# Integration volume

1. May 12, 2010

### thereddevils

1. The problem statement, all variables and given/known data

Find the volume of solid generated when the shaded area is rotated through pi radians about the y-axis . The graphs are y=x^2 and y=2-x^2 .

2. Relevant equations

3. The attempt at a solution

i did everything and found the answer to be 16/15 pi but the answer given is pi .

$$V=\pi \int^{1}_{0}x^4 dx+\pi \int^{2}_{1}(2-x^2)^2 dx$$

any mistakes in my set up ?

2. May 12, 2010

### phyzguy

When I set it up, I got:

$$V=\int_0^1\pi x\int_{x^2}^{2-x^2} dy dx = \int_0^1\pi x(2-2x^2)dx = 2\pi[\frac{x^2}{2}-\frac{x^4}{4}]_0^1 = \frac{\pi}{2}$$

However, this gives pi/2, not pi. Are you sure pi is the right answer? Anyone else?

3. May 12, 2010

### thereddevils

ok , i think i see my mistake , its with respect to y .