Solução_Calculo_Stewart_6e

F.
9. A cross-section is a washer with inner radius y 2
and outer radius 2y, so its area is
A(y) =π(2y) 2 − π(y 2 ) 2 = π(4y 2 − y 4 ).
V =
2
0
TX.10
2
(4y 2 − y 4 ) dy
0
=
64
π 15
1 − √ 2
x
1 − 2 √
x + x
.
1
−3x + x 2 +2 √
x dx
0
=
π
6
π/3
=2π 4π
− √ 3 − 0
3
0
1
π(1 − y 2 ) 2 dy
0
π 15
A(y) dy = π
= π 4
3 y3 − 1 5 y5 2
0 = π 32
3 − 32 5
11. A cross-section is a washer with inner radius 1 − √ x and outer radius 1 − x, so its area is
A(x) =π(1 − x) 2 − π
= π (1 − 2x + x 2 ) −
= π −3x + x 2 +2 √ x
V =
1
0
A(x) dx = π
1
= π − 3 2 x2 + 1 3 x3 + 4 3 x3/2 = π − 3 + 5 2 3
0
SECTION 6.2 VOLUMES ¤ 271
13. A cross-section is a washer with inner radius (1 + sec x) − 1=secx and outer radius 3 − 1=2, so its area is
A(x) =π 2 2 − (sec x) 2 = π(4 − sec 2 x).
π/3
π/3
V = A(x) dx = π(4 − sec 2 x) dx
−π/3
−π/3
=2π
π/3
0
=2π 4x − tan x
=2π 4π
3 − √ 3
1
15. V = π(1 − y 2 ) 2 dy =2
−1
=2π
1
0
(4 − sec 2 x) dx [by symmetry]
(1 − 2y 2 + y 4 ) dy
=2π y − 2 3 y3 + 1 5 y5 1
0
=2π · 8
15 = 16