ODDREV14 - Review Exercises for Chapter 14 215 Review...

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Unformatted text preview: Review Exercises for Chapter 14 215 Review Exercises for Chapter 14 1. F x, y, z z 3 2 xi j 2k 3. f x, y, z F x, y, z 8x2 16x xy yi z2 xj 2z k 3 4 x 4 y 5. Since M y 1 y2 N x, F is not conservative. x 6xy 2 y3 7y 3x 2 and N U y 6x 2 y 3y 2 7, g x which suggests h y y3 7y, N x, F is conservative. From M U 7. Since M y 12xy partial integration yields U 3x 2 y 2 x3 h y and U 3x 2 y 2 gx x3, and U x, y 3x 2 y 2 x3 y3 7y C. 9. Since M y 4x N , x P M 1 . z x F is not conservative. 11. Since M y 1 y 2z N , x M z 1 yz2 P , x N z x y 2z2 P , y F is conservative. From M we obtain U x yz x2i 2x P y f y, z , U x yz g x, z , U x yz h x, y ⇒ f x, y, z x yz K U x 1 ,N yz U y x ,P y 2z U z x yz2 13. Since F (a) div F (b) curl F y 2j 2y z2 k: 2z N i z P x M j z N x M k y 0i 0j 0k 0 15. Since F (a) div F (b) curl F cos y y cos x i x cos y cos x sin x xy sin y x sin y j x yz k: y sin x xz i yz j sin y cos x k xz i yz j 216 Chapter 14 Vector Analysis ln x 2 2x x2 2x x2 (b) curl F 2x x2 y2 2y y2 2y k y2 x2 1 y2 i ln x 2 2y y2 y2 j 1 17. Since F arcsin x i x y 2 j y z2 k: 1 (a) div F 2xy 2yz 1 x2 (b) curl F z2 i y2k 19. Since F (a) div F z k: 21. (a) Let x x2 C t, y t, y 2 ds 1 ≤ t ≤ 2, then ds 2 2 dt. 22 t3 3 2 2t 2 2 dt 1 62 1 (b) Let x x2 C 4 cos t, y y 2 ds 4 sin t, 0 ≤ t ≤ 2 , then ds 2 4 dt. 16 4 dt 0 128 23. x cos t x2 C t sin t, y y 2 ds sin t 2 t cos t, 0 ≤ t ≤ 2 , cos t t sin t 2 2 dx dt t cos t, t cos t 2 dy dt t sin t 2 sin t t 2 cos2 t t 2 sin2 t dt 0 t3 t dt 0 2 25. (a) Let x 2t, y 2x C 2 1 2 3t, 0 ≤ t ≤ 1 1 1 y dx x 3y dy 0 7t 2 3 sin t dt, dy 2 7t 3 dt 0 35t dt 35 2 (b) x 3 cos t, y 2x C 3 sin t, dx x 3y dy 3 cos t dt, 0 ≤ t ≤ 2 9 sin t cos t dt 18 y dx 9 0 27. C 2x xt yt 2x C y ds, r t 3a 3a a cos3 t i a sin3 t j, 0 ≤ t ≤ 2 cos2 t sin t sin2 t cos t 2 y ds 0 2a cos3 t a sin3 t xt 2 y t 2 dt 9 a2 5 29. f x, y C: y rt rt rt 5 sin x y 3x from 0, 0 to 2, 6 ti i 3t j, 0 ≤ t ≤ 2 3j 10 Lateral surface area: 2 2 f x, y ds C2 0 5 sin t 3t 10 dt 10 0 5 sin 4t dt 10 41 4 cos 8 32.528 R eview Exercises for Chapter 14 31. d r F F C 217 2t i t5 i dr 3t 2 j dt t 4 j, 0 ≤ t ≤ 1 1 33. d r F F C 2 sin t i 2 cos t i 2 2 cos t j 2 sin t j k dt t k, 0 ≤ t ≤ 2 2 5t 6 d t 0 5 7 dr 0 t dt 2 35. Let x F F C t, y t dr 2 t, z 2t 2 i 2 2t 2, 2t2 tj 4t 3 2 ≤ t ≤ 2, d r 2t k 64 3 i j 4 t k dt. 32 2 4t2 dt 37. For y For y xy dx C x 2, r1 t 2x, r2 t x2 ti 2 y 2 dy t 2 j, 0 ≤ t ≤ 2 ti 4 xy dx C1 y y = 2x 4 3 2t j, 0 ≤ t ≤ 2 x2 32 4 3 y 2 dy C2 (2, 4) C2 y = x2 C1 x 1 2 3 4 xy dx x2 y 2 dy 2 100 3 1 39. F xi y j is conservative. 