Newtonian potential at (0, 0, – a) I found this problem in the book Advanced Calculus, written by Friedman.
"Newtonian potential at (0, 0, – a) due to a mass with constant densinty $\sigma$ on the hemisphere S: $x^2 + y^2 + z^2 = a^2$, $z \geq 0$, is
$$\int\int_S \frac{\sigma}{[x^2+y^2+(z+a)^2]^{1/2}}dS$$
Compute U."
I wrote $z=\sqrt{a^2-x^2-y^2}$, and the hemisphere as $(x, y, \sqrt{a^2-x^2-y^2})$, then $dS=\sqrt{1+{z_x}^2+{z_y}^2}dxdy=\dfrac{a}{\sqrt{a^2-x^2-y^2}}dxdy$.
So, my initial integral became 
$$\int_{0}^{a}\int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \frac{\sigma a}{\sqrt{a^2-x^2-y^2}[2a^2+2a\sqrt{a^2-x^2-y^2}]^{1/2}}dxdy$$
I don't even know how to start integrating this thing. So I decided to use polar coordinates, so I got:
$$\int_{0}^{2\pi}\int_{0}^{a} \frac{\sigma ar}{\sqrt{a^2-r^2}[2a^2+2a\sqrt{a^2-r^2}]^{1/2}}drd\theta,$$
which wasn't so much better...
If someone could tell me where I went wrong, or give a full solution, or just a hint, I'd be really grateful.
Thanks!

The surface integral,
$$\Phi(0,0,-a)=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S,$$
is most easily done using spherical coordinates: $\langle x,y,z\rangle=\langle a\cos{\varphi}\sin{\theta}, a\sin{\varphi}\sin{\theta}, a\cos{\theta}\rangle$, where $\varphi\in[0,2\pi]$ and $\theta\in[0,\frac{\pi}{2}]$. With these coordinates,
$$\begin{align}
x^2+y^2+(z+a)^2&=a^2\cos^2{\varphi}\sin^2{\theta}+a^2\sin^2{\varphi}\sin^2{\theta}+(a\cos{\theta}+a)^2\\
&=a^2\sin^2{\theta}+a^2(\cos{\theta}+1)^2\\
&=a^2\sin^2{\theta}+a^2(\cos^2{\theta}+2\cos{\theta}+1)\\
&=a^2(\sin^2{\theta}+\cos^2{\theta}+2\cos{\theta}+1)\\
&=2a^2(\cos{\theta}+1),
\end{align}$$
which implies, $\left[x^2+y^2+(z+a)^2\right]^{1/2}=\sqrt{2}\,a\sqrt{\cos{\theta}+1}$.
The surface element on the surface of the hemisphere is $\mathrm{d}S=a^2\sin{\theta}\,\mathrm{d}\theta\mathrm{d}\varphi$, so putting it all together the surface integral is:
$$\begin{align}
\Phi(0,0,-a)&=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S\\
&=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\frac{\sigma a^2\sin{\theta}}{\sqrt{2}\,a\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\mathrm{d}\varphi\\
&=\frac{a\sigma}{\sqrt{2}}\int_{0}^{2\pi}\mathrm{d}\varphi\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\\
&=\sqrt{2}\,\pi a\sigma\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\\
&=\sqrt{2}\,\pi a\sigma\left[2(\sqrt{2}-1)\right]\\
&=(4-2\sqrt{2})\pi a\sigma.
\end{align}$$
