EVNREV14 - 442 Chapter 14 zi y2 i curl F x z 1 j y 0 Vector...

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Unformatted text preview: 442 Chapter 14 zi y2 i curl F x z 1 j y 0 Vector Analysis yk 22. F x, y, z S: x 2 k z y N i j Letting N k, curl F 0 and S curl F N dS 0. 24. curl F measures the rotational tendency. See page 1084. 26. f x, y, z (a) x yz, g x, y, z k x yz k 2 sin t j dr xyk 0 k, 0 ≤ t ≤ 2 0 z, S: z 4 x2 y2 g x, y, z f x, y, z g x, y, z rt 2 cos t i f x, y, z g x, y, z C (b) f x, y, z g x, y, z f g yz i k i yz 0 x x2 1 xz j j xz 0 y2 4 i k xy 1 4 x x2 y2 xz i y x2 2 yz j N dS 4 y2 j k y x2 S 2 4 N dS y2 dA x 2z 4 x2 4 y2 2 x2 y2 dA 2 x2 dA f x, y, z S g x, y, z y 2z 4 x2 y2 4 y2 S 2 0 2 0 2 0 2 x2 y2 dA 4 x2 y2 2r 2 cos 2 4 sin 2 r2 2 0 r d dr dr 0 2r 3 1 sin 2 4 r2 2 Review Exercises for Chapter 14 2. F x, y 5 4 3 2 x −1 −2 −3 −4 −5 2 4 i y 2y j 4. f x, y, z F x, y, z x 2eyz 2xeyz i xeyz 2i x 2ze yz j xz j x 2 ye y z k xyk Review Exercises for Chapter 14 6. Since M y 1 x2 N x, F is conservative. From M Ux y x 2 and N Uy yields U yx h y and U yx g x which suggests that U x, y yx C. 8. Since M y 6y 2 sin 2x N x, F is conservative. From M Ux 2y3 sin 2x and N we obtain U y3 cos 2x h y and U y3 1 cos 2x g x which suggests that h y y3, g x U x, y y3 1 cos 2x C. 10. Since M y M z N z 4x 2z 6y N , x P , x P y 12. Since M y sin z N , x M z y cos z P , x 443 1 x, partial integration U y 3y 2 1 C, and cos 2x , F is not conservative. F is not conservative. 14. Since F (a) div F (b) curl F xy 2 j 2 xy 2 xz j x2 2x 0 (b) curl F z x 2 k; x2 y 2k 16. Since F (a) div F (b) curl F sin2 y j: 20. Since F (a) div F 3x 3 2i z i x z x2 yi 1 3j z j y z y2 1 j x 1 y 5 k z 2 k: 2z 2z j z 3x k: 18. Since F (a) div F (b) curl F yi x 2 sin y cos y z2 1 x2 1 y2 1 i y 22. (a) Let x 5t, y xy ds 4t, 0 ≤ t ≤ 1, then ds 1 41 dt. 4 3 2 y 20t 2 0 41 dt C 20 41 3 dt 2 5 dt dt 2 (b) C1: x C2: x C3: x t, y 4 0, y 0, 0 ≤ t ≤ 4, ds 4t, y 2 xy ds C 0 (0, 2) y = − 1x + 2 2 C2 (4, 0) x 3 4 2t, 0 ≤ t ≤ 1, ds t, 0 ≤ t ≤ 2, ds 4 1 C3 C1 Therefore, 0 dt 0 8t 16 5 t2 2 dx dt 8t 2 2 5 dt 0 0 dt t3 3 1 0 85 . 