Integrate this function over the volume of a sphere sphere

[Silversonic]

Homework Statement

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

Homework 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

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?

 

[Thaakisfox]

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

 

[Silversonic]

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

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 into

$(r^{2} + R^{2} - 2Rrcos(\theta))^{3/2}$ 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?

 

[Thaakisfox]

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

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