Solução_Calculo_Stewart_6e

F.
23. V = 2 3 − 3
2
x
(6 − 3x − 2y) dy dx
0 0
SECTION 16.3 TX.10 DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 15.3 ¤ 229
= 2
0
6y − 3xy − y
2 y =3− 3 2 x
dx
= 2
0 6(3 −
3
= 2
0
y =0
x) − 3x(3 − 3 x) − (3 − 3 x)2 dx
2 2 2
9
4 x2 − 9x +9 dx = 3
4 x3 − 9 2 x2 +9x 2
=6− 0=6
0
25.
V = 2
−2
4
x 2 x 2 dy dx
= 2
−2 x2 y y=4
y=x 2 dx = 2
−2 (4x2 − x 4 ) dx
= 4
3 x3 − 1 x5 2
= 32 − 32 + 32 − 32 = 128
5 −2 3 5 3 5 15
27.
1
√ 1 − x 2 1
y
2
V =
ydydx=
0 0
0 2
1
1 − x 2
=
dx = 1
0 2
2 x −
1
x3 1
= 1
3 0 3
y =
√1 − x 2
y =0
dx
29. From the graph, it appears that the two curves intersect at x =0and
at x ≈ 1.213. Thus the desired integral is
D xdA≈ 1.213 3x − x
2
xdydx= 1.213
0 x 4 0
xy
y =3x − x
2
y = x 4
= 1.213
(3x 2 − x 3 − x 5 ) dx = x 3 − 1 0 4 x4 − 1 x6 1.213
6 0
≈ 0.713
31. The two bounding curves y =1− x 2 and y = x 2 − 1 intersect at (±1, 0) with 1 − x 2 ≥ x 2 − 1 on [−1, 1]. Withinthis
region, the plane z =2x +2y +10is above the plane z =2− x − y,so
V = 1
1−x
2
(2x +2y +10)dy dx − 1
−1 x 2 −1 −1
= 1
−1
1−x
2
x 2 −1
1−x
2
x 2 −1
(2x +2y +10− (2 − x − y)) dy dx
(2 − x − y) dy dx
= 1
1−x
2
(3x +3y +8)dy dx =
y=1−x
2
1
3xy + 3 −1 x 2 −1 −1
2 y2 +8y dx
y=x 2 −1
= 1
−1 3x(1 − x 2 )+ 3 (1 − 2 x2 ) 2 +8(1− x 2 ) − 3x(x 2 − 1) − 3 2 (x2 − 1) 2 − 8(x 2 − 1) dx
= 1
−1 (−6x3 − 16x 2 +6x +16)dx = − 3 2 x4 − 16
3 x3 +3x 2 +16x 1
−1
= − 3 2 − 16
3 +3+16+ 3 2 − 16 3 − 3+16= 64 3
dx