# The area of interest - volume integral

I have two objects:

$$x^{2}+(y-\frac{1}{2}a)^{2}=\frac{1}{4}a^{2}$$ $$x^{2}+y^{2}+z^{2}=a^{2}$$

And I need to find the volume of their intersection...My question is only about finding the area of interest here. I've been thinking about dividing the intersecting cylinder and then multiplying the integral by 2.

So the XY plane would look like this (the bigger one is the sphere, the smaller cylinder - sorry for using paint ;)):

The area of interest of the top piece of the cylinder would be:

$0<r<a$ (if I remember correctly the radius should be calculated from (0,0,0))
$0<\phi<2\pi$
$0<z<\sqrt{r^{2}-a^{2}}$

And now, I'm not sure about the z variable...I know it should be restricted with the sphere function...but what should I do with the $a^{2}$? I mean, what's next?

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I think you may want the two-dimensional area of intersection of the sphere and the cylinder, say in the plane $\,z=0\,$ ... – DonAntonio Aug 3 '13 at 17:22
Do you mean using double integral for this? But won't this be the same thing? – khernik Aug 3 '13 at 17:26
Look at a cross-section above z=0, say z=0.5a. What does it look like then? – Mark Ping Aug 5 '13 at 15:03