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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.

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  • $\begingroup$ 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). $\endgroup$
    – Thomas
    Feb 14 at 10:47
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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.

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