How to calculate $ \intop_{\partial(B_{R}\left(0\right))}\frac{5z^{3}-12z}{|\left(\tau_{1},\tau_{2},\tau_{3}\right)-\left(x,y,z\right)|^{3}}dS_{2} $ Conside a ball with center at $ (0,0,0) $ and radius $ 2 $ in $ \mathbb{R}^3 $.
Now let $ \tau=\left(\tau_{1},\tau_{2},\tau_{3}\right)\in\mathbb{R}^{3} $ be a constant point in the space. How can I calculate
$ \intop_{\partial(B_{2}\left(0\right))}\frac{5z^{3}-12z}{|\left(\tau_{1},\tau_{2},\tau_{3}\right)-\left(x,y,z\right)|^{3}}dS_{2} $
Where $ dS_{2} $ means integrate by the 2 dimesional surface area of the sphere.
Background
Im trying to solve the following Laplace equation:
$ \begin{cases}
u_{xx}+u_{yy}+u_{zz}=0 & x^{2}+y^{2}+z^{2}<4\\
u\left(x,y,z\right)=5z^{3}-12z & x^{2}+y^{2}+z^{2}=4
\end{cases}$
Eearlier, I found a formula for homogenous Laplace equation's in a ball in $\mathbb{R}^n $ (using the right Green function), and it is given by:
$ u\left(x\right)=\frac{1}{R|S^{N-1}|}\intop_{\partial\left(B_{R}\left(0\right)\right)}\frac{\left(R^{2}-|x|^{2}\right)}{|x-y|^{N}}\varphi\left(y\right)dS_{N-1}\left(y\right) $
Where$ x $ here in $ \mathbb{R}^N$,  $ S^{N-1}| $ is the $N-1 $ dimensional area of a unit sphere in $ \mathbb{R}^N$ ,$ \varphi $ is the function for the boundary condition of the Laplace equation $$ \begin{cases}
\varDelta u\left(x\right)=0 & inside\thinspace\thinspace the\thinspace\thinspace ball\\
u\left(x\right)=\varphi\left(x\right) & for\thinspace\thinspace\thinspace x\thinspace\thinspace\thinspace in\thinspace\thinspace\thinspace the\thinspace\thinspace\thinspace boundary\thinspace\thinspace\thinspace of\thinspace\thinspace\thinspace the\thinspace\thinspace\thinspace\thinspace ball
\end{cases} $$
and $ dS_{N-1}(x) $ means integration by the $N-1$ dimesnional surface of the boudary.
I cant see how to calculate this integral (I wrote it without the constatnts from the formula). I will highly appriciate any help.
Thanks in advance.
 A: I slightly missed the mark in the comments. I was thinking of section 3.3.2 in Griffiths' Introduction to Electrodynamics.
There, he states that the azimuthally symmetric separable solutions of $\Delta u = 0$ have the form
$$
u(r, \theta) = \sum_{l = 0}^\infty \left(A_l r^l + \frac{B_l}{r^{l + 1}} \right) P_l (\cos \theta)
$$
where $P_l$ is the $l$-th Legendre polynomial.
On the other hand, we have (remembering that $z = r\cos\theta$ in spherical coordinates):
\begin{align}
u(4, \theta) &= 5z^3  - 12 z = 5(4\cos\theta)^3  - 12 (4\cos\theta)  
\\&= 16(20\cos^3 \theta - 3 \cos\theta) = 16( 8 P_3(\cos\theta) + 9P_1(\cos\theta)) 
\\&= 128  P_3(\cos\theta) + 144 P_1(\cos\theta)
\end{align}
We set $B_l = 0$ for $r < 4$, otherwise the solutions would blow up at the origin. We have
$$
128  P_3(\cos\theta) + 144 P_1(\cos\theta) = u(4, \theta) = \sum_{l = 0}^\infty \left(A_l (4)^l \right) P_l (\cos \theta)
$$
So that $A_1 = 36$ and $A_3 = 2$. That is
$$
u(r, \theta) = 36r\cos\theta + 2 r^3\frac{5\cos^3 \theta -3 \cos\theta}{2}
$$
for $r \leq 4$.
For $r > 4$, I don't know what boundary conditions you've enforced, but taking $u \to 0$ as $r \to \infty$ is common. In that case, we need $A_l = 0$. Playing the same game, we find $B_1 = 2304$ and $B_3 = 32768$. Thus
$$
u(r, \theta) = \frac{2304}{r^2}\cos\theta + 32768 \frac{5\cos^3 \theta -3 \cos\theta}{2 r^4}
$$
for $r \geq 4$.
In summary
$$
u(r, \theta) = 
\begin{cases}
36r\cos\theta +  r^3(5\cos^3 \theta -3 \cos\theta) & r \leq 4 \\
\frac{2304}{r^2}\cos\theta + 16384\left( \frac{5\cos^3 \theta -3 \cos\theta}{ r^4}\right) & r > 4
\end{cases}
$$
You can check that $\Delta u = 0$ for this function and that $u$ matches the boundary conditions. By uniqueness, this must be the solution.

The final expression for $u(r, \theta)$ above is sufficiently complicated that I think evaluating your integral explicitly is more or less hopeless. However, I'd be happy to be proven wrong.
