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 s
x2 + y 2 = r 2 V = =
r (r 2 r h
x2 = r 2 y 2 y 2 ) dy = r 2 y r 3
3
y3 3
r
r h
r3
r 2 (r h)
(r h)3 3
2 1 = 3 r 3 3 (r h) 3r 2 (r h)2 1 = 3 cfw_2r 3 (r h)[3r 2 (r 2 2rh + h2 )] 1 = 3 cfw_2r 3 (r h)
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 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
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
8+x x The curves intersect when 2 + x = =2+ 4 4 2 x x x= x= 4 16 16x x2 = 0 x(16 x) = 0 x = 0 or 16, so x 8+x 16 16 d x = 0 (2 + x ) 2 + dx A = 0 (2 + x ) 4 4 16 1 2 1 128 256 32 16 =0 = x x dx = x3/
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 rotati