# Nearly impossible integration problem!

• Oct 13th 2009, 08:33 PM
MarkOfEternity
Nearly impossible integration problem!
How do I find the volume common to two spheres, each with radius r, if the center of each sphere lies on the surface of the other sphere?
• Oct 13th 2009, 09:59 PM
Chris L T521
Quote:

Originally Posted by MarkOfEternity
How do I find the volume common to two spheres, each with radius r, if the center of each sphere lies on the surface of the other sphere?

This doesn't make sense. It would make more sense if you mean to say:

Quote:

How do I find the volume common to two spheres, each with radius r, if the center of one sphere lies on the surface of the other sphere?
• Oct 13th 2009, 10:05 PM
Matt Westwood
No, because if they have the same radius, if the centre of one is on the surface of the other, the centre of the other will be on the surface of the first.
• Oct 13th 2009, 11:02 PM
MarkOfEternity
Agreed. Since the radius of each sphere is similar, then the outside of each sphere must touch the center point of the other. In any case, I've been trying to pound out an answer. I found something online that says I should try $\displaystyle dv = 2 * [pi] x^2 dy$, but I can't figure out how they came to that conclusion or really even what to do with it.
$\displaystyle V=\int_1^2 \pi r^2 dx=\int_1^2 \pi (\sqrt{4-x^2})^2 dx=\int_1^2 \pi (4-x^2)dx$.