Solução_Calculo_Stewart_6e

F.
242 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
3
21. V =
−3
√ 9−x 2 5−y
√
− 9−x 2 1
3
√ 9−x 2
3
dz dy dx =
(5 − y − 1) dy dx =
4y −
−3 −
√9−x 1 y2 y=
2 2
−3
= 3
8 √ 9 − x
−3 2 dx =8 √
x
2 9 − x2 + 9 sin−1
x 3
2 3 −3
=8 9
2 sin−1 (1) − 9 2 sin−1 (−1) =36 π
2 − − π 2
=36π
Alternatively, use polar coordinates to evaluate the double integral:
3
√ 9−x 2
2π
3
(4 − y) dy dx = (4 − r sin θ) rdrdθ
−3 −
√9−x 2 0 0
= 2π
0 2r 2 − 1 3 r3 sin θ r=3
dθ r=0
= 2π
(18 − 9sinθ) dθ
0
=18θ +9cosθ
2π
0
=36π
23. (a) The wedge can be described as the region
D = (x, y, z) | y 2 + z 2 ≤ 1, 0 ≤ x ≤ 1, 0 ≤ y ≤ x
=
(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤
1 − y 2
So the integral expressing the volume of the wedge is
D dV = 1 x
√ 1 − y 2
dz dy dx.
0 0 0
(b) A CAS gives 1 x
√ 1 − y 2
dz dy dx = π − 1 .
0 0 0 4 3
(Or use Formulas 30 and 87 from the Table of Integrals.)
25. Here f(x, y, z) =
using trigonometric substitution or
Formula 30 in the Table of Integrals
1
and ∆V =2· 4 · 2=16, so the Midpoint Rule gives
ln(1 + x + y + z)
B f(x, y, z) dV ≈ l
m
n
i=1 j=1 k=1
TX.10
f x i , y j , z k
∆V
√
y=−
= 16[f(1, 2, 1) + f(1, 2, 3) + f(1, 6, 1) + f(1, 6, 3)
+ f(3, 2, 1) + f(3, 2, 3) + f(3, 6, 1) + f(3, 6, 3)]
9−x 2
√9−x 2 dx
=16 1
ln 5 + 1
ln 7 + 1
ln 9 + 1
ln 11 + 1
ln 7 + 1
ln 9 + 1
ln 11 + 1
ln 13
≈ 60.533
27. E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x, 0 ≤ y ≤ 2 − 2z},
the solid bounded by the three coordinate planes and the planes
z =1− x, y =2− 2z.