Say we have a point charge $+q$ at the origin from which it emanates an electric field $$\mathbf E=\frac{kq}{r^2}\hat{\mathbf r}\ .$$ Now suppose we want to calculate the flux of this electric field through a surface defined by the intersection of a cone $z^2=x^2+y^2$ and the plane $z=z_0>0$, namely, $$ \mathcal S:\left\{ \begin{align} &x^2+y^2\leq z_0^2\\ &z=z_0 \end{align} \right.$$ with normal vector $\mathbf n=\hat{\mathbf z}$.
Then, in spherical coordinates, we'd find this: $$\Phi=\oint_{\mathcal S}\mathbf E\cdot\mathrm d\mathbf S=\int_0^{2\pi}\int_0^{\pi/4}\frac{kq}{r^2}\underbrace{(\hat{\mathbf r}\cdot\hat{\mathbf z})}_{\cos\theta}r^2\sin\theta\,\mathrm d\theta\,\mathrm d\Phi=kq\underbrace{\int_0^{2\pi}\mathrm d\Phi}_{2\pi}\underbrace{\int_0^{\pi/4}\sin\theta\cos\theta\,\mathrm d\theta}_{1/4}=\frac{\pi}{2}kq.$$ The thing is when I do the same in cylindrical coordinates I get something different: $$\begin{align} \Phi=\oint_{\mathcal S}\mathbf E\cdot\mathrm d\mathbf S&=\int_0^{2\pi}\int_0^{z_0}\frac{kq}{\rho^2+z_0^2}\underbrace{(\hat{\mathbf r}\cdot\hat{\mathbf z})}_{\cos\theta}\rho\,\mathrm d\rho\,\mathrm d\Phi\\ &=2\pi kq\int_0^{z_0}\frac{\rho\cos\theta}{\rho^2+z_0^2}\,\mathrm d\rho\\ &=2\pi kqz_0\underbrace{\int_0^{z_0}\frac{\rho}{(\rho^2+z_0^2)^{3/2}}\,\mathrm d\rho}_{\frac{2-\sqrt 2}{2z_0}}=(2-\sqrt 2)\pi kq\ (?), \end{align}$$ where we used the fact $$\cos\theta=\frac{z_0}{\sqrt{z_0^2+\rho^2}}.$$ So where did I commmit a mistake?
The first is quite wrong. The element of surface area on the plane $z=z_0$ is certainly not the element of surface area on the sphere. Note that we have $r\cos\theta=z_0$, so $\rho=z_0\tan\theta$ and $\rho\,d\rho=z_0^2\tan\theta\sec^2\theta\,d\theta$. When you multiply by $\cos\theta/r^2$, you will get the integral of $\sin\theta$, I believe.