4r cos 4r 3 cos 3 4 r dr d 1 2 0 2 2r 2 0 d 2 0 cos 2 2 4 cos 3 4 sin 3 2 d 2 0 2 9 4 1 sin 2 2 18 4 4 19 2 4 19. f x, y, z S: z x2 x2 y2 z2 y2 ≤ 4 2 4 4 x2 y 2, x 2 f x, y, z dS S 2 x2 x2 y2 x2 x2 y2 y2 x2 2 1 x2 y2 y2 x x2 dy dx y2 2 y x2 y2 2 dy dx 2 4 4 4 4 2 x2 x2 2 2 2 x2 x2 x2 2 x2 2 y2 y 2 dy dx 2 2 2 0 0 r 2 dr d 2 0 r3 3 2 d 0 16 3 2 0 32 3 Section 14.6 x2 y2 y2 z2 Surface Integrals 205 21. f x, y, z S: x 2 9, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9 9 y2 3 9 y2 dy y y 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9. z2 1 3 Project the solid onto the yz-plane; x 3 9 f x, y, z dS S 0 3 0 0 9 9 y2 z2 3 9 y 9 3 9 y 3 3 2 y2 y2 9z 0 2 dz dy z3 3 0 9 9 0 3 dz dy 2 dy 0 0 324 0 972 arcsin 972 0 2 486 23. F x, y, z S: x y 3z i z x i N dS 4j yk 1 y 1 (first octant) y j z k 1 1 0 1 0 1 0 1 x x x x G x, y, z G x, y, z F S 1 y = −x + 1 R x 1 F R G dA 0 1 0 1 0 1 3z 4 y dy dx 31 x y 4 y dy dx 1 3x y2 2y dy dx y 0 1 3xy dx 0 2 1 0 1 x 3x 1 4 3 x 1 x dx 2 0 2x2 dx 25. F x, y, z S: z 9 xi x2 x2 2xi N dS yj zk 4 2 y y 2, 0 ≤ z y2 z 2y j F R x2 + y2 ≤ 9 R x 2 4 G x, y, z G x, y, z F S 9 k G dA R −4 −2 −2 2x 2 2x 2 R 2y 2 2y 2 y2 z dA x2 y 2 dA −4 9 x2 R 2 0 2 0 3 9 dA r2 0 9 r dr d 9r 2 2 3 r4 4 d 0 243 2 206 Chapter 14 4i 2 Vector Analysis 3j 5k y ≤4 2 y 27. F x, y, z S: z x x 2 + y2 ≤ 4 y, x 2 2 x2 y 2 G x, y, z G x, y, z F S z 2y j F k G dA R 2 0 2 0 2 0 2 −1 1 R x 2xi N dS R 1 −1 8x 6y 5 dA 8r cos 0 6r sin 2r 3 sin 5 r dr d 52 r 2 10 d 2 2 83 r cos 3 64 cos 3 64 sin 3 d 0 16 sin 16 cos 10 20 0 29. F x, y, z 4xy i z2 j yz k 0, x 1, y 0, y 1, z 0, z 1 1 z S: unit cube bounded by x S1: The top of the cube N k, z F S1 1 1 1 N dS 0 0 y 1 dy dx 1 2 1 x 1 y S2: The bottom of the cube N F S2 S3: The front of the cube N i, x F S3 k, z N dS 0 1 0 1 1 1 1 y 0 dy dx 0 0 N dS 0 0 4 1 y dy dz 2 S4: The back of the cube N i, x F S4 0 1 1 S5: The right side of the cube N 4 0 y dy dx 0 0 S5 j, y F 1 1 1 N dS 0 N dS 0 0 z 2 dz dx 1 3 S6: The left side of the cube N j, y F S6 0 1 1 N dS 0 0 z 2 dz dx 1 3 Therefore, F S N dS 1 2 0 2 0 1 3 1 3 5 . 2 g x, y , is defined as 31. The surface integral of f over a surface S, where S is given by z n f x, y, z dS S lim →0 i f xi , yi , z i 1 Si . (page 1061) See Theorem 14.10, page 1061. 