12 x 2 23 y 3 4, 8 2 0, 0 Work 1 16 2 x 2 yz 23 8 3 1, 4, 3 2 8 3 3 12 42 41. C 2xyz dx x 2z dy x 2 y dz 0, 0, 0 1 43. (a) C y 2 dx 2xy dy 0 1 1 3 t2 0 1 t 2 3 1 21 3t 1 2 3t 2 4t t dt 1 dt 2t 14t 7t 2 9t 2 0 5 dt 1 3t 3 4 5t 0 15 t 1 2t dt (b) C y 2 dx 2xy dy 1 4 t1 t 1 4 2t t dt 15 t2 1 (c) F x, y Hence, F C y 2i 2 xy j f where f x, y xy 2. dr 42 2 11 2 15 2 2 2 45. C y dx 2x dy 0 0 2 1 dy dx 0 2 dx 4 47. C xy 2 dx x 2y dy R 2xy 2xy dA 0 218 Chapter 14 Vector Analysis 1 x 1 49. C xy dx x 2 dy 0 x 2 x dy dx 0 x2 x3 dx 1 12 51. r u, v 0≤u≤ sec u cos v i 3 1 2 tan u sin v j 2u k 6 z , 0≤v≤2 −4 2 4 x −2 2 4 y 53. (a) 3 −4 4 x −2 −3 z (b) 3 −4 −4 −3 4 y x 4 3 2 −1 −2 −3 z −4 2 3 4 y (c) 3 2 −4 −3 4 x 3 −2 −3 z (d) 3 −2 2 −3 −4 −4 −3 3 4 y x 4 3 −2 −3 1 z −4 −2 2 3 4 y The space curve is a circle: r u, (e) ru rv ru rv 3 cos v sin u i 3 sin v cos u i 3 cos v cos u j 3 sin v sin u j cos v k 4 32 cos u i 2 32 sin u j 2 2 k 2 i 3 cos v sin u 3 sin v cos u 3 cos2 v cos u i 3 cos2 v cos u i j k 3 cos v cos u 0 3 sin v sin u cos v 3 cos2 v sin u j 3 cos2 v sin u j 9 cos4 v sin2 u 9 cos v sin v sin2 u 9 cos v sin v k 81 cos2 v sin2 v 9 cos v sin v cos2 u k ru rv 9 cos4 v cos2 u 9 cos4 v 81 cos2 v sin2 v Using a Symbolic integration utility, 2 4 2 ru 0 rv du dv 14.44 (f) Similarly, 4 0 0 2 ru rv dv du 4.27 R eview Exercises for Chapter 14 55. S: r u, v ru u, v rv u, v ru ru ru rv x S 219 u cos v i cos v i u sin v j sin v j 3 u 2u k 12 u k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 2 −3 3 x z −3 u sin v i i cos v u sin v u y dS 0 2 0 u cos v j j k sin v 3 2u u cos v 0 3 2 3 −2 y 2u 3 u cos v i 2u 3 u sin v j uk 2u 2 1 u cos v u sin v u 2u 3 2 2 1 du dv 0 2 cos v 0 sin v u2 2u 3 2 1 dv du 0 57. F x, y, z x2 i xy j zk 3y 4z 12 Q: solid region bounded by the coordinates planes and the plane 2x Surface Integral: There are four surfaces for this solid. z 0 N k, F N z, S1 0 dS 0 y 0, N j, F N xy, S2 0 dS 0 x 0, N i, 2i F N x 2, S3 0 dS 1 4 0 9 dA 16 29 dA 4 2x 3y 4z 12, N 1 4 1 4 1 4 1 6 3j 4k , dS 29 3xy 4z dA 1 F S4 N dS 2x 2 R 6 0 6 4 0 (2x 3) 2x 2 12 3 x2 2x 3xy 12 2x 2 3y dy dx 12 3 x3 3 12x 2 2x 12 3 6 2x 2 0 6 3x 12 2x 2 3 24x 2x (12 0 2x 12 1 6 3x x4 4 2x 2x 3 12 2x 2 3 2 dx x3 0 36 dx x 3y) 4 36x 0 66 Divergence Theorem: Since div F 6 (12 0 (12 0 6 2x) 3 2x) 3 1 3x 1, Divergence Theorem yields div F dV 0 Q 6 0 1 dz dy dx 3y 3x 1 4 1 4 1 4 1 12 2x 4 32 y 2 2x dy dx 2x 3 (12 0 3x 0 6 1 12y 2xy dx 12 3 1 3x 4 64 2x 3 12 2x 2 3 35x3 3 2 3x 0 6 0 1 4 12 2x dx 6 23 3x 3 35x 2 96x 36 dx 48x 2 36x 0 66. 