Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.3 VOLUMESBYCYLINDRICALSHELLS ¤ 279
9. V = 2
2πy(1 + 1 y2 ) dy =2π 2
(y + 1 y3 ) dy =2π 1
2 y2 + 1 y4 2
4 1
=2π (2 + 4) − 1
+ 1
2 4 =2π 21
4 =
21
π 2
8
11. V =2π y( 3
y − 0) dy
0
8
=2π y 4/3 3
dy =2π y7/3 8
7
0
0
= 6π 7 (87/3 )= 6π 7 (27 )= 768
7 π
13. Theheightoftheshellis2 − 1+(y − 2) 2 =1− (y − 2) 2 =1− y 2 − 4y +4 = −y 2 +4y − 3.
V =2π 3
1 y(−y2 +4y − 3) dy
=2π 3
1 (−y3 +4y 2 − 3y) dy
=2π − 1 4 y4 + 4 3 y3 − 3 y2 3
2 1
=2π − 81 +36− 27
4 2 − −
1
+ 4 −
3
4 3 2
=2π
8
3 =
16
π 3
15. The shell has radius 2 − x, circumference 2π(2 − x), and height x 4 .
V = 1
2π(2 − 0 x)x4 dx
=2π 1
0 (2x4 − x 5 ) dx
=2π 2
5 x5 − 1 x6 1
6 0
=2π 2
− 1
5 6 − 0 =2π 7
30 =
7
π 15