# D e f 7 e is to the left of f case 2 6x 7 3

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Unformatted text preview: 3) = 4/3, G z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9) G 21. x = y = z from the symmetry of the region, V = πa3 /6, ¯¯¯ √2 2 2 √ √ a2 −x2 a −x −y a2 −x2 1a 1a x= ¯ x dz dy dx = x a2 − x2 − y 2 dy dx V00 V00 0 = 1 V π /2 a r2 0 0 a2 − r2 cos θ dr dθ = 6 (πa4 /16) = 3a/8; centroid (3a/8, 3a/8, 3a/8) πa3 Exercise Set 16.6 592 22. x = y = 0 from the symmetry of the region, V = 2πa3 /3 ¯¯ √2 2 2 √ √ a2 −x2 a −x −y a2 −x2 1a 1a 12 z= ¯ (a − x2 − y 2 )dy dx z dz dy dx = √ √ V −a − a2 −x2 0 V −a − a2 −x2 2 = 2π 1 V a 0 3 12 (a − r2 )r dr dθ = (πa4 /4) = 3a/8; centroid (0, 0, 3a/8) 2 2πa3 0 a a a (a − x)dz dy dx = a4 /2, y = z = a/2 from the symmetry of density and ¯¯ 23. M = 0 0 0 a 1 region, x = ¯ M a a x(a − x)dz dy dx = (2/a4 )(a5 /6) = a/3; 0 0 0 mass a4 /2, center of gravity (a/3, a/2, a/2) a √ a2 −x2 −a √ − a2 −x2 24. M = h (h − z )dz dy dx = 0 1 M and region, z = ¯ z (h − z )dV = 1 22 πa h , x = y = 0 from the symmetry of density ¯¯ 2 2 (πa2 h3 /6) = h/3; πa2 h2 G 22 mass πa h /2, center of gravity (0, 0, h/3) 1 1−y 2 1 yz dz dy dx = 1/6, x = 0 by the symmetry of density and region, ¯ 25. M = −1 y= ¯ 0 0 1 M y 2 z dV = (6)(8/105) = 16/35, z = ¯ 1 M G yz 2 dV = (6)(1/12) = 1/2; G mass 1/6, center of gravity (0, 16/35, 1/2) 9−x2 3 1 xz dz dy dx = 81/8, x = ¯ 26. M = 0 0 y= ¯ 0 1 M 1 M x2 z dV = (8/81)(81/5) = 8/5, G xyz dV = (8/81)(243/8) = 3, z = ¯ G 1 M xz 2 dV = (8/81)(27/4) = 2/3; G mass 81/8, center of gravity (8/5, 3, 2/3) 1 1 k (x2 + y 2 )dy dx = 2k/3, x = y from the symmetry of density and region, ¯¯ 27. (a) M = 0 x= ¯ 0 1 M kx(x2 + y 2 )dA = 3 (5k/12) = 5/8; center of gravity (5/8, 5/8) 2k R (b) y = 1/2 from the symmetry of density and region, ¯ 1 1 M= kx dy dx = k/2, x = ¯ 0 0 center of gravity (2/3, 1/2) 1 M kx2 dA = (2/k )(k/3) = 2/3, R 593 Chapter 16 28. (a) x = y = z from the symmetry of density and region, ¯¯¯ 1 1 1 k (x2 + y 2 + z 2 )dz dy dx = k , M= 0 0 0 1 M x= ¯ kx(x2 + y 2 + z 2 )dV = (1/k )(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12) G (b) x = y = z from the symmetry of density and region, ¯¯¯ 1 1 1 k (x + y + z )dz dy dx = 3k/2, M= 0 0 0 1 M x= ¯ kx(x + y + z )dV = 2 (5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9) 3k G π 1/(1+x2 +y 2 ) sin x dV = 29. V = dz dy dx = 0.666633, 0 G 1 x= V 0 0 xdV = 1.177406, y = 1 V G ydV = 0.353554, z = 1 V G zdV = 0.231557 G 30. (b) Use cylindrical coordinates to get 2π r dz dr dθ = π ln(1 + a2 ), 0 G 1 V z= 1/(1+r 2 ) a dV = V= zdV = 0 0 a2 2(1 + a2 ) ln(1 + a2 ) G (c) ¯ lim+ z = a→0 1 ; lim z = 0; solve z = 1/4 for a to obtain a ≈ 1.980291. ¯ 2 a→+∞ 31. Let x = r cos θ, y = r sin θ, and dA = r dr dθ in formulas (10) and (11). 