33. See the definition, page 1067. See Theorem 14.11, page 1067. Section 14.6 35. (a) 4 −6 −6 6 −4 6 y z Surface Integrals 4 cos 2u i 207 (b) If a normal vector at a point P on the surface is moved around the Möbius strip once, it will point in the opposite direction. (c) r u, 0 4 sin 2u j This is a circle. z 4 x −2 2 x 2 y −4 (d) (construction) (e) You obtain a strip with a double twist and twice as long as the original Möbius strip. 37. z m x2 y 2, 0 ≤ z ≤ a k dS S k R 1 x x k x2 2 2 y x 2 2 y y2 2 y 2 dA k R 2 dA 2 k a2 Iz S k x2 2 a y 2 dS R 2 dA 2k 0 0 r 3 dr d a2 2 2ka 4 2 4 a 2m 2 z 2k a 4 2 39. x 2 y2 2k a 2 a 2, 0 ≤ z ≤ h 1 a2 x2 h x, y, z y ± Project the solid onto the xz-plane. Iz 4 S h a x2 x2 0 0 h a 0 y 2 1 dS x a a y 4 4a 3 0 h a2 1 a2 x a x2 a 0 x2 dx dz 4a 3 1 x a2 x2 2 0 2 dx dz 4a 3 0 arcsin dz 2 h 2 a 3h 41. S: z 16 x2 0.5z k y 2, z ≥ 0 F x, y, z F N dS S R F gx x, y i gy x, y j 0.5 16 k dA R 0.5 z k 2x i 2y j k dA 0.5 z dA R 2 4 R x2 y 2 dA 2 0.5 0 0 16 r 2 r dr d 0.5 0 64 d 64 208 Chapter 14 Vector Analysis Section 14.7 Divergence Theorem 1 dA. 1. Surface Integral: There are six surfaces to the cube, each with dS z 0, N k, F N z 2, S1 a a 0 dA a 2 dA S2 0 a 2 dx dy 0 0 z a, N k, F N z 2, a4 x 0, N i, F N 2x, S3 0 dA 0 a a x a, N i, F N 2x, S4 2a dy dz 0 0 2a dy dz 2a3 y 0, N j, F N 2y, S5 0 dA 0 a a y a, N F j, N dS F a4 N 2a 3 2y, S6 2a dA 0 0 2a dz dx 2a 3 Therefore, s 2a 3 a 4. Divergence Theorem: Since div F a a 0 a 2z, the Divergence Theorem yields a a div F dV 0 Q 0 2z dz dy dx 0 0 a 2 dy dx a 4. 3. Surface Integral: There are four surfaces to this solid. z 0, N 0 dS S1 z 6 k, F 0 j, F 6 N z y 0, N z dS S2 N 6 0 z 2y z, dS 6 dA z2 0 dx dz 6 3 y z dx dz 0 6z dz dz dy 2y 2 dy 2x 5y 6 x 18. 9 36 x x 0, N y dS S3 i, F 3 0 N 6 0 2y y 2x, dS 3 dA 6y y dz dy 0 x 2y z 2x 6, N 5y F i 2j k ,F 6 3 6 0 N 2y 3z , dS 6 dA 3 3z dz dy 0 18 36 9 45 11y dx dy 0 90 90y 20y 2 dy 45 S4 Therefore, s N dS 0 Divergence Theorem: Since div F dV Q 1, we have 1 Area of base 3 Height 1 96 3 18. Volume of solid Section 14.7 5. Since div F 2x 2y a Divergence Theorem 209 2z, we have a 0 a a div F dV 0 Q a 0 0 2x 2ax 0 2y 2ay 2z dz dy dx a a a 2 dy dx 0 2a 2x 2a 3 dx a 2x 2 2a 3x 0 3a 4. 7. Since div F 2x 2x 2 xyz 2 xyz a 2 0 0 2 div F dV Q Q a 0 a 0 2 0 2 0 2 xyz dV 0 2 2 sin cos sin sin cos 2 sin ddd 2 0 5 sin cos sin3 a cos 5 ddd 2 1 2 5 sin cos d d 0 sin2 2 2 d 0 0. 9. Since div F 3 dV Q 3, we have 3 Volume of sphere 3 4 3 2 3 32 . 11. Since div F 1 2y 4 3 3 1 9 2y, we have y2 4 3 4 2y dV 0 Q 9 y2 2y dx dy dz 0 3 4y 9 y 2 dy dz 0 4 9 3 y 232 3 dz 3 0. 13. Since div F 3x 2 x2 6 4 0 0 4 0 y 4x 2, we have 6 0 4 6 4 x2 dV 0 Q 4x 2 dz dy dx 0 4x 2 4 y dy dx 0 32x 2 dx 2304. 