Calculate $\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$ where $a^2+b^2+c^2<1$. 
Let $a^2+b^2+c^2 < 1$ and $S^2$ be unit sphere in $R^3$. Calculate $$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$$

Let $(x,y,z)=(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$. By definition,
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{\sqrt{(\cos\theta \cos\phi-a)^2+(\cos\theta \sin\phi-b)^2+(\sin\theta-c)^2}}\sin\theta d\theta d\phi$$
It is too complicate.
This is calculus exam problem that I took yesterday.  Is there any good idea?
 A: taking my suggested simplification a little further, since $y^2+z^2=1-x^2$ on the surface of the sphere, we obtain:
$$
I = 2\pi \int_{-1}^1 \frac{1}{\sqrt{1-2xr+r^2}}dx
$$
A: Let $r=1,\vec{A}=(x,y,z)=r(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$ and
$r'=\sqrt{a^2+b^2+c^2}<1$, $\vec{B}=(a,b,c)=r'(\cos\theta' \cos\phi',\cos\theta' \sin\phi', \sin\theta')$, then we can expand the $\frac{1}{|\vec{A}-\vec{B}|}$ potential as:
$$\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}=\frac{1}{|\vec{A}-\vec{B}|}=\frac{1}{\sqrt{r^2+(r')^2-2r r'\cos\gamma}}$$
$$=\sum_{0}^{\infty}\frac{r'^l}{r^{l+1}}P_l(\cos\gamma)\tag{1}$$
where $\gamma$ is the angle between $\vec{A}$ and $\vec{B}$.
We can also expand the Legendre polynomial $P_l(\cos\gamma)$ in terms of spherical harmonics $Y_{lm}(\theta,\phi)$ as:
$$P_l(\cos\gamma)=\frac{4\pi}{2l+1}\sum_{m=-l}^{+l}Y_{lm}(\theta,\phi)Y_{lm}^{*}(\theta',\phi') \tag{2}$$
Since
$$\int_{S^2}Y_{lm}(\theta,\phi)dS = \int_{0}^{2\pi}\int_{0}^{\pi}Y_{lm}(\theta,\phi)\sin\theta d\theta d\phi=2\sqrt{\pi}\delta_{l0}\delta_{m0}\tag{2}$$
and 
$$Y_{00}^{*}(\theta',\phi')=\frac{1}{2\sqrt{\pi}}\tag{3}$$
we obtain:
$$\int_{S^2}\frac{1}{|\vec{A}-\vec{B}|}dS = 4\pi \tag{4}$$
A: Let $(x,y,z)=(a+\cos\theta \cos\phi,b+\cos\theta \sin\phi,c+ \sin\theta)$
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_0^{2\pi}\int_0^{\pi}\frac{1}{\sqrt{(\cos\theta\cos\phi)^2+(\cos\theta \sin\phi)^2+(\sin\theta)^2}}\sin\theta d\theta d\phi=\int_0^{2\pi}\int_0^{\pi}\sin\theta d\theta d\phi\\=4\pi$$
