Solução_Calculo_Stewart_6e

F.
272 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
17. y = x 2 ⇒ x = √ y for x ≥ 0. The outer radius is the distance from x = −1 to x = y and the inner radius is the
distance from x = −1 to x = y 2 .
V =
= π
1
= π
0
1
0
2
π y − (−1) − y 2 − (−1) 2
dy = π
y +2
y +1− y 4 − 2y 2 − 1 dy = π
1
2 y2 + 4 3 y3/2 − 1 5 y5 − 2 3 y3 1
0
1
0
1
= π 1
+ 4 − 1 − 2
2 3 5 3 =
29
π 30
0
y +1
2
− (y 2 +1) 2
dy
y +2 y − y 4 − 2y 2 dy
19. R 1 about OA (the line y =0): V =
21. R 1 about AB (the line x =1):
V =
1
0
A(y) dy =
= π 1 − 3 2 + 3 5
1
0
=
π
10
1
0
A(x) dx =
0
1
0
π(x 3 ) 2 dx = π
1
0
1 1
x 6 dx = π
7 x7 = π
0
7
π 1 − 3 2
1
y dy = π (1 − 2y 1/3 + y 2/3 ) dy = π
y − 3 2 y4/3 + 3 y5/3 1
5
23. R 2 about OA (the line y =0):
1
1
√
2
V = A(x) dx = π(1) 2 − π x dx = π
0
25. R 2 about AB (the line x =1):
1
V =
0
= π
A(y) dy =
2
3 y3 − 1 5 y5 1
0
0
1
0
1
0
(1 − x) dx = π x − 1 x2 1
= π
2
1 − 1 0 2 =
π
2
π(1) 2 − π(1 − y 2 ) 2 1
dy = π 1 − (1 − 2y 2 + y 4 ) 1
dy = π (2y 2 − y 4 ) dy
= π 2
3 − 1 5
=
7
15 π
27. R 3 about OA (the line y =0):
1
1
√
2
V = A(x) dx = π x − π(x 3 ) 2 dx = π
0
0
0
1
0
(x − x 6 ) dx = π 1
2 x2 − 1 x7 1
= π 1
− 1
7 0 2 7 =
5
π. 14
Note: Let R = R 1 + R 2 + R 3 .IfwerotateR about any of the segments OA, OC, AB,orBC, we obtain a right circular
cylinder of height 1 and radius 1. Its volume is πr 2 h = π(1) 2 · 1=π. As a check for Exercises 19, 23, and 27, we can add the
answers, and that sum must equal π. Thus, π + π + 5π =
2+7+5
7 2 14 14 π = π.
0
0