Solução_Calculo_Stewart_6e

F.
228 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
7.
9.
11.
D y2 dA = 1
y
−1 −y−2 y2 dx dy = 1
−1 xy
2 x=y
dy = 1
x=−y−2 −1 y2 [y − (−y − 2)] dy
= 1
−1 (2y3 +2y 2 )dy = 1
2 y4 + 2 y3 1
= 1 + 2 − 1 + 2 = 4
3 −1 2 3 2 3 3
D xdA= π
sin x
xdydx= π
0 0 0 [xy]y=sin x
y=0
dx =
π
x sin xdx 0
= −x cos x +sinx π
= −π cos π +sinπ +0− sin 0 = π
0
D y2 e xy dA = 4
0
=
y
0 y2 e xy dx dy = 4
0 ye
xy x=y
dy =
4
ye y2 − y dy
x=0 0
4
1
2 ey2 − 1 2 y2 = 1 2 e16 − 8 − 1 +0= 1 2 2 e16 − 17
2
0
integrate by parts
with u = x, dv =sinxdx
13.
1
x
2
0
0
x cos ydydx= 1
0
y = x
2
x sin y dx = 1
x sin y =0 0 x2 dx = − 1 cos x2 1
= 1 (1 − cos 1)
2 0 2
15. 2
1
2y−1
2−y
y 3 dx dy =
2
1
xy 3 x=2y−1
dy =
x=2−y
2
= 2
1 (3y4 − 3y 3 ) dy = 3
5 y5 − 3 4 y4 2
1
= 96 5 − 12 − 3 5 + 3 4 = 147
20
1
[(2y − 1) − (2 − y)] y 3 dy
17. 2
√ 4−x 2
(2x − y) dy dx
−2 −
√4−x 2 √
2
=
2xy − 1 y2 y= 4−x 2
2
√4−x dx
2
−2
y=−
= 2
√
−2 2x 4 − x2 − 1 2 4 − x
2
+2x √
4 − x 2 + 1 2 4 − x
2
dx
= 2
4x √
4 − x
−2 2 dx = − 4 3 4 − x
2 3/2
2
=0
−2
[Or, note that 4x √ 4 − x 2 is an odd function, so 2
−2 4x √ 4 − x 2 dx =0.]
19. V = 1
x
(x +2y) dy dx
0 x 4
= 1
0 xy + y
2 y=x
dx = 1
y=x 4 0 (2x2 − x 5 − x 8 ) dx
= 2
3 x3 − 1 6 x6 − 1 9 x9 1
0 = 2 3 − 1 6 − 1 9 = 7 18
21. V = 2 7 − 3y
xy dx dy = 2
1
1 1 1 2 x2 y x =7− 3y
dy
x =1
= 1 2
2
1 (48y − 42y2 +9y 3 ) dy
= 1 2
24y 2 − 14y 3 + 9 y4 2
= 31
4 1 8