Here's how I would do it. Take one of the spheres to be centered at (0,0,0) and the other at (0,0,1). The first sphere is given by x^2+ y^2+ z^2= 1 and the second by x^2+ y^2+ (z-1)^2= 1=. Their intersection, then, is where x^2+ y^2+ z^2= x^2+ y^2+ (z-1)^2 or [math]z^2= (z-1)^2= z^2- 2z+ 1[/itex] so z= 1/2. Now, just calculate the volume of the portion of x^2+ y^2+ z^2= 1 above z= 1/2. Now just "translate" the figure down 1/2l That is, let z'= z- 1/2 so that z= z'+ 1/2 and the equations are x^2+ y^2+ (z'+ 1/2)^2= 1 and z'= 0. It would probably be simplest to do that in cylindrical coordinates. That way, x^2+ y^2+ (z'+ 1/2)^2= 1 becomes r^2+ (z+1/2)^2= 1. Since z> 0, z'> 1/2 and z'= 1/2+ \sqrt{1- r^2}.
The volume is that portion is
\int_{\theta= 0}^{2\pi}\int_{r= 0}^1 (1/2+ \sqrt{1- r^2})r drd\theta[/itex] and the volume of the entire intersection is twice that.
