# Finding the volume of a 3-dimensional spherical intersection

## Main Question or Discussion Point

If two spheres of radius 1 intersect each other so the surface of each sphere passes through the other’s center how could you find the exact volume of the intersection? It doesn't matter what kind of method, it could be double or triple integration, geometry, polar, cylindrical or spherical coordinates.

HallsofIvy
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 $z'= 1/2+ \sqrt{1- r^2}$.