Solução_Calculo_Stewart_6e

F.
274 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
39. V = π
π
0
CAS
= 11 8 π2
sin 2 x − (−1) 2
− [0 − (−1)]
2 dx
41. π π/2
cos 2 xdxdescribes the volume of the solid obtained by rotating the region R = (x, y) | 0 ≤ x ≤ π 0 2
, 0 ≤ y ≤ cos x
43. π
of the xy-plane about the x-axis.
1
0
(y 4 − y 8 ) dy = π
1
0
(y 2 ) 2 − (y 4 ) 2 dy describes the volume of the solid obtained by rotating the region
R = (x, y) | 0 ≤ y ≤ 1,y 4 ≤ x ≤ y 2 of the xy-plane about the y-axis.
45. There are 10 subintervals over the 15-cm length, so we’ll use n =10/2 =5for the Midpoint Rule.
V = 15
0
A(x) dx ≈ M 5 = 15−0
5
[A(1.5) + A(4.5) + A(7.5) + A(10.5) + A(13.5)]
= 3(18 + 79 + 106 + 128 + 39) = 3 · 370 = 1110 cm 3
47. (a) V = 10
π [f(x)] 2 dx ≈ π 10 − 2
2 4 [f(3)] 2 +[f(5)] 2 +[f(7)] 2 +[f(9)] 2
≈ 2π (1.5) 2 +(2.2) 2 +(3.8) 2 +(3.1) 2 ≈ 196 units 3
(b) V = 4
π (outer radius) 2 − (inner radius) 2 dy
0
≈ π 4 − 0
4 (9.9) 2 − (2.2) 2 + (9.7) 2 − (3.0) 2 + (9.3) 2 − (5.6) 2 + (8.7) 2 − (6.5) 2
≈ 838 units 3
49. We’ll form a right circular cone with height h and base radius r by
revolving the line y = r x about the x-axis.
h
V = π
h
0
r
h x 2dx
= π
h
= π r2
h 2 1
3 h3
= 1 3 πr2 h
0
r 2
h 2 x2 dx = π r2
h 2 1
3 x3 h
0
Another solution: Revolve x = − r y + r about the y-axis.
h
h
V = π − r 2dy
h
h y + r = ∗ r
2
π
h 2 y2 − 2r2
h y + r2 dy
0
0
r
2
h
= π
3h 2 y3 − r2
h y2 + r y 2 = π 1
3 r2 h − r 2 h + r 2 h = 1 3 πr2 h
0
∗ Or use substitution with u = r − r h y and du = − r dy to get
h
0
π u 2 − h
r r du = −π h 0 1
r 3 u3 = −π h − 1
r
r 3 r3 = 1 3 πr2 h.