# How to setup/evaluate a triple integral to show an interesting result in physics?

I know this isn't the physics forum, but the task i'm struggling with is purely mathematical.

Let $A$ be a sphere centered at origin with radius $R$ and assume $a \geq R$.

Show that:$$\iiint\limits_{A}\frac{dA}{\sqrt{x^2 + y^2 + (z - a)^2}} \: = \: \frac{4\pi R^3}{3a}.$$

My work so far is;

Using spherical coordinates $A$ becomes: $$A = \big\{(\phi, \theta, p): 0\leq \phi \leq\pi,\: 0\leq\theta\leq2\pi,\:0\leq p \leq R\big\}.$$

With spherical substitution $x = p\sin(\phi)\cos(\theta), \:y = p\sin(\phi)\sin(\theta),\: z = p\cos(\phi)$, the expression under the radical should become:$$x^2 + y^2 + (z - a)^2 = p^2\sin^2(\phi) + (p\cos(\phi) - a)^2 = p^2 - 2ap\cos(\phi) + a^2.$$

Setting up the integral we get (remember the Jacobian!):$$\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}\int\limits_{0}^{R}\frac{p^2\sin(\phi)}{\sqrt{p^2 - 2ap\cos(\phi) + a^2}}\:dp\:d\theta\:d\phi.$$

I find this hard to integrate or am i just rusty?:) Any hints or tips would be more than welcome, and as usual don't calculate and show it all as i would i like to that myself!

This result is important in physics where it can be used to show that the gravitational force from a homogeneous sphere is the same as if all the mass was gathered in the center.

• Have you tried translating to $(0,0,a)$ before converting to spherical coordinates? The bounds of integration will get messy, but the integrand itself will be much simpler. – amd Mar 8 '16 at 18:52