3 dy dt 2 24. x t sin t, y 2 1 t 0 2 cos t, 0 ≤ t ≤ 2 , sin t 1 cos t 2 1 cos t, sin t t sin t 2 1 3 2 2 cos t dt 2 2 x ds C sin t 2 dt 0 2 0 2 t1 cos t sin t 1 cos t dt 2 cos t 32 0 2 0 t1 cos t dt 2 0 t1 cos t dt 8 444 Chapter 14 Vector Analysis 26. x cos t 2x C t sin t, y y dx x sin t t sin t, 0 ≤ t ≤ 2 2 , dx t cos t dt, dy 6t 2 cos t cos2 t t t cos t 1 sin t dt 3 dt 1.01 3y dy 0 sin t cos t 5t 2 sin2 t 2t 28. r t xt x2 C ti t2j t 3 2 k, 0 ≤ t ≤ 4 2t, z t 4 1, y t y2 31 t 2 t2 2 z2 ds 0 t4 t3 1 4t 2 9 t dt 4 2080.59 30. f x, y C: y rt rt rt 12 2 x y x from 0, 0 to 2, 4 ti i t 2 j, 0 ≤ t ≤ 2 2t j 1 4t 2 Lateral surface area: 2 f x, y ds C 0 12 t t2 1 4t 2 dt 41.532 32. dr F F C 4 sin t i 4 cos t 2 3 cos t j dt 4 cos t 3 sin t j, 0 ≤ t ≤ 2 12t 7 sin2 t 2 2 3 sin t i 12 0 dr 7 sin t cos t dt 24 0 34. x dr F F C 2 t, y i 4 dr 0 2 j t, z 4t t 2, 0 ≤ t ≤ 2 2t k dt 4t t 2 4t t2 i 2 dt t2 2 4t 2 2t 2 t2 2 2 tj 0k t 2t 0 36. Let x dr F F C 2 sin t, y 2 cos t i 2 cos t, z 2 sin t j 2 sin t k 8 sin t 4 sin2 t, 0 ≤ t ≤ 8 sin t cos t k dt . 0i dr 4j 16 sin2 t cos t dt 8 cos t 0 16 3 sin t 3 16 0 38. C F rt F C dr 2 cos t dr 4 C 2x y dx 2y 2 sin t x dy 2 t cos t j, 0 ≤ t ≤ 2 t sin t i 2 4 Review Exercises for Chapter 14 2000 5280 t k, 0 ≤ t ≤ 2 2 25 tk 33 445 40. r t 10 sin t i 10 sin t i 10 cos t j 10 cos t j F dr F C 20k 10 cos t i 2 10 sin t j 500 dt 33 25 k 33 250 mi 33 ton dr 0 42. C y dx x dy 1 dz z ,y a1 y dx. 4, 4, 4) xy ln z 0, 0, 1 16 ln 4 44. x a 1 2 sin cos ,0 ≤ ≤2 y (a) A x dy C Since these equations orient the curve backwards, we will use A 1 2 1 2 a2 2 a2 2 2 C1 C2 2a 2π a y dx x dy 1 2 x a2 1 0 2 cos 1 cos a2 sin sin d 0 0 0d 1 0 2 2 cos cos2 sin a2 6 2 sin2 d 2 0 2 cos sin d 3 a2. (b) By symmetry, x y 1 2A a. From Section 14.4, y 2 dx 1 2A 2 a3 1 0 cos 2 1 cos d C 1 a3 5 2 3 a2 5 a 6 2 2 a a2 a2 x2 46. C xy dx x2 y 2 dy 0 2 0 2x 2x dx 0 x dy dx 4 48. C x2 y 2 dx 2xy dy a a x2 4y dy dx 0 dx a 0 1 1 (1 x2 33 2 50. C y 2 dx x 4 3 dy 1 1 1 1 1 x2 3 3 2 41 x 3 1 1 3 2y dy dx 332 4 13 x y y2 3 81 x 3 82 x 7 3 x2 dx x2 332 1 1 x2 x2 332 dx 16 1 35 1 3 352 x2 352 1 0 446 Chapter 14 Vector Analysis u k 6 52. r u, v e u4 cos v i e u4 sin v j 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 z 2 2 x 2 y 54. S: r u, v ru u, v ru u, v ru ru rv rv z dS S u i i i 1 1 j j vi u vj sin v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ cos v k cos v i cos v j 2k j k 10 1 cos v 2 cos2 v 2 4 4 du dv 2 6 2 ln 6 6 2 2 sin v 2 cos2 v 0 0 56. (a) z z aa 0 ⇒ x2 z x2 y2 a2 y2 , 0 ≤ z ≤ a2 a2 a x2 y2 a2 z (b) S: g x, y x, y m S k x2 y2 a a y e x, y, z dS k x2 R x y2 y2 1 2 a 1 1 x2 gx2 gy2 d A a2y2 dA x y2 2 k R x2 a2 R a2x2 x y2 2 k k a2 k a2 2 k a2 3 y2 dA 1 0 2 0 r2 dr