Finding the volume of a 3-dimensional spherical intersection

[volleyball21]

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

The volume is that portion is
[tex]\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.[/tex]

 

[volleyball21]

Okay I am following but wouldn't the bounds for r be r=0 and r=sqrt(3/4)? When you find z=1/2 you also need to find the distance from the center at z=1/2 to the outer boundary where the surfaces of the spheres intersect because this is less than 1. Am I correct on this?