Solução_Calculo_Stewart_6e

F.
234 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
27. The given solid is the region inside the cylinder x 2 + y 2 =4between the surfaces z = 64 − 4x 2 − 4y 2
and z = − 64 − 4x 2 − 4y 2 .So
V =
64 − 4x2 − 4y 2 −
−
64 − 4x 2 − 4y 2 dA =
2 64 − 4x 2 − 4y 2 dA
x 2 + y 2 ≤ 4
=4 2π
0
x 2 +y 2 ≤ 4
2
√
0 16 − r2 rdrdθ=4 2π
dθ 2
r √ 16 − r
0 0 2 dr =4 θ 2π
0
=8π √
− 1 3 (12 3/2 − 16 2/3 )= 8π 3 64 − 24 3
− 1 3 (16 − r2 ) 3/2 2
0
29.
3
−3
√ 9−x 2
sin(x 2 + y 2 )dy dx =
0
π
3
0
0
sin r 2 rdrdθ
= π
0 dθ 3
0 r sin r 2 dr =[θ] π 0
−
1
2 cos r 2 3
0
= π − 1 2
(cos 9 − 1) =
π
(1 − cos 9)
2
31.
π/4
√ 2
(r cos θ + r sin θ) rdrdθ = π/4
(cos θ +sinθ) dθ √ 2
r 2 dr
0 0 0 0
=[sinθ − cos θ] π/4 1
3 r3√ 2
=
√2
0
2 − √ 2
2 − 0+1
· 1
3
0
2
√
2 − 0
=
2 √ 2
3
33. The surface of the water in the pool is a circular disk D with radius 20 ft. If we place D on coordinate axes with the origin at
35.
the center of D and define f(x, y) to be the depth of the water at (x, y), then the volume of water in the pool is the volume of
the solid that lies above D = (x, y) | x 2 + y 2 ≤ 400 and below the graph of f(x, y). We can associate north with the
positive y-direction, so we are given that the depth is constant in the x-direction and the depth increases linearly in the
y-direction from f(0, −20) = 2 to f(0, 20) = 7. The trace in the yz-plane is a line segment from (0, −20, 2) to (0, 20, 7).
The slope of this line is
7 − 2
= 1 , so an equation of the line is z − 7= 1 (y − 20) ⇒ z = 1 y + 9
20 − (−20) 8 8 8 2
.Sincef(x, y) is
independent of x, f(x, y) = 1 y + 9 . Thus the volume is given by 8 2
f(x, y) dA, which is most conveniently evaluated
D
using polar coordinates. Then D = {(r, θ) | 0 ≤ r ≤ 20, 0 ≤ θ ≤ 2π} and substituting x = r cos θ, y = r sin θ the integral
becomes
2π
0
20
0
1 r sin θ + 9 2π
8 2 rdrdθ= 1
0 24 r3 sin θ + 9 r2 r =20
4 r =0
Thus the pool contains 1800π ≈ 5655 ft 3 of water.
1
1/ √ 2
x
√1 − x 2 xy dy dx +
=
π/4
2
= 15
4
0 1
π/4
0
√ 2 x
1
r 3 cos θ sin θdrdθ=
0
= − 1000 cos θ + 900θ 2π
= 1800π
3 0
2
xy dy dx +
π/4
sin θ cos θdθ= 15 sin 2 θ
4 2
0
√ 4 − x 2
xy dy dx
√
2 0
r
4
r =2
4 cos θ sin θ dθ
π/4
0
= 15
16
r =1
dθ = 2π
1000
sin θ +900 dθ
0 3