Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 16 REVIEW ET CHAPTER 15 ¤ 261
17.
D
y
1
1+x dA = 2
= 1 2
√ x
0 0
1
0
y
1
1+x dy dx = 1 1 2 1+x y2 y= √ x
dx
2 2 y=0
x
1+x 2 dx = 1
4 ln(1 + x2 ) 1
0 = 1 4 ln 2
0
19.
D ydA= 2
8−y
2
ydxdy
0 y 2
= 2
0 y x x=8−y 2
x=y 2 dy = 2
0 y(8 − y2 − y 2 ) dy
= 2
0 (8y − 2y3 ) dy = 4y 2 − 1 2 y4 2
0 =8
21.
D
x 2 + y 2 3/2
dA =
π/3
3
0 0
(r 2 ) 3/2 rdrdθ
23.
25.
27.
E xy dV = 3
x
x+y
xy dz dy dx = 3
x xy z z=x+y
0 0 0 0 0 z=0
= 3
0
= 5 6
E y2 z 2 dV = 1
−1
x
0 (x2 y + xy 2 ) dy dx = 3
0
3
0 x4 dx = 1
6 x5 3
0 = 81
2 =40.5
= 2π
0
= 2π
0
E yz dV = 2
−2
29. V = 2
0
= 16
5
√ √
1−y2
−
1−y 2 1 − y
2 − z
2
=
= π 3
π/3
1
2 x2 y 2 + 1 xy3 y=x
dx = 3
3 y=0 0
0
y 2 z 2 dx dz dy = 1
−1
1
0 (r2 cos 2 θ)(r 2 sin 2 θ)(1 − r 2 ) rdrdθ= 2π
0
0
dθ
3
0
3 5
5 = 81π
5
r 4 dr = θ 3
π/3 1
0
5 r5 0
dy dx = 3
x
xy(x + y) dy dx
0 0
1
2 x4 + 1 x4 dx
3
√ 1−y 2
√1−y 2 y2 z 2 (1 − y 2 − z 2 ) dz dy
−
1
0
1
4 sin2 2θ(r 5 − r 7 ) dr dθ
1
(1 − cos 4θ) 1
8 6 r6 − 1 r8 r=1
dθ = 1
8 r=0 192 θ −
1
sin 4θ 2π
4
= 2π = π 0 192 96
√ 4−x 2
0
π
0 sin3 θdθ= 16
5
4
1 (x2 +4y 2 ) dy dx = 2
y yz dz dy dx = 2
√ 4−x 2 1
0 −2 0 2 y3 dy dx = π 2
0 0
0
− cos θ +
1
3 cos3 θ π
= 64
0 15
x 2 y + 4 3 y3 y=4
y=1 dx = 2
0 (3x2 +84)dx = 176
1
2 r3 (sin 3 θ) rdrdθ
31.
V = 2 y
(2−y)/2
dz dx dy = 2 y
0 0 0 0 0 1 −
1
y dx dy
2
= 2
0 y −
1
y2 dy = 2 2 3