Has anyone published a formula for the volume of the intersection of two balls

Question

Math people:

I have Googled this question and searched Math Stack Exchange, and not found an answer. Given $r_1, r_0, t >0$, $r_0 \leq r_1$ I have found a formula for the volume of the intersection of two three-dimensional balls with radii $r_0$ and $r_1$ and centers separated by $t$. It is obvious what it is if $t \leq r_1 - r_0$ or $t \geq r_1 + r_0$. For $r_1 -r_0 \leq t \leq r_1 + r_0$, it is polynomial (cubic) in $r_0$ and $r_1$ and rational in $t$, with powers of $t$ going from $t^{-1}$ to $t^3$, if I remember correctly. I found it using Calc 3 techniques and some help from Maple to simplify some nasty expressions.

Someone has to have done this before. Has anyone ever seen a formula for this? This is part of a research project and I don't want to take credit for it if someone has done it before.

Stefan (STack Exchange FAN)

Answer 1

Mathworld has the formula (http://mathworld.wolfram.com/Sphere-SphereIntersection.html), and cites Kern and Bland 1948, but I haven't checked if the book actually contains the formula.

Several sources have the formula for the volume of a spherical cap (from which you can get the volume of the spherical intersection), for instance the Handbook of Mathematics and Computational Science.

In any case, in your position I would treat the volume of intersection as "well known."

Answer 2

The region you ask about is the union of two spherical caps. The volume of intersection of two balls, even of different radii, can be worked out from the sum of those two volumes.

That is, the bounding spheres intersect in a circle (if at all, the case of one sphere nested in another being trivial), and the plane that circle lies in cuts off a spherical cap from each ball. Add their volumes and you have the volume of intersection.

The higher-dimensional cases are discussed in this previous question.