Electric potential due to circular disk Relevant diagram is available here.
The circular disk of radius $a$ lies in the $xy$ plane and carries surface charge density of
$\sigma (s, \phi) = s^{2}cos\phi $,
where $(s,\phi)$ are in cylindrical co-ordinates.
The problem is to find potential at a point which is slightly displaced from the $z$ axis at position $ r = z \hat z + \delta s \hat s = z\hat z + \delta x \hat x + \delta x \hat y$, and
$ r' = s cos\phi \hat x + s sin\phi \hat y  $ 
Since potential is given by
$V(r) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{1}{\bf {r} - \bf {r'}} dq$
Here, $dq = \sigma da = \sigma dl_{s} dl_{\phi} = \sigma sdsd\phi = s^3 cos\phi dsd\phi$
and 
${r-r'} =  z\hat z + \delta x\hat x + \delta x \hat y - (s cos\phi \hat x + s sin\phi \hat y) $
Therefore,
$|r-r'| = \sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)} $.
It can be assumed that $(\delta x)^2 = 0$ since $\delta x$ is infinitesimally small.
This means that the final integral for potential is given by
$V(r) = \frac{1}{4 \pi \epsilon_0} \int_S \frac{s^3 cos\phi}{\sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)}} dsd\phi$.
Any suggestions on how to proceed with evaluating this integral would be very much appreciated.
Thanks.
 A: The problem is that you're ignoring the angular dependence of your probe point $\mathbf r$, and that is messing with your integral. If your probe point has cylindrical coordinates $(s,\phi,z)$ and your integration variables are $(s',\phi')$, then the distance between the two is
$$\frac 1 {|\mathbf r-\mathbf r'|}=\frac{1}{\sqrt{s^2-2ss'\cos(\phi-\phi')+s'^2+(z-z')^2}}$$
by the cosine rule (draw it!). If you put this into your integral it will no longer vanish.
(A few pointers on the new integral: the new dependence on $\phi'$ is a bit more complicated. The standard practice is to change variables to $\varphi=\phi'-\phi$. This will leave a simpler denominator, and a factor of $\cos(\varphi+\phi)=\cos(\varphi)\cos(\phi)-\sin(\varphi) \sin(\phi)$ on the numerator. One of the two terms will vanish and the other will yield to a change of variables to $u=\cos(\varphi)$.)
A: Looks like you calculate $\boldsymbol{r}-\boldsymbol{r'}$ incorrectly: your $\boldsymbol{r'}$ does not depend on $\phi$, whereas you should integrate over the entire disk.