15. F x, y, z div F y F S xyi 4 N dS 4y j x xz k div F dV Q 3 0 3 0 3 3 0 3 0 0 0 2 Q y x 4 dV sin 0 2 3 0 sin sin 3 cos 4 2 sin 2 ddd sin2 sin sin2 sin2 cos 4 2 sin ddd 2 sin2 cos 3 sin 4 sin 0 dd 8 0 3 0 2 sin dd 8 0 3 2 2 cos 0 d 16 3 33 16 0 d 144 . 0 210 Chapter 14 Vector Analysis 17. Using the Divergence Theorem, we have curl F S Q N dS i div curlF dV j y z2 2x 2 6yz k z 2xz 6y i 2z 2z j 4x 4x k 6y i curl F x, y, z 4xy div curl F Therefore, Q x 0. 0. div curl F d V 19. See Theorem 14.12, page 1073. 21. Using the triple integral to find volume, we need F so that div F M x N y P z 1. y j, or F z k. i 1 fx i 1 fx i 1 x dy dz S S Hence, we could have F For dA dy dz consider F x i, F x i, x f y, z , then N fy j fy2 j fx2 fy j fx2 fz k fz2 fz k fz2 k fy2 and dS 1 fy2 fx2 fx2 fz2 dy dz. fz2 dz d x. fy2 d x dy. For dA dz dx consider F y j, y f x, z , then N and dS 1 For dA dx dy consider F z k, z f x, y , then N and dS 1 Correspondingly, we then have V S F N dS y dz dx S z dx dy. 23. Using the Divergence Theorem, we have S curl F N dS Q div curl F dV. Let F x, y, z curl F div curl F Therefore, S Mi P y 2P xy Nj Pk P x 2P yx N i z 2N xz M j z 2M yz N x 2N zx M k y 2M zy 0. curl F N dS Q 0 dV 0. 25. If F x, y, z F S xi N dS yj z k, then div F div F d V Q Q 3. 3 dV 3V. 27. S f D N g dS S fg N dS div f g dV Q Q f div g f g dV Q f 2g f g dV Section 14.8 Stokes’s Theorem 211 Section 14.8 1. F x, y, z 2y i curl F 2y x z Stokes’s Theorem zi j y xyz x yz j k z ez xyi j yz 2k ezk 3. F x, y, z 2z i i 4x2 j j y 4x2 arctan x k k z arctan x 2 curl F x 2z 2 2 1 1 x2 j 8xk 5. F x, y, z ex y2 i ey j y z2j x yz k i curl F ex xz zx 2 k z z2 z2 z2 x y2 ey 2 xyz i yz j yz j 2ye x 2ye x 2 2 2ze y 2e y 2 2 y2 k y2 k i 7. In this case, M Line Integral: y F C z, N dr x z, P y dx x x dy y and C is the circle x 2 y2 1, z 0, dz 0. C Letting x cos t, y y dx x dy sin t, we have dx 2 sin t d t, dy 2. z2 1. cos t d t and sin 2 t 0 cos 2 t d t x2 y2 C Double Integral: Consider F x, y, z Then N Since z2 1 x2 y 2, z x 2x 2z F F 2x i 2y j 2z k 2 x2 y2 z2 xi yj z k. x z , and z y y z , dS 1 x2 z2 y2 dA z2 1 dA . z Now, since curl F curl F S 2k, we have N dS R 2z 1 dA z 2 dA R 2 Area of circle of radius 1 2. 212 Chapter 14 Vector Analysis z 6 4 9. Line Integral: From the accompanying figure we see that for C1: z C2: x C3 : y Hence, C (0, 0, 6) 0, dz 0, dx 0, dy F dr 0 0 0. 