Solução_Calculo_Stewart_6e

F.
288 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
15. (a) A cross-section is a washer with inner radius x 2 and outer radius x.
V = 1
π (x) 2 − (x 2 ) 2 dx = 1
0 0 π(x2 − x 4 ) dx = π 1
3 x3 − 1 x5 1
= π 1
− 1
5 0 3 5 =
2
π 15
(b) A cross-section is a washer with inner radius y and outer radius y.
V =
1
π 2
0 y − y
2
dy = 1
π(y − 0 y2 ) dy = π 1
2 y2 − 1 y3 1
= π 1
− 1
3 0 2 3 =
π
6
(c) A cross-section is a washer with inner radius 2 − x and outer radius 2 − x 2 .
V = 1
0 π (2 − x 2 ) 2 − (2 − x) 2 dx = 1
0 π(x4 − 5x 2 +4x) dx = π 1
5 x5 − 5 3 x3 +2x 2 1
0 = π 1
5 − 5 3 +2 = 8
15 π
17. (a) Using the Midpoint Rule on [0, 1] with f(x) =tan(x 2 ) and n =4,weestimate
A =
1
0 tan(x2 ) dx ≈ tan 1 1
2
+tan
3
2
+tan
5
2
+tan
7
2
≈ 1 (1.53) ≈ 0.38
4 8
8
8
8
4
(b) Using the Midpoint Rule on [0, 1] with f(x) =π tan 2 (x 2 ) (fordisks)andn =4,weestimate
V =
1
tan f(x) dx ≈ 1 π 2 1
2
+tan 2 3
2
+tan 2 5
2
+tan 2 7
2
≈ π (1.114) ≈ 0.87
0 4 8
8
8
8
4
19.
π/2
0
2πx cos xdx= π/2
0
(2πx)cosxdx
The solid is obtained by rotating the region R = (x, y) | 0 ≤ x ≤ π 2 , 0 ≤ y ≤ cos x about the y-axis.
21.
π
0
π(2 − sin x)2 dx
The solid is obtained by rotating the region R = {(x, y) | 0 ≤ x ≤ π, 0 ≤ y ≤ 2 − sin x} about the x-axis.
23. Takethebasetobethediskx 2 + y 2 ≤ 9. ThenV = 3
−3 A(x) dx,whereA(x 0) is the area of the isosceles right triangle
whose hypotenuse lies along the line x = x 0 in the xy-plane. The length of the hypotenuse is 2 √ 9 − x 2 and the length of
each leg is √ 2 √ 9 − x 2 . A(x) = 1 2
√
2
√
9 − x
2 2
=9− x 2 ,so
V =2 3
A(x) dx =2 3
(9 − 0 0 x2 ) dx =2 9x − 1 x3 3
=2(27− 9) = 36
3 0
25. Equilateral triangles with sides measuring 1 x meters have height 1 x sin 4 4 60◦ = √ 3
8
x. Therefore,
A(x) = 1 · 1 x · √3
x = √ 3
2 4 8 64 x2 . V = 20
A(x) dx = √
3 20
x 2 dx = √
3 1 x3 20
= 8000 √ 3
= 125 √ 3
m 3 .
0 64 0 64 3 0 64 · 3 3
27. f(x) =kx ⇒ 30 N = k(15 − 12) cm ⇒ k =10N/cm = 1000 N/m. 20 cm − 12 cm =0.08 m ⇒
W = 0.08
kx dx =1000 0.08
xdx=500 x 2 0.08
= 500(0.08) 2 =3.2 N·m =3.2 J.
0 0 0