Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 241
11. Here E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ √ x, 0 ≤ z ≤ 1+x + y},so
E 6xy dV = 1
0
= 1
0
√ x
1+x+y
6xy dz dy dx = 1
0 0 0
√ x
0
6xyz
z=1+x+y
z=0
dy dx = 1
√ x
0
3xy 2 +3x 2 y 2 +2xy 3 y= √ x
y=0
dx = 1
0 (3x2 +3x 3 +2x 5/2 ) dx =
6xy(1 + x + y) dy dx
0
1
x 3 + 3 4 x4 + 4 7 x7/2 = 65
28
0
13. E is the region below the parabolic cylinder z =1− y 2 and above the
square [−1, 1] × [−1, 1] in the xy-plane.
E x2 e y dV = 1
1
1−y
2
x 2 e y dz dy dx
−1 −1 0
= 1
1
−1 −1 x2 e y (1 − y 2 ) dy dx
= 1
−1 x2 dx 1
−1 (ey − y 2 e y ) dy
= 1
x3 1
3 −1 e y − (y 2 − 2y +2)e y 1
−1
integrate by
parts twice
= 1 3 (2)[e − e − e−1 +5e −1 ]= 8 3e
15. Here T = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y},so
T x2 dV = 1
1−x
1−x−y
x 2 dz dy dx = 1
1−x
0 0 0 0
= 1
0
= 1
0
= 1
0
1−x
(x 2 − x 3 − x 2 y)dy dx = 1
0 0
x 2 (1 − x) − x 3 (1 − x) − 1 2 x2 (1 − x) 2 dx
1
2 x4 − x 3 + 1 2 x2 dx = 1
10 x5 − 1 4 x4 + 1 6 x3 1
0
= 1
10 − 1 4 + 1 6 = 1 60
x 2 (1 − x − y)dy dx
0
x 2 y − x 3 y − 1 2 x2 y 2 y=1−x
dx
y=0
17. The projection E on the yz-plane is the disk y 2 + z 2 ≤ 1. Using polar
19. The plane 2x + y + z =4intersects the xy-plane when
2x + y +0=4 ⇒ y =4− 2x,so
E = {(x, y, z) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 4 − 2x, 0 ≤ z ≤ 4 − 2x − y} and
V = 2
0
= 2
0
= 2
0
4−2x
0
4−2x−y
dz dy dx = 2
4−2x
(4 − 2x − y) dy dx
0 0 0
4y − 2xy −
1
y2 y=4−2x
dx
2 y=0
4(4 − 2x) − 2x(4 − 2x) −
1
(4 − 2x)2 dx
2
= 2
0 (2x2 − 8x +8)dx = 2
3 x3 − 4x 2 +8x 2
= 16
0 3
coordinates y = r cos θ and z = r sin θ,weget
xdV =
4
xdx dA = 1 E D 4y 2 +4z 2 2 D 4 2 − (4y 2 +4z 2 ) 2 dA
=8 2π 1 (1 − 0 0 r4 ) rdrdθ=8 2π
dθ 1
(r − 0 0 r5 ) dr
=8(2π) 1
2 r2 − 1 r6 1
= 16π
6 0 3