Electric field of semi-sphere I have to find the electrical field in the center (of the base) of a semi spherical shell of radius R. The total charge Q (Q > 0) is uniform on the intern surface of the semi sphere. Here's a scheme:

And here's the solution:

I found the first equality ($E_{z}$ = the first integral), but then I was stuck to integrate other the intern surface of the semi-sphere. 
Could someone explain me how it's done? I don't get the second equality.
Many thanks,
 A: Let's go through this step by step:
The electric field point away from a single charge q distance r away is:
$E = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{R^2}$
However since we are dealing with charge spread over a hemisphere we must integrate over the surface charge density $\sigma_{q} = \frac{Q}{2\pi R^2}$ Furthermore, we know that charges opposite each other will cancel, so we must times put $cos(\theta)$ in the integral
So we have:
$$E = \frac{1}{4\pi\epsilon_{0}R^2}\iint_{S(A)} cos(\theta)\frac{Q}{2\pi R^2}dA = \frac{Q}{8\pi^2\epsilon_{0}R^4}\iint_{S(A)}\cos(\theta)dA$$
We then decompose the surface integral into two regular itegrals.
Quantitatively this is expressed in spherical coordinates as $dA = R^2sin(\theta)d\theta d\phi$
Alternatively it can be thought of qualitatively as a surface of revolution using spherical cordinates as described below:
We first integrate a quarter circle $\theta$ from $0$ to $\frac{\pi}{2}$, then we take that quarter circle and integrate a surface of revolution around the z axis with $\phi$ from $0$ to $2\pi$ which is just $2\pi$. Since the distance ("height") of the surface from z is given by $sin(\theta)$ we multiply by that in the integral as well (because the "lower" it is the less it will revolve). This sine term goes in the second integral, not the first, a) because it is also a function of theta and b) because if put $sin(\phi)$ in the first integral the answer would be zero. Since the surface of the sphere scales with $R^2$ we must also times multiply our answer by that.
Hence either way we have:
$$E = \frac{Q}{8\pi^2\epsilon_{0}R^4}R^2\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\sin(\theta)cos(\theta)d\theta d\phi = \frac{Q}{8\pi\epsilon_{0}R^2}$$
A: Since $Q$ is uniform, the charge differential $dq$ is given by $$dq = \frac{Q}{\text{Area}(\Sigma)}d\sigma = \frac{Q}{2\pi R^2}d\sigma.$$ Thus $$dE_z = dE\cos\theta = \frac{1}{4\pi \varepsilon_0} \frac{dq}{r^2}\cos\theta = \frac{1}{4\pi \varepsilon_0} \frac{Q}{2\pi R^2}\frac{d\sigma}{R^2}\cos\theta.$$ Now we can write $$E_z = \int_{\Sigma} dE_z = \frac{1}{4\pi \varepsilon_0}\int_\Sigma \frac{Q}{2\pi R^2}\frac{d\sigma}{R^2}\cos\theta.$$ In spherical coordinates, $d\sigma = R^2\sin \theta\, d\theta\, d\phi$. Since $\phi$ ranges from $0$ to $2\pi$ and $\theta$ ranges from $0$ to $\pi/2$, we have $$\frac{1}{4\pi \varepsilon_0} \int_\Sigma \frac{Q}{2\pi R^2}\frac{d\sigma}{R^2}\cos\theta = \frac{Q}{8\pi^2\varepsilon_0 R^2} \int_0^{2\pi} d\phi \int_0^{\pi/2} \sin\theta \cos\theta\, d\theta.$$
