Electric field due to uniformly charged disk I have a uniformly charged disk, radius $R$, with surface charge density $\sigma$. I want to find the electric field along the axis through the centre of the disk (which I've called the $x$ axis). 
I am aware that one method is to consider small rings and integrate from $0$ to $R$, but that is not how I approached the problem and so I would like to know how to do it my way. 
I considered a small section on the disk (I don't know how to formulate its charge), but it had position vector $r \hat{r}$ in cylindrical coordinates. Then I got stuck on how to express the charge of that small section. I think it is $dq = \sigma dz dr$ (again in cylindrical coordinates), but I am not sure. 
Any help is appreciated! 
 A: If a small piece of your disk is located at 
$$
x=r\cos\theta\, ,\qquad y=r\sin\theta,\qquad z=0
$$
then the density would be $dq=\sigma dA=\sigma r dr d\theta$ where $\theta$ is the polar angle on the disk since $dA=r dr d\theta$ is the area of a small piece of your disk, with radius $r$ and arclength $rd\theta$.
For a point located on the $\hat z$ axis at $Z\ne 0$, this small amount of charge will produce the infinitesimal field
$$
d\vec E=\frac{dq}{4\pi\epsilon_0} \frac{Z\hat z- r\cos\theta\hat x - r\sin\theta\hat y}{(r^2+Z^2)^{3/2}}\, .
$$
The full field is obtained by integrating the infinitesimal fields.
A: $$dq=\sigma dA$$
Thus, your question reduces to writing the surface element on a horizontal disk in polar coordinates.
If I move an infinitesimal distance $dr$ along $\mathbf{\hat r}$, I have traced an infinitesimal line segment.
I want to make an infinitesimal rectangle, so what should I choose for the other side? I can move in the direction of $\mathbf{\hat \theta}$.
So if I trace an infinitesimal angle $d\theta$, what length will I have traced? If the angle is small enough and in radians, the arc length (which will be like a straight line segment if the angle subtending it is small enough) is given by $dl = r d\theta$ (you can verify it by considering a triangle with two equal sides. Write the length of the other side as a function of the angle which is facing it. If the angle shrinks toward zero, the formula can be seen.).
So I have an infinitesimal rectangle with sides $dr$ and $rd\theta$. My surface element will be $$dA = dr \cdot rd\theta = rdrd\theta$$
