Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.3 VOLUMESBYCYLINDRICALSHELLS ¤ 281
27. V = 1
0 2πx √ 1+x 3 dx. Letf(x) =x √ 1+x 3 .
Then the Midpoint Rule with n =5gives
1
0
f(x) dx ≈
1−0
5
[f(0.1) + f(0.3) + f(0.5) + f(0.7) + f(0.9)]
≈ 0.2(2.9290)
Multiplying by 2π gives V ≈ 3.68.
29.
3
0 2πx5 dx =2π 3
0 x(x4 ) dx. The solid is obtained by rotating the region 0 ≤ y ≤ x 4 , 0 ≤ x ≤ 3 about the y-axis using
cylindrical shells.
1
31. 2π(3 − y)(1 − 0 y2 ) dy. The solid is obtained by rotating the region bounded by (i) x =1− y 2 , x =0,andy =0or
(ii) x = y 2 , x =1,andy =0about the line y =3using cylindrical shells.
33. From the graph, the curves intersect at x =0and x = a ≈ 0.56,
with √ x +1>e x on the interval (0,a). So the volume of the solid
obtained by rotating the region about the y-axis is
V =2π √
a
x 0 x +1 − e x dx ≈ 0.13.
35. V =2π
π/2
0
CAS
= 1 32 π3
π
2 − x sin 2 x − sin 4 x dx
37. Use shells:
V = 4
2 2πx(−x2 +6x − 8) dx =2π 4
2 (−x3 +6x 2 − 8x) dx
=2π − 1 4 x4 +2x 3 − 4x 2 4
2
=2π[(−64 + 128 − 64) − (−4+16− 16)]
=2π(4) = 8π
39. Use shells:
V = 4
2π[x − (−1)][5 − (x +4/x)] dx
1
=2π 4
(x + 1)(5 − x − 4/x) dx
1
=2π 4
1 (5x − x2 − 4+5− x − 4/x) dx
=2π 4
1 (−x2 +4x +1− 4/x) dx =2π − 1 3 x3 +2x 2 + x − 4lnx 4
1
=2π − 64 +32+4− 4ln4 − − 1 +2+1− 0
3 3
=2π(12 − 4ln4)=8π(3 − ln 4)