Recall that the electric field on a surface is given by \begin{eqnarray*} \EE(\rr) = \int \frac{1}{4\pi\epsilon_0} \frac{\sigma(\rrp)(\rr-\rrp)\,dA}{|\rr-\rrp|^3} \end{eqnarray*} Let's find the electric field due to a charged disk, on the axis of symmetry.

In cylindrical coordinates, we have that \begin{eqnarray*} \rr - \rrp = z\,\zhat - r'\,\rhat\Prime \end{eqnarray*} (where we write $\rhat\Prime$ to emphasize that this basis is associated with $\rrp$). The integral becomes \begin{eqnarray*} \EE(z) = \Int_0^{2\pi}\Int_0^R \frac{\sigma}{4\pi\epsilon_0} \frac{(z\,\zhat-r'\,\rhat\Prime)\,r'\,dr'\,d\phi'} {(z^2 + r'^2)^{3/2}} \end{eqnarray*} It is important to note that $\rhat\Prime$ can not be pulled out of the integral, since it is not constant. Quite the opposite, by symmetry, this integral must vanish! Explicitly, writing \begin{eqnarray*} \rhat\Prime = r'\cos\phi'\,\ii + r'\sin\phi'\,\jj \end{eqnarray*} and then integrating will indeed yield zero. The remaining term is \begin{eqnarray*} \EE(z) &=& \Int_0^{2\pi}\Int_0^R \frac{\sigma}{4\pi\epsilon_0} \frac{z\,r'\,dr'\,d\phi'} {(z^2 + r'^2)^{3/2}} \> \zhat \\ &=& - \frac{\sigma\,\zhat}{4\pi\epsilon_0} \frac{2\pi z}{\sqrt{z^2+r'^2}} \Bigg|_0^R = \frac{2\pi\sigma\,\zhat}{4\pi\epsilon_0} \left( \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} \right) \end{eqnarray*}

Recall that the electric field of a uniform disk is given along the axis by \begin{eqnarray*} \EE(z) = \frac{2\pi\sigma}{4\pi\epsilon_0} \left( \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} \right)\,\zhat \end{eqnarray*} where of course $\frac{z}{\sqrt{z^2}}=\pm1$ depending on the sign of $z$. (The notation ${\rm sgn}(z)$ is often used to represent the sign of $z$, in order to simplify expressions like $\frac{z}{\sqrt{z^2}}$.) In the limit as $R\to\infty$, one gets the electric field of a uniformly charged plane, which is just \begin{eqnarray*} \EE(z) = \hbox{sgn}(z) \> \frac{\sigma}{2\epsilon_0}\,\zhat \end{eqnarray*} which is valid everywhere, as any point can be thought of as being on the axis.