Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.2 VOLUMES ¤ 273
29. R 3 about AB (the line x =1):
1
1
V = A(y) dy = π(1 − y 2 ) 2 − π 1 − 3 2
y dy = π
0
1
0
1
0
(1 − 2y 2 + y 4 ) − (1 − 2y 1/3 + y 2/3 ) dy
= π (−2y 2 + y 4 +2y 1/3 − y 2/3 ) dy = π
− 2 3 y3 + 1 5 y5 + 3 2 y4/3 − 3 y5/3 1
5
= π − 2 + 1 + 3 − 3 3 5 2 5
0
0
Note: See the note in Exercise 27. For Exercises 21, 25, and 29, we have π + 7π + 13π =
3+14+13
10 15 30 30 π = π.
31. V = π
π/4
0
(1 − tan 3 x) 2 dx
=
13
30 π
33. V = π
= π
π
0
π
0
(1 − 0) 2 − (1 − sin x) 2 dx
1 2 − (1 − sin x) 2 dx
√ 8
2
35. V = π [3 − (−2)] 2 − y2 +1− (−2) dy
− √ 8
√
2 2
2
= π 5 2 − 1+y2 +2 dy
−2 √ 2
37. y =2+x 2 cos x and y = x 4 + x +1intersect at
x = a ≈ −1.288 and x = b ≈ 0.884.
V = π
b
a
[(2 + x 2 cos x) 2 − (x 4 + x +1) 2 ]dx ≈ 23.780