2π a(1+sin θ ) r dr dθ = 3πa2 /2, 32. x = 0 from the symmetry of the region, A = ¯ 0 y= ¯ 1 A 2π a(1+sin θ ) r2 sin θ dr dθ = 0 0 0 2 (5πa3 /4) = 5a/6; centroid (0, 5a/6) 3πa2 π /2 sin 2θ r dr dθ = π/8, 33. x = y from the symmetry of the region, A = ¯¯ 0 x= ¯ 1 A π /2 0 sin 2θ r2 cos θ dr dθ = (8/π )(16/105) = 0 0 128 ; centroid 105π 128 128 , 105π 105π 34. x = 3/2 and y = 1 from the symmetry of the region, ¯ ¯ x dA = xA = (3/2)(6) = 9, ¯ R y dA = y A = (1)(6) = 6 ¯ R ¯ 35. x = 0 from the symmetry of the region, πa2 /2 is the area of the semicircle, 2π y is the distance ¯ ¯¯ traveled by the centroid to generate the sphere so 4πa3 /3 = (πa2 /2)(2π y ), y = 4a/(3π ) Exercise Set 16.7 594 12 πa 2 36. (a) V = 2π a + 4a 3π = 1 π (3π + 4)a3 3 √ 2 (b) the distance between the centroid and the line is 2 √ 4a 1√ 2 12 πa a+ = V= 2π 2π (3π + 4)a3 2 2 3π 6 a+ 4a 3π so 37. x = k so V = (πab)(2πk ) = 2π 2 abk ¯ 38. y = 4 from the symmetry of the region, ¯ 8−x2 2 A= dy dx = 64/3 so V = (64/3)[2π (4)] = 512π/3 −2 x2 1 πab2 when it is revolved about the x-axis, the area of the 3 1 1 1 1 ab (2π y ), y = b/3. A cone of volume πa2 b is generated when the ¯¯ region is ab so πab2 = 2 3 2 3 1 1 ab (2π x), x = a/3. The centroid is (a/3, b/3). ¯¯ region is revolved about the y -axis so πa2 b = 3 2 39. The region generates a cone of volume a b y 2 δ dy dx = 40. Ix = 0 0 a b Iz = 0 0 a 1 δab3 , Iy = 3 b x2 δ dy dx = 0 0 1 (x2 + y 2 )δ dy dx = δab(a2 + b2 ) 3 2π a 2π a r3 sin2 θ δ dr dθ = δπa4 /4; Iy = 41. Ix = 0 13 δa b, 3 0 r3 cos2 θ δ dr dθ = δπa4 /4 = Ix ; 0 0 Iz = Ix + Iy = δπa4 /2 EXERCISE SET 16.7 2π √ 1−r 2 1 1. 2π 1 zr dz dr dθ = 0 0 0 π /2 0 0 r2 cos θ 1 (1 − r2 )r dr dθ = 2 π /2 0 0 0 π /2 π /2 π /2 r3 sin θ dr dθ = 0 0 0 π /2 1 π /2 ρ3 sin φ cos φ dρ dφ dθ = 3. 0 0 0 π /4 a sec φ 2π 0 0 2π π /4 ρ2 sin φ dρ dφ dθ = 4. 0 0 0 2π 3 0 2π 9 0 0 r2 0 13 a sec3 φ sin φ dφ dθ = 3 2π r(9 − r2 )dr dθ = 0 0 1 cos4 θ sin θ dθ = 1/20 4 1 sin φ cos φ dφ dθ = 4 3 r dz dr dθ = 5. V = 0 1 dθ = π/4 8 cos θ r sin θ dz dr dθ = 2. 2π 0 π /2 0 2π 0 81 dθ = 81π/2 4 1 dθ = π/16 8 13 a dθ...
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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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