Solução_Calculo_Stewart_6e

F.
262 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
33. Using the wedge above the plane z =0and below the plane z = mx and noting that we have the same volume for m0 (so use m>0), we have
V =2 a/3 √ a 2 −9y 2
mx dx dy =2 a/3 1
0 0 0 2 m(a2 − 9y 2 ) dy = m a 2 y − 3y 3 a/3
= m 1
0
3 a3 − 1 a3 = 2 9 9 ma3 .
35. (a) m = 1
1−y
2
ydxdy= 1
(y − 0 0 0 y3 ) dy = 1 − 1 = 1 2 4 4
(b) M y = 1 1 − y
2
xy dx dy = 1 1
y(1 − 0 0 0 2 y2 ) 2 dy = − 1 (1 − 12 y2 ) 3 1
= 1 , 0 12
M x = 1 1 − y
2
y 2 dx dy = 1
0 0 0 (y2 − y 4 ) dy = 2 .Hence(x, y) = 1
, 8
15 3 15 .
(c) I x = 1 1−y
2
y 3 dx dy = 1
0 0 0 (y3 − y 5 ) dy = 1 , 12
I y = 1 1−y
2
yx 2 dx dy = 1 1
y(1 − 0 0 0 3 y2 ) 3 dy = − 1 (1 − 24 y2 ) 4 1
= 1 , 0 24
I 0 = I x + I y = 1 8 , y 2 = 1/12
1/4 = 1 3
⇒ y = 1 √
3
,andx 2 = 1/24
1/4 = 1 6
⇒ x = 1 √
6
.
37. The equation of the cone with the suggested orientation is (h − z) = h a
x2 + y 2 , 0 ≤ z ≤ h. ThenV = 1 3 πa2 h is the
volume of one frustum of a cone; by symmetry M yz = M xz =0;and
√
h−(h/a) x 2 +y 2 2π
a
M xy =
zdzdA=
0
0 0 0
x 2 +y 2 ≤a 2
= πh2
a 2
a
Hence the centroid is (x, y, z) = 0, 0, 1 4 h .
0
(h/a)(a−r)
rz dz dr dθ = π
(a 2 r − 2ar 2 + r 3 ) dr = πh2 a
4
a 2 2 − 2a4
3 + a4
= πh2 a 2
4 12
a
0
r h2
a 2 (a − r)2 dr
39. 3
√ 9−x 2
(x
0 −
√9−x 3 + xy 2 ) dy dx =
2
41. From the graph, it appears that 1 − x 2 = e x at x ≈−0.71 and at
x =0,with1 − x 2 >e x on (−0.71, 0). So the desired integral is
D y2 dA ≈ 0
1−x
2
y 2 dy dx
−0.71 e x
0 [(1 − −0.71 x2 ) 3 − e 3x ] dx
= 1 3
= 1 3
x − x 3 + 3 5 x5 − 1 7 x7 − 1 3 e3x 0
−0.71 ≈ 0.0512
3
0
√ 9−x 2
x(x
−
√9−x 2 + y 2 ) dy dx
2
= π/2
−π/2
3
0 (r cos θ)(r2 ) rdrdθ
= π/2
−π/2 cos θdθ 3
0 r4 dr
= sin θ π/2 1 r5 3
−π/2 5
0 =2· 1
5
(243) =
486
5
=97.2