26 use 0 0 and x y to get 27 show that opposite sides

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Unformatted text preview: = πa3 /3 6 595 Chapter 16 2π √ 9−r 2 2 6. V = 2 2π 2 r dz dr dθ = 2 0 9 − r2 dr dθ r 0 0 0 0 √ 2 (27 − 5 5) 3 = √ dθ = 4(27 − 5 5)π/3 2π 0 7. r2 + z 2 = 20 intersects z = r2 in a circle of radius 2, 2π √ 20−r 2 2 V= 2π 2 r dz dr dθ = 0 0 20 − r2 − r3 )dr dθ (r r2 0 0 √ 2π 4 (10 5 − 19) 3 = √ dθ = 8(10 5 − 19)π/3 0 8. z = hr/a intersects z = h in a circle of radius a, 2π a h 2π a h (ar − r2 )dr dθ = a r dz dr dθ = V= 0 hr/a 0 2π π /3 0 0 2π 4 π /3 ρ2 sin φ dρ dφ dθ = 9. V = 0 0 2π 0 π /4 0 0 2π 2 π /4 ρ2 sin φ dρ dφ dθ = 10. V = 0 0 1 0 0 2π 0 12 a h dθ = πa2 h/3 6 32 64 sin φ dφ dθ = 3 3 2π dθ = 64π/3 0 √ 7 7 sin φ dφ dθ = (2 − 2) 3 6 2π √ dθ = 7(2 − 2)π/3 0 11. In spherical coordinates the sphere and the plane z = a are ρ = 2a and ρ = a sec φ, respectively. They intersect at φ = π/3, 2π π /3 a sec φ 0 0 2π 0 π /3 = 0 = 2π 0 13 a 2 2π 13 a sec3 φ sin φ dφ dθ + 3 π /2 2π π /2 π/3 0 dθ = 11πa3 /3 2π π /2 ρ sin φ dρ dφ dθ = π/4 π /2 0 0 a2 −r 2 a 13. 0 π /2 π π /2 1 0 e−ρ ρ2 sin φ dρ dφ dθ = 3 0 π /2 √ 8 π /4 15. 0 0 2π π 3 ρ3 sin φ dρ dφ dθ = 81π 0 0 2π √ dθ = 9 2π 0 a 16 a 12 0 π /2 cos2 θ dθ = πa6 /48 0 1 (1 − e−1 ) 3 π 0 π /2 sin φ dφ dθ = (1 − e−1 )π/3 0 √ ρ4 cos2 φ sin φ dρ dφ dθ = 32(2 2 − 1)π/15 0 16. √ 92 9 sin φ dφ dθ = 2 (a2 r3 − r5 ) cos2 θ dr dθ 0 = 14. π/4 r3 cos2 θ dz dr dθ = 0 0 83 a sin φ dφ dθ 3 0 2 0 0 2π 3 12. V = 0 2a ρ2 sin φ dρ dφ dθ π/3 0 4 dθ + a3 3 0 2π 0 π /2 ρ2 sin φ dρ dφ dθ + V= Exercise Set 16.7 π /2 596 √ 1 2 37 cos θ cos φ dφ dθ = 18 36 π /4 18. 0 0 2π √ a2 −r 2 a 19. (a) V = 2 0 0 π cos37 θ dθ = 0 4,294,967,296 √ 2 ≈ 0.008040 755,505,013,725 r dz dr dθ = 4πa3 /3 0 2π π /2 a ρ2 sin φ dρ dφ dθ = 4πa3 /3 (b) V = 0 0 0 √ √ 4−x2 2 4−x2 −y 2 xyz dz dy dx 20. (a) 0 0 0 √ 4−x2 2 1 1 xy (4 − x2 − y 2 )dy dx = 2 8 = 0 0 π /2 √ 4−r 2 2 0 0 π /2 2 = 0 π /2 x(4 − x2 )2 dx = 4/3 0 r3 z sin θ cos θ dz dr dθ (b) 0 2 π /2 0 π /2 8 13 (4r − r5 ) sin θ cos θ dr dθ = 2 3 sin θ cos θ dθ = 4/3 0 2 ρ5 sin3 φ cos φ sin θ cos θ dρ dφ dθ (c) 0 0 0 π /2 π /2 = 0 2π 3 0 2π 3 3 27 1 r(3 − r)2 dr dθ = 2 8 (3 − z )r dz dr dθ = 21. M = 0 r 0 2π a 0 h 22. M = 0 2π a 1 12 kh r dr dθ = ka2 h2 2 4 k zr dz dr dθ = 0 0 2π 0 π 0 0 a 2π π kρ3 sin φ dρ dφ dθ = 23. M = 0 0 2π 0 π 0 0 2π 2 24. M = π ρ sin φ dρ dφ dθ = 0 0 π /2 8 32 sin3 φ cos φ sin θ cos