d a3 d 3 1 0 1 a3 Review Exercises for Chapter 14 447 58. F x, y, z xi yj zk 3y 4z 12 (0, 0, 3) z Q: solid region bounded by the coordinate 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 (0, 4, 0) y x y 0, N j, F N y, S2 0 dS 0 (6, 0, 0) x 2x 3y 0, 4z N i, 12, N 1 4 1 4 2i F N x, S3 0 dS 1 1 4 0 9 dA 16 29 dA 4 3j 4k , dS 29 3y 4z dy dx 6 N S4 F dS 2x R 6 0 (12 0 2x) 3 12 dy dx 3 0 4 2x dx 3 3 4x x2 3 6 36 0 Triple Integral: Since div F div F dV Q Q 3, the Divergence Theorem yields. 3 Volume of solid 3 1 Area of base Height 3 1 643 2 36. 3 dV 60. F x, y, z x zi y zj x2 k y 2z 12 (0, 0, 6) z S: first octant portion of the plane 3x Line Integral: C1: y C2: x C3: z F C 0, 0, 0, x C dy dx dz 0, 0, 0, y 3x 2 5 x 2 z z y 12 2 12 2 12 z dy x2 6 dx 3 2 3x y , dz dz dy 3 dx 2 1 dy 2 3 dx x (4, 0, 0) (0, 12, 0) y , 3x, x 2 dz dx dr z dx 12 32 x 2 x C1 0 4 y C2 12 2 y 4 dy C3 x 36 dx 12 8 3x 3 dx 12 0 3 y 2 z k 6 dy 0 10x Double Integral: G x, y, z G x, y, z curl F i 12 3x 2 3 i 2 y 1 j 2 2x N dS 1j 4 12 0 3x 4 curl F S x 0 1 dy dx 0 3x 2 15x 12 dx 8 448 Chapter 14 Vector Analysis Problem Solving for Chapter 14 2. (a) z dS T 1 x2 z x x2 y2 z i x z x 1 xi Flux S 2 2 y2, z x z y 2 1 x x2 y2 1 zk , z y 1 x2 1 y2 y x2 y2 1 dA xi k yi 25 z2 32 25 x i yj zk N z j y z y y2 1 i 2 1 y x2 xi j yj k zk 1 x2 y2 x x2 yj kT 1 x2 y2 k y2 N dS yj zk dA r dr d 50 k xi yj zk 1 x2 dA k R 25 x i 1 y2 k R 2 25 1 x2 1 0 y2 r2 25k 0 1 1 (b) r u, v ru rv ru ru Flux rv rv sin u cos v, sin u sin v, cos u cos u cos v, cos u sin v, sin u sin u sin v, sin u cos v, 0 sin2 u cos v, sin2 u sin v, sin u cos u sin2 v sin u 2 2 sin u cos u cos2 v 25k 0 0 sin u du dv 50 k 4. r t rt ds Iy C t2 2 2t 3 , t, 2 3 t, 1, 1 1 x2 y2 C 2 2t1 t t z2 z2 y2 2 ,rt 1 1 t 1 1 dt ds 0 1 t4 4 t2 83 t dt 9 83 t dt 9 t 2 dt 49 180 5 9 23 60 Ix Iz C ds 0 1 x2 ds 0 t4 4 Problem Solving for Chapter 14 2 449 6. 1 x dy 2C y dx 2 0 1 sin 2t cos t 2 sin t cos 2t dt 2 2 3 Hence, the area is 4 3. 8. F x, y f x, y Work 3x2 y 2 i 2 x 3 y j is conservative. x3y2 potential function. f 2, 4 f 1, 1 8 16 1 127 10. Area rt rt F F dr 2 ab a cos t i a sin t i b sin t j, 0 ≤ t ≤ 2 b cos t j 1 a cos t j 2 1 ab cos2 t dt 2 ab 1 ab 2 1 b sin t i 2 1 ab sin2 t 2 F 0 W dr 1 ab 2 2 Same as area. ...
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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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