# Triple Integration of a Strange Cylinder

1. Feb 8, 2009

### Rocker222

Hey all,

If anyone has some hints on how to do this one it would be much appreciated:

Find the volume of the region given by x2 + y2 ≤ a2, 0 ≤ z ≤ x.

So I've gone to cylindrical polars, and threw in the Jacobian, r. If I integrate with my bounds being:

0 to a, 0 to pi, 0 to rcos(theta) of r dz d(theta) dr I'm getting 0 which can't be. What is it that I'm missing here. Any hint would be great.

Last edited: Feb 9, 2009
2. Feb 9, 2009

### HallsofIvy

Staff Emeritus
So you are integrating
[tex]\int_{r=0}^a\int_{\theta= 0}^\pi\int_{z= 0}^{r cos(\theta)} r dzd\theta dr[/itex]?

If you look closely at this situation you will see that this is symmetric about the x axis and so, with z= x as a boundary, half the volume is below the z= 0 plane, half above. Since your integral treats the volume below the z= 0 plane as negative, those cancel. The true "volume" is twice the volume above the z-axis. You get the volume above the z= 0 plane by assuring that x is positive: by taking $\theta$ from $-\pi/2$ to $\pi/2$.

3. Feb 9, 2009

### Rocker222

Thank you very much. That's just the bit I needed. Sorry about not using the Latex code. I'm new to this site.