θ dφ dθ = 3 3 1 0 0 sin θ cos θ dθ = 4/3 0 2π dθ = 27π/4 0 2π dθ = πka2 h2 /2 0 1 14 ka sin φ dφ dθ = ka4 4 2 3 sin φ dφ dθ = 3 2 2π dθ = πka4 0 2π dθ = 6π 0 25. x = y = 0 from the symmetry of the region, ¯¯ 2π √ 2−r 2 1 0 z= ¯ 2π 1 r dz dr dθ = V= 1 V r2 0 2π 0 √ 2−r 2 1 (r zr dz dr dθ = 0 centroid r2 0 0, 0, √ 2 − r2 − r3 )dr dθ = (8 2 − 7)π/6, 0 √ 6 (7π/12) = 7/(16 2 − 14); (8 2 − 7)π √ 7 √ 16 2 − 14 26. x = y = 0 from the symmetry of the region, V = 8π/3, ¯¯ z= ¯ 1 V 2π 2 2 zr dz dr dθ = 0 0 r 3 (4π ) = 3/2; centroid (0, 0, 3/2) 8π 597 Chapter 16 27. x = y = z from the symmetry of the region, V = πa3 /6, ¯¯¯ z= ¯ π /2 1 V π /2 a 6 (πa4 /16) = 3a/8; πa3 ρ3 cos φ sin φ dρ dφ dθ = 0 0 0 centroid (3a/8, 3a/8, 3a/8) 2π π /3 4 ρ2 sin φ dρ dφ dθ = 64π/3, 28. x = y = 0 from the symmetry of the region, V = ¯¯ 0 z= ¯ 2π 1 V π /3 4 ρ3 cos φ sin φ dρ dφ dθ = 0 0 0 0 0 3 (48π ) = 9/4; centroid (0, 0, 9/4) 64π π /2 r2 2 cos θ r dz dr dθ = 3π/2, 29. y = 0 from the symmetry of the region, V = 2 ¯ 0 π /2 2 x= ¯ V r2 2 cos θ r2 cos θ dz dr dθ = 0 2 z= ¯ V 0 0 π /2 r2 2 cos θ rz dz dr dθ = 0 0 0 π /2 π /2 2 cos θ zr dz dr dθ = 0 = 4 (π ) = 4/3, 3π 4 (5π/6) = 10/9; centroid (4/3, 0, 10/9) 3π 4−r 2 2 cos θ 30. M = 0 0 16 3 0 0 0 0 1 r(4 − r2 )2 dr dθ 2 π /2 (1 − sin6 θ)dθ = (16/3)(11π/32) = 11π/6 0 π /2 π /3 π /2 2 π /3 ρ2 sin φ dρ dφ dθ = 31. V = π/6 0 0 √ = 2( 3 − 1)π/3 2π π /4 π/6 0 2π 1 π /4 ρ3 sin φ dρ dφ dθ = 32. M = 0 0 0 0 0 4√ 8 sin φ dφ dθ = ( 3 − 1) 3 3 √ 1 1 sin φ dφ dθ = (2 − 2) 4 8 π /2 dθ 0 2π dθ = (2 − √ 2)π/4 0 33. x = y = 0 from the symmetry of density and region, ¯¯ 2π 1−r 2 1 M= 0 z= ¯ 0 0 2π 1 M 0 (r2 + z 2 )r dz dr dθ = π/4, 1−r 2 1 z (r2 +z 2 )r dz dr dθ = (4/π )(11π/120) = 11/30; center of gravity (0, 0, 11/30) 0 0 2π r 1 34. x = y = 0 from the symmetry of density and region, M = ¯¯ zr dz dr dθ = π/4, 0 z= ¯ 2π 1 M 0 0 r 1 z 2 r dz dr dθ = (4/π )(2π/15) = 8/15; center of gravity (0, 0, 8/15) 0 0 0 35. x = y = 0 from the symmetry of density and region, ¯¯ 2π π /2 a kρ3 sin φ dρ dφ dθ = πka4 /2, M= 0 z= ¯ 1 M 0 0 2π π /2 a kρ4 sin φ cos φ dρ dφ dθ = 0 0 0 2 (πka5 /5) = 2a/5; center of gravity (0, 0, 2a/5) πka4 Exercise Set 16.7 598 36. x = z = 0 from the symmetry of the region, V = 54π/3 − 16π/3 = 38π/3, ¯¯...
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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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