# Y c 4 10 d2 d3 a1 1 x d1 b 2 8 7 let a5 2 b

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Unformatted text preview: y= ¯ = π 1 V π 3 ρ3 sin2 φ sin θ dρ dφ dθ = 0 0 π 1 V 2 π 1 V π 0 65 sin2 φ sin θ dφ dθ 4 0 3 65π sin θ dθ = (65π/4) = 195/152; centroid (0, 195/152, 0) 8 38π 0 2π π R 2π 3 π δ0 e(ρ/R) ρ2 sin φ dρ dφ dθ = 37. M = 0 0 0 0 0 4 1 (e − 1)R3 δ0 sin φ dφ dθ = π (e − 1)δ0 R3 3 3 38. (a) The sphere and cone intersect in a circle of radius ρ0 sin φ0 , √2 2 θ2 ρ0 sin φ0 ρ0 −r θ1 ρ0 sin φ0 r r cot φ0 0 θ2 = θ1 = θ2 r dz dr dθ = V= θ1 ρ2 − r2 − r2 cot φ0 dr dθ 0 0 1 13 ρ0 (1 − cos3 φ0 − sin3 φ0 cot φ0 )dθ = ρ3 (1 − cos3 φ0 − sin2 φ0 cos φ0 )(θ2 − θ1 ) 3 30 13 ρ (1 − cos φ0 )(θ2 − θ1 ). 30 (b) From part (a), the volume of the solid bounded by θ = θ1 , θ = θ2 , φ = φ1 , φ = φ2 , and 1 1 1 ρ = ρ0 is ρ3 (1 − cos φ2 )(θ2 − θ1 ) − ρ3 (1 − cos φ1 )(θ2 − θ1 ) = ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 ) 0 0 3 3 30 so the volume of the spherical wedge between ρ = ρ1 and ρ = ρ2 is 1 1 ∆V = ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 ) − ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 ) 32 31 13 (ρ − ρ3 )(cos φ1 − cos φ2 )(θ2 − θ1 ) 1 32 = (c) d cos φ = − sin φ so from the Mean-Value Theorem cos φ2 − cos φ1 = −(φ2 − φ1 ) sin φ∗ where dφ d3 φ∗ is between φ1 and φ2 . Similarly ρ = 3ρ2 so ρ3 − ρ3 = 3ρ∗2 (ρ2 − ρ1 ) where ρ∗ is between 2 1 dρ ∗ 3 3 ρ1 and ρ2 . Thus cos φ1 −cos φ2 = sin φ ∆φ and ρ2 −ρ1 = 3ρ∗2 ∆ρ so ∆V = ρ∗2 sin φ∗ ∆ρ∆φ∆θ. 2π a h 2π a h r2 δ r dz dr dθ = δ 39. Iz = 0 0 2π 0 a r3 dz dr dθ = 0 0 0 h 2π a (r2 cos2 θ + z 2 )δr dz dr dθ = δ 40. Iy = 0 0 0 2π 0 1 (hr3 cos2 θ + h3 r)dr dθ 3 1 π 14 π4 a h cos2 θ + a2 h3 dθ = δ a h + a2 h3 4 6 4 3 =δ 0 2π a2 h 2π a2 h r2 δ r dz dr dθ = δ 41. Iz = a1 0 0 0 π r3 dz dr dθ = a 2π 0 a1 0 2π π 0 0 1 δπh(a4 − a4 ) 2 1 2 a (ρ2 sin2 φ)δ ρ2 sin φ dρ dφ dθ = δ 42. Iz = 0 1 δπa4 h 2 ρ4 sin3 φ dρ dφ dθ = 0 0 0 8 δπa5 15 599 Chapter 16 EXERCISE SET 16.8 1. ∂ (x, y ) = ∂ (u, v ) 1 3 3. ∂ (x, y ) = ∂ (u, v ) cos u sin u 4. ∂ (x, y ) = ∂ (u, v ) 5. x = 4 −5 2. − sin v cos v 2(v 2 − u2 ) (u2 + v 2 )2 − 4uv (u2 + v 2 )2 ∂ (x, y ) = ∂ (u, v ) 1/u v 0 u 2/9 −1/9 8. x = u3/2 /v 1/2 , y = v 1/2 /u1/2 ; ∂ (x, y, z ) = ∂ (u, v, w) 3 1 0 1 0 1 10. ∂ (x, y, z ) = ∂ (u, v, w) 1−v v − vw vw 0 −2 1 = −1 − 16uv 5/9 2/9 = 1 9 =1 1 √√ 2 2 u+v 1 −√√ 2 2 v−u √ √ √ ∂ (x, y ) = u + v/ 2, y = v − u/ 2; ∂ (u, v ) 9. 