Solução_Calculo_Stewart_6e

F.
230 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
33. The solid lies below the plane z =1− x − y
or x + y + z =1and above the region
D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}
in the xy-plane. The solid is a tetrahedron.
35. The two bounding curves y = x 3 − x and y = x 2 + x intersect at the origin and at x =2,withx 2 + x>x 3 − x on (0, 2).
Using a CAS, we find that the volume is
V =
2
x
2 + x
0
x 3 − x
zdydx=
2
x
2 + x
0
x 3 − x
(x 3 y 4 + xy 2 ) dy dx = 13,984,735,616
14,549,535
37. The two surfaces intersect in the circle x 2 + y 2 =1, z =0and the region of integration is the disk D: x 2 + y 2 ≤ 1.
1
√
Using a CAS, the volume is (1 − x 2 − y 2 1−x 2
) dA =
(1 − x
D
−1 −
√1−x 2 − y 2 ) dy dx = π
2 2 .
39. Because the region of integration is
D = {(x, y) | 0 ≤ y ≤ √ x, 0 ≤ x ≤ 4} = (x, y) | y 2 ≤ x ≤ 4, 0 ≤ y ≤ 2
we have 4
√ x
f(x, y) dy dx = f(x, y) dA = 2
4
f(x, y) dx dy.
0 0 D 0 y 2
41. Because the region of integration is
D = (x, y) | − 9 − y 2 ≤ x ≤
9 − y 2 , 0 ≤ y ≤ 3
= (x, y) | 0 ≤ y ≤ √ 9 − x 2 , −3 ≤ x ≤ 3
f(x, y) dx dy =
we have
3
√ 9−y 2
f(x, y) dA
0 −
√9−y 2 D
3
√ 9−x 2
=
f(x, y) dy dx
−3 0
43. Because the region of integration is
D = {(x, y) | 0 ≤ y ≤ ln x, 1 ≤ x ≤ 2} = {(x, y) | e y ≤ x ≤ 2, 0 ≤ y ≤ ln 2}
45.
we have
2
ln x
ln 2
2
f(x, y) dy dx = f(x, y) dA =
1 0
D
0
f(x, y) dx dy
e y
1
3
3
x/3
3 y=x/3
e x2 dx dy = e x2 dy dx = e x2 y dx
y=0
0
3y
0
0
0
=
3
0
x
3
e x2 dx = 1 6 ex2 3
0
= e9 − 1
6