Solução_Calculo_Stewart_6e

F.
5. A =
2
=
0
sin
πx
2
− 2 π cos πx
2
− (x 2 − 2x) dx
− 1 3 x3 + x 2 2
0
= 2
π − 8 3 +4 − − 2 π − 0+0 = 4 3 + 4 π
TX.10
CHAPTER 6 REVIEW ¤ 287
7. Using washers with inner radius x 2 and outer radius 2x,wehave
V = π 2
0 (2x) 2 − (x 2 ) 2 dx = π 2
0 (4x2 − x 4 ) dx
= π 4
3 x3 − 1 x5 2
= π 32
−
32
5 0 3 5
=32π · 2
15 = 64
15 π
9. V = π
3 (9
−3 − y 2 ) − (−1) 2 − [0 − (−1)] 2 dy
=2π 3
0 (10 − y 2 ) 2 − 1 dy =2π 3
(100 − 0 20y2 + y 4 − 1) dy
=2π 3
0 (99 − 20y2 + y 4 ) dy =2π 99y − 20
3 y3 + 1 5 y5 3
0
=2π 297 − 180 + 243
5
=
1656
5 π
11. The graph of x 2 − y 2 = a 2 is a hyperbola with right and left branches.
Solving for y gives us y 2 = x 2 − a 2 ⇒ y = ± √ x 2 − a 2 .
We’ll use shells and the height of each shell is
√
x2 − a 2 − − √ x 2 − a 2 =2 √ x 2 − a 2 .
The volume is V = a+h
a
2πx · 2 √ x 2 − a 2 dx. Toevaluate,letu = x 2 − a 2 ,
so du =2xdxand xdx= 1 2
du. Whenx = a, u =0,andwhenx = a + h,
u =(a + h) 2 − a 2 = a 2 +2ah + h 2 − a 2 =2ah + h 2 .
2ah+h
2
√
2ah+h
2
1 2
Thus, V =4π
u
0
2 du =2π
3 u3/2 = 4
0
3 π 2ah + h 2 3/2
.
13. A shell has radius π 2 − x, circumference 2π π
2 − x , and height cos 2 x − 1 4 .
y =cos 2 x intersects y = 1 4 when cos2 x = 1 4
cos x = ± 1 2
[ |x| ≤ π/2] ⇔ x = ± π 3 .
V =
π/3
−π/3
π
2π
2 − x cos 2 x − 1
dx
4
⇔