1 4v 4u −1 = 4/(u2 + v 2 )2 5 1 2 ∂ (x, y ) 2 u + v, y = − u + v; = 9 9 9 9 ∂ (u, v ) √ ∂ (x, y ) = ∂ (u, v ) = cos u cos v + sin u sin v = cos(u − v ) 2(v 2 − u2 ) (u2 + v 2 )2 4uv 2 + v 2 )2 (u 6. x = ln u, y = uv ; 7. x = = −17 ∂ (x, y ) = ∂ (u, v ) 3u1/2 2v 1/2 − − 1/2 v 2u3/2 1 √√ 2 2 u+v 1 √√ 2 2 v−u u3/2 2v 3/2 1 = 1 =√ 2 − u2 4v 1 2v 2u1/2 v 1/2 =5 −u u − uw uw 0 −uv uv = u2 v ∂ (x, y, z ) = 11. y = v, x = u/y = u/v, z = w − x = w − u/v ; ∂ (u, v, w) 12. x = (v + w)/2, y = (u − w)/2, z = (u − v )/2, ∂ (x, y, z ) = ∂ (u, v, w) 1/v 0 −1/v 0 1/2 1/2 −u/v 2 1 u/v 2 1/2 0 −1/2 0 0 1 = 1/v 1/2 −1/2 0 =− 1 4 Exercise Set 16.8 600 y 13. y 14. (3, 4) 4 (0, 2) 3 2 1 x x (0, 0) (–1, 0) (0, 0) 2 (4, 0) 3 (1, 0) y 15. 1 3 y 16. (0, 3) 2 (2, 0) -3 x 1 3 x -3 17. x = 1 2 2 1 ∂ (x, y ) 11 1 u + v, y = − u + v, =; 5 5 5 5 ∂ (u, v ) 55 3 1 u dAuv = v 5 1 1 1 ∂ (x, y ) 11 1 u + v, y = u − v, =− ; 2 2 2 2 ∂ (u, v ) 22 veuv dAuv = 4 1 S 18. x = 2 1 2 S 3 u du dv = ln 3 v 2 1 4 1 veuv du dv = 1 0 14 (e − e − 3) 2 ∂ (x, y ) = −2; the boundary curves of the region S in the uv -plane are ∂ (u, v ) 1 u 1 sin u cos vdAuv = 2 sin u cos v dv du = 1 − sin 2 v = 0, v = u, and u = 1 so 2 2 0 0 19. x = u + v , y = u − v , S √ 1 ∂ (x, y ) = − ; the boundary curves of the region S in ∂ (u, v ) 2u 1 4 32 1 dAuv = uv 2 v du dv = 21 the uv -plane are u = 1, u = 3, v = 1, and v = 4 so 2u 21 1 20. x = v /u, y = uv so, from Example 3, S 21. x = 3u, y = 4v, ∂ (x, y ) = 12; S is the region in the uv -plane enclosed by the circle u2 + v 2 = 1. ∂ (u, v ) 2π r2 dr dθ = 96π 0 S 22. x = 2u, y = v, 1 12 u2 + v 2 (12) dAuv = 144 Use polar coordinates to obtain 0 ∂ (x, y ) = 2; S is the region in the uv -plane enclosed by the circle u2 + v 2 = 1. Use ∂ (u, v ) e−(4u polar coordinates to obtain S 2 +4v 2 ) 2π 1 (2) dAuv = 2 0 0 re−4r dr dθ = (1 − e−4 )π/2 2 601 Chapter 16 23. Let S be the region in the uv -plane bounded by u2 + v 2 = 1, so u = 2x, v = 3y, ∂ (x, y ) = ∂ (u, v ) x = u/2, y = v/3, 1 6 1/2 0 0 1/3 π /2 1 6 sin(u2 + v 2 )du dv = 1 r sin r2 dr dθ = 0 S 0 24. u = x/a, v = y/b, x = au, y = bv ; 25. x = u/3, y = v/2, z = w, = 1/6, use polar coordinates to get π (− cos r2 ) 24 ∂ (x, y ) = ab; A = ab ∂ (u, v ) 2π 1 = 0 π (1 − cos 1) 24 1 r dr dθ = πab 0 0 ∂ (x, y, z ) = 1/6; S is the region in uvw-space enclosed by the sphere ∂ (u, v, w) u2 + v 2 + w2 = 36 so 2π 1 u2 1 dVuvw = 96 54 6 (ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ 0 S = π 0 2π 1 54 0 π 6 ρ4 sin3 φ cos...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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