1 1 1 x,so that dy = 2 x1/2 dx = dx = dx. 2y 2x Thus, 85 cos xdx = 85 cos y (2y dy ) = 170 y cos y dy . Now use integration by parts with u = y, dv = cos y dy, du = dy, v = sin y to get y cos y dy y sin y sin y dy = y sin y + cos y + C1 , = so 8cos = 170y
The cross-section of the base corresponding to the coordinate y has length x = 2 y. The corresponding equilateral triangle with side s has area 3 3 2 = (2 y ) . Therefore, A(y ) = s2 4 4 3 2 2 2 V = 0 A(y ) dy = 0 (2 y ) dy 4 32 3 2 1 2 4y 2y 2 + 3 y 3 0
A cross-section is a washer with inner radius 1 y 2 and outer radius 2 2y , so its area is A(y ) = (2y )2 ( 1 y 2 )2 = (4y 2 1 y 4 ). 2 4 4 4 V = 0 A(y ) dy = 0 (4y 2 1 y 4 ) dy 4 =
43 y 3
1 54 y0 20
=
256 3
256 5
=
512 15
4
3
2
1
0
2
4 x
6
8
Let R = 3r . A cross-section is shaded in the diagram. 2 A(x) = (2y )2 = 2 R2 x2 , so V =
R R
A(x) dx = 2
R 0
R 0
1 = 8 R2 x 3 x3
=8
4(r 2 x2 ) dx
23 R 3
=
16 3 R 3
= 144r 3
Volumes by Cylindrical Shells
Some volume problems are very difcult to handle by the methods of Section 6.2. For instance, lets consider the problem of nding the volume of the solid obtained by rotating about the y-axis the region bounded by y 2 x 2 x 3 a