Solução_Calculo_Stewart_6e

F.
282 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
41. Use disks: x 2 +(y − 1) 2 =1 ⇔ x = ± 1 − (y − 1) 2
V = π
2
0
2
2
1 − (y − 1)
2 dy = π (2y − y 2 ) dy
= π y 2 − 1 3 y3 2
0 = π 4 − 8 3
=
4
3 π
43. Use shells:
V =2 r
2πx √ r
0 2 − x 2 dx
= −2π r
0 (r2 − x 2 ) 1/2 (−2x) dx
r
= −2π · 2
3 (r2 − x 2 ) 3/2 0
45. V =2π
6.4 Work
= − 4 3 π(0 − r3 )= 4 3 πr3
r
0
=2πh
− x3
3r + x2
2
x
− h
r x + h dx =2πh
r
=2πh r2
0
6 = πr2 h
3
1. W = Fd = mgd =(40)(9.8)(1.5) = 588 J
3. W =
b
a
9
f(x) dx =
0
10
10
(1 + x) dx =10 2
TX.10
0
r
− x2
0 r + x dx
1
1
− 1
100 400 =
25
24 m =10.8 cm 2500
1
−
u du [u =1+x, du = dx] =10 1 10
=10 − 1
2 u
+1 =9ft-lb
10
1
5. The force function is given by F (x) (in newtons) and the work (in joules) is the area under the curve, given by
8 F (x) dx = 4
F (x) dx + 8
F (x) dx = 1 (4)(30) + (4)(30) = 180 J.
0 0 4 2
7. 10 = f(x) =kx = 1 k [4 inches = 1 foot], so k =30lb/ft and f(x) =30x. Now6 inches = 1 3 3 2
foot, so
W = 1/2
30xdx= 15x 2 1/2
= 15 ft-lb.
0 0 4
9. (a) If 0.12
kx dx =2J, then 2= 1
kx2 0.12
0 2
= 1 2
0 2
k(0.0144) = 0.0072k and k = = 2500
0.0072 9
≈ 277.78 N/m.
Thus, the work needed to stretch the spring from 35 cm to 40 cm is
0.10
0.05
2500
xdx= 1250
x2 1/10
= 1250
9 9 1/20 9
(b) f(x) =kx,so30 = 2500
270
9
x and x =
≈ 1.04 J.