2 C3 C2 2 (0, 3, 0) C1 y x yz d x C y dy y dy C2 0 z dz x 4 (4, 0, 0) y dy C1 3 z dz C3 6 z dz 0 y dy 0 3 y dy 0 z dz 6 z dz 0. Double Integral: curl F Considering F x, y, z N Thus, curl F S xyj 4y xz k 2z 12, then 29 d A. 3x F F 3i 4j 2k and dS 29 N dS R 4 0 4 0 4 4xy ( 3x 0 (12 0 3x) 4 2xz dy dx 12) 4 4 xy 8xy 0 dx 0 2x 6 3x 2 2y 3 x 2 dy dx 12x dy dx 0. \ \ 11. Let A N 0, 0, 0 , B U U V V 1, 1, 1 and C 2 i 2k 22 i 2 0, 2, 0 . Then U k . x AB i j k and V AC 2 j. Thus, Surface S has direction numbers curl F S 1, 0, 1, with equation z 1 2 2 dA R 0 and dS 2 d A. Since curl F 1, b 2 1. 3i j 2 k, we have N dS R dA Area of triangle with a 13. F x, y, z z2 i i j x2 j k z y2 z 2y j N dS y 2 k, S: z 4 x2 y 2, 0 ≤ z curl F x z2 x2 2x i y x2 y2 2yi 2z j 2xk G x, y, z G x, y, z 4 k 2 4 4 4 4 x2 curl F S 4xy R 4yz 2x d A 2 2 2 2 x2 x2 4xy 4xy x2 4y 4 16y 0 x2 4x 2y y2 4y 3 2x d y d x 2x d y d x 4x 4 2 x2 dx S ection 14.8 15. F x, y, z z2 i i curl F x z2 G x, y, z G x, y, z z yj j y y 4 x x2 F dS R Stokes ’s Theorem 213 x z k, S: z k z xz x2 y2 i zj 4 x2 y2 4 curl F y 2 4 y x2 yz x2 y2 y2 j k y4 4 x2 x2 y2 dA y2 2 2 4 4 x2 S 4 dA R y dy dx x2 0 17. F x, y, z ln x 2 i y2 i x arctan j y j y k k z 1 2 sin 2 in the first octant. 1 1y x2 y2 y x2 y2 k 2y x2 y2 k curl F x 1 2 ln x 2 y2 arctan x y S: z 9 2x 2x 2i 3y over one petal of r 3y 3j N dS R 2 0 2 0 2 G x, y, z G x, y, z z k 9 curl F S 2y x2 y2 2 sin 2 0 dA 2r sin r dr d r2 2 sin dr d 4 sin cos 0 8 sin 2 cos d 0 8 sin 3 3 2 0 8 3 19. From Exercise 10, we have N 2x i 1 xz dA k and dS 4x2 a a 1 4x 2 d A . Since curl F a xyj a5 . 4 x z k , we have curl F S N dS R x 3 dy dx 0 0 0 a x 3 dx ax4 4 a 0 21. F x, y, z i i j j y 1 2k k z 2 curl F S 23. See Theorem 14.13, page 1081. curl F x 1 0 Letting N k, we have N dS 0. 214 Chapter 14 Vector Analysis 25. (a) C fg dr S curl f g f g i x i f g j y j y xfg 2g N d S (Stoke’s Theorem) f g k z k z yfg f y g z f x f x f z g y g z g y i z f 2g fg curl f g fg f x yz f f f y g z j f y g y dr S 2g zy f f f x 2g f z zx 2g g y f z f y f z i g x g x g j x j k f x g y f y g k x xz 2g xy yx g z i f x g x Therefore, C k f z g z f g fg dr S curl f g f f f fg C N dS S f g N d S. (b) C ff N d S (using part a.) 0. dr C 0 since f (c) C fg gf dr gf N dS S dr g f S f S g g f g N d S (using part a.) N dS 0 f S N dS 27. Let C 1 2 since C and ai C C bj r c k, then dr 1 2 curl C S r N dS 1 2 2C S N dS S C N dS r i a x j b y k c z bz cy i az cx j ay bx k i curl C r bz x cy cx j y az ay k z bx 2 ai bj ck 2C. ...
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