# Gravitational potential inside and outside a spherical shell

Let $$\Phi$$ denote gravitational potential. This is an example from a book (Classical dynamics, Thornton-Marion). We assume a homogeneous spherical shell. The writer says: We integrate over $$d\phi$$ in the expression for the potential. Thus

$$\Phi=-G\int_V\frac{\rho(r')}{r}dv'=-2\pi\rho G\int_b^ar'^2dr'\int_0^{\pi}\frac{\sin\theta}{r}d\theta.$$

I don't follow this. It seems at first there is a coordinate transform at work, but then again we have two different functions being integrated over two different bounds - instead of a single composition map. And that's assuming this is an iterated integral, and not a product of two different integrals. Can someone explain the steps taken here? I can provide more information from the example if necessary.

• You should probably provide more information, as the meaning of your equation is unclear. What is $r$ for instance? You may also want to have a look at hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html for the potential of a spherical shell (in that case the electric potential, which is equivalent to your problem however). Feb 14 at 10:47

$$$$dV'=r'^2\sin\theta dr'd\theta d\phi$$$$
where $$\theta\in[0,\pi]$$, $$\phi\in[0,2\pi]$$.
Then he integrates over $$\phi$$, thereby getting a factor of $$2\pi$$. Moreover, since the density is constant, he puts it outside the integral.