220 Chapter 14 cos y Vector Analysis y cos x i sin x x sin y j x yz k 59. F x, y, z S: portion of z y 2 over the square in the xy-plane with vertices 0, 0 , a, 0 , a, a , 0, a z 1 Line Integral: Using the line integral we have: C1: y C2: x C3: y C4: x F C 0, dy 0 C3 0, dx a, dy a, dx cos y C 0, 0, 0, z z z y 2, a2, y 2, dz dz dz 2y dy 0 2y dy sin x x sin y dy xyz dz x C4 C1 a C2 a y dr y cos x dx dx C1 a C2 0 0 C3 cos a a cos x dx C4 a sin a a sin y dy a ay3 2y dy dx 0 a cos a x cos a a cos x dx 0 0 sin a a a sin y dy 0 2ay 4 dy a 0 a a a sin x a y sin a a sin a z 1 a cos y 0 2a 2a6 5 y5 5 a cos a a sin a a cos a a 2a6 5 Double Integral: Considering f x, y, z N Hence, a a y 2, we have: xz i y z j. f f 2y j k , dS 1 4y 2 4y 2 dA, and curl F a a a curl F S N dS 0 0 2y 2z dy dx 0 0 2y 4 dy dx 0 2a5 dx 5 2a6 . 5 Problem Solving for Chapter 14 1. (a) T N dS Flux S 25 x2 xi 1 1 x2 y2 1 z2 32 xi yi zk x2 k dy dx N dS x2 z2 3 y2 1 1 1 y2 3 y2 32 32 kT 25k R 12 x2 1 y2 x2 2 1 x2 1 12 12 x2 z y2 x2 z2 y2 32 dA dy dx 25k 12 0 12 1 x2 z2 3 2 1 12 x2 1 2 dy dx dx 1 x2 z2 3 2 1 x2 12 25k 12 0 1 x2 1 25k 0 1 2 2 dy 12 1 x2 12 25k 25k 2 6 —CONTINUED— P roblem Solving for Chapter 14 1. —CONTINUED— (b) r u, v ru ru T rv x2 v2 T Flux 0 3 221 cos u, v, sin u sin u, 0, cos u , rv cos u, 0, 25 y2 25 13 rv 1 2 3 2 0, 1, 0 sin u xi yj vj zk sin u k cos2 u 25k sin2 u 2 6 25 1 z2 32 cos u i 25 13 32 ru v2 v2 2 v2 32 25k 1 du dv 3. r t rt Ix C 3 cos t, 3 sin t, 2t 3 sin t, 3 cos t, 2 , r t 2 13 4t2 4t2 13 dt 13 dt 1 3 1 3 13 13 32 32 2 y2 x2 C z2 z2 ds 0 2 9 sin2 t 9 cos2 t 0 2 27 27 Iy ds 2 Iz C x2 y2 ds 0 9 cos2 t 9 sin2 t 13 dt 18 13 5. 1 x dy 2C y dx 1 2 2 a 0 2 sin sin a sin sin2 2 cos d 1 a1 2 cos 2d cos a1 d cos d 12 a 2 12 a 2 cos2 0 2 sin 0 3 a2 Hence, the area is 3 a2. 7. (a) r t rt W (b) r t rt W F t j, 0 ≤ t ≤ 1 j 1 1 F C dr 0 ti j j dt 0 dt 1 t 1 dr t2 i 2t i 1 t j, 0 ≤ t ≤ 1 j 2t 0 1 t2 i 2t 2t 4 t2 2t t t2 t2 2 1j 2t3 13 15 1 t2 2t i 1 dt j dt 1 0 1 t4 1 dt t4 0 —CONTINUED— 222 Chapter 14 Vector Analysis 7. —CONTINUED— (c) r t rt F dr ct c1 ct c2t 4 W dW dc d 2W dc 2 9. v r F C t2 i 2t i t2 t j, 0 ≤ t ≤ 1 j t c1 c2t 1 c 6 5 2 2t 2 ct 2 1 c2 t ct t2 1 2 11 2c2t 2 12 c 30 dr 1 6 1 c 15 0⇒c c 1 >0 15 a1, a2, a3 a2z a 3 y, 5 minimum. 2 x, y, z a1z a 3 x, a 1 y 2v a2 x curl v r 2 a 1, 2 a 2, 2 a 3 By Stoke’s Theorem, v C r dr S curl v 2v S r N dS N dS. 11. F x, y M M y M x, y i 3mxy y2 5 N x, y j x2 m y2 y2 52 3xy i 2y 2 x2 j x2 2 3mxy x 2 y2 72 72 52 3mxy 3mx x 2 52 x 2 y2 2y x2 x2 x2 y2 x2 12y 2 72 x2 y2 y2 2x 5 y2 52 3mx 5y 2 3 mx x 2 4 y 2 x2 y2 7 2 52 N N x m 2y 2 x 2 x2 y2 5 2 m 2y 2 mx x2 mx x2 x2 y2 y2 m 2y 2 52 x 2 72 x2 x2 y2 y2 2 52 2 mx 2y 2 3x 2 72 3mx x2 x2 y2 4y2 72 Therefore, N x M and F is conservative. y ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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