# Volume of solids rotating about two axises

1. Apr 2, 2012

### PirateFan308

1. The problem statement, all variables and given/known data
Find the volumes of the solids revolution obtained by rotating the region about the x-axis and the y-axis.

$y=2x-x^2, y=0$

3. The attempt at a solution
I know how to get the volume of a function that is rotating around one axis, but the "y=0" is confusing me. Because $y=2x-x^2$ is a parabola (with a max at (1,1)), so when I picture it, it looks like a squished donut (with the hole having no area), where a cross sectional area of the donut is shaped like a football with the area being 4/3 (integral of $f(x)=2x-x^2$ from 0 to 2). The outer radius will be 2 and the inner radius will be 0.

Is this correct, or am I completely off track? Also, what does the y=0 mean? Thanks

2. Apr 2, 2012

### Staff: Mentor

y=0 is the lower boundary of the region.

As I read it, this is actually two problems: 1) Find the volume when the region is revolved around the x-axis. 2) Find the volume when the region is revolved around the y-axis.

3. Apr 2, 2012

### Staff: Mentor

Also, you should sketch each of the solids of revolution. When you revolve the region around the x-axis, you get something that looks a little like a football. When you revolve the region around the y-axis, you get something like the upper half of a bagel (what you described as a squished donut).

For the two shapes, you'll need to choose what your typical volume element is - either a disk or a shell. In neither case is the outer radius fixed.

4. Apr 2, 2012

Thanks!