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 xaxis, 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.
 Spring '14
 The Land

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