Integrate this function over the volume of a sphere sphere

1. Dec 23, 2011

Silversonic

1. The problem statement, all variables and given/known data

Integrate $\nabla^{2}$ ($\frac{1}{|\underline{r} - \underline{r}'|}$) over the volume of a sphere using the divergence theorem.

2. Relevant equations

$\nabla^{2}$ ($\frac{1}{|\underline{r} - \underline{r}'|}$) = -4$\pi$$\delta^{(3)}$$(\underline{r} - \underline{r}')$ (i.e. it's a dirac delta function, this just tells me that the answer I should get if I integrate over the volume is -4$\pi$)

$\nabla^{}$ ($\frac{1}{|\underline{r} - \underline{r}'|}$) = $-\frac{(\underline{r} - \underline{r}')}{|\underline{r} - \underline{r}'|^{3}}$

$\underline{r} = (x,y,z)$

$\underline{r'} = (x',y',z')$ (note that the dashes do not mean the differential of, just a different point in space from x,y,z).

And the divergence theorem

3. The attempt at a solution

$\int_{V'}$$\nabla^{2}$ $(\frac{1}{|\underline{r} - \underline{r}'|} )dV'$

=
$\int_{S'}$$\nabla$ $(\frac{1}{|\underline{r} - \underline{r}'|} ) \circ d\underline{S}'$ (by the divergence theorem)

$d\underline{S}' = \hat{\underline{n}} dS'$

$dS' = R^{2}sin(\theta)d\theta d\phi$

I guess R should be the radius of the sphere. In which case I choose r' = R?

= $\int_{S'}-\frac{(\underline{r} - \underline{r}')}{|\underline{r} - \underline{r}'|^{3}} \circ d\underline{S}'$

Then, putting r' = R

$\int_{S'}-\frac{(\underline{r} - \underline{R})}{|\underline{r} - \underline{R}|^{3}} \circ \hat{\underline{n}} dS'$

=

$\int^{2\pi}_{0} \int^{\pi}_{0} -(\frac{(\underline{r} - \underline{R})}{|\underline{r} - \underline{R}|^{3}} \circ \hat{\underline{n}}) R^{2}sin(\theta)d\theta d\phi$

Am I right so far? This looks like a complete mess, and I'm not sure what to do with the dot product of the vector and the normal vector?

2. Dec 24, 2011

Thaakisfox

Whats the direction of the radius vector compared to the normal vector on the surface of the sphere? ;)

3. Dec 25, 2011

Silversonic

Cheers. Clearly they are in the same direction, so I can make the simplification $\underline{R} \circ \hat{\underline{n}} = R$

Then maybe with $\underline{r} \circ \hat{\underline{n}}$ I can turn it into $rcos(\theta)$

Thus $- (\underline{r} - \underline{R}) \circ \hat{\underline{n}}$ becomes

$R - rcos(\theta)$

Then what though? I guess I could turn the denominator in to

$(r^{2} + R^{2} - 2Rrcos(\theta))^{3/2}$ by the cosine rule, but then I get, overall;

$\int^{2\pi}_{0} \int^{\pi}_{0} (\frac{R - rcos(\theta)}{(r^{2} + R^{2} - 2Rrcos(\theta))^{3/2}}) R^{2}sin(\theta)d\theta d\phi$

But how would I do that integral with respect to theta?

4. Dec 25, 2011

Thaakisfox

yes very good. Now put $x=\cos\theta$. And separate the integrands.

Note that: $$\frac{d}{dx}(r^2+R^2-2Rrx)^{-1/2}) =\frac{-Rr}{(r^2+R^2-2Rrx)^{3/2}}$$