Gravitational potential inside and outside a spherical shell

Question

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.

Answer 1

He is just using spherical coordinates. The volume element in these coordinates is

\begin{equation} dV'=r'^2\sin\theta dr'd\theta d\phi \end{equation}

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.