Solução_Calculo_Stewart_6e

F.
270 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
6.2 Volumes
1. A cross-section is a disk with radius 2 − 1 2 x, so its area is A(x) =π 2 − 1 2 x2 .
V =
= π
2
1
2
1
A(x) dx =
2
4 − 2x +
1
4 x2 dx
= π 4x − x 2 + 1
12 x3 2
1
= π 8 − 4+ 8
12
= π 1+ 7 12
1
=
19
12 π
π 2 − 1 2 x2 dx
−
4 − 1+
1
12
3. A cross-section is a disk with radius 1/x, so its area is
A(x) =π(1/x) 2 .
V =
2
1
A(x) dx =
2
1
2 1
π dx
x
2
1
= π
−
1 x dx = π 1 2 2 x
1
= π − 1 2 − (−1) = π 2
5. A cross-section is a disk with radius 2 y, so its area is
A(y) =π 2 2.
y
V =
9
0
A(y) dy =
9
=4π 1
y2 9
=2π(81) = 162π
2 0
0
π 2 2
9
y dy =4π
7. A cross-section is a washer (annulus) with inner
radius x 3 and outer radius x, so its area is
A(x) =π(x) 2 − π(x 3 ) 2 = π(x 2 − x 6 ).
V =
1
0
A(x) dx =
1
0
π(x 2 − x 6 ) dx
= π 1
3 x3 − 1 7 x7 1
0 = π 1
3 − 1 7
=
4
TX.10
ydy
0
π 21