Solução_Calculo_Stewart_6e

F.
244 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
31.
If D 1 , D 2 ,andD 3 are the projections of E on the xy-, yz-, and xz-planes, then
D 1 =
D 2 =
(x, y) | −2 ≤ x ≤ 2,x 2 ≤ y ≤ 4 =
(y, z) | 0 ≤ y ≤ 4, 0 ≤ z ≤ 2 − 1 y =
2
(x, y) | 0 ≤ y ≤ 4, − y ≤ x ≤
y ,
(y, z) | 0 ≤ z ≤ 2, 0 ≤ y ≤ 4 − 2z ,and
D 3 =
(x, z) | −2 ≤ x ≤ 2, 0 ≤ z ≤ 2 − 1 2 x2 = (x, z) | 0 ≤ z ≤ 2, − √ 4 − 2z ≤ x ≤ √
4 − 2z
Therefore E =
(x, y, z) | −2 ≤ x ≤ 2, x 2 ≤ y ≤ 4, 0 ≤ z ≤ 2 − 1 y 2
=
(x, y, z) | 0 ≤ y ≤ 4, − y ≤ x ≤
y, 0 ≤ z ≤ 2 − 1 y 2
= (x, y, z) | 0 ≤ y ≤ 4, 0 ≤ z ≤ 2 − 1 y, − y ≤ x ≤
y
2
= (x, y, z) | 0 ≤ z ≤ 2, 0 ≤ y ≤ 4 − 2z, − y ≤ x ≤
y
= (x, y, z) | −2 ≤ x ≤ 2, 0 ≤ z ≤ 2 − 1 2 x2 , x 2 ≤ y ≤ 4 − 2z
= (x, y, z) | 0 ≤ z ≤ 2, − √ 4 − 2z ≤ x ≤ √
4 − 2z, x 2 ≤ y ≤ 4 − 2z
Then
E f(x, y, z) dV = 2
4
2−y/2
f(x, y, z) dz dy dx = 4
√ y
2−y/2
−2 x 2 0 0 − √ f(x, y, z) dz dx dy
y 0
= 4
0
= 2
−2
2−y/2
0
2 − x
2 /2
0
√ y
− √ f(x, y, z) dx dz dy = 2
y 0
4−2z
4−2z
f(x, y, z) dy dz dx = 2
√ 4−2z
x 2 0 − √ 4−2z
0
√ y
− √ f(x, y, z) dx dy dz
y
4−2z
f(x, y, z) dy dx dz
x 2