Imagine we have a parametic surface given by: $$ \Phi(r, \theta) = \begin{pmatrix} r \cos{\theta} \\ r \sin{\theta} \\ f(r) \\ \end{pmatrix} $$ $$ r \in \mathbb R^+, \; \; \; \ \theta \in [0, 2\pi] $$ $f$, will only depend on $r$, for simplicity.
Finding the tangent vectors to the surfaces: $$ \vec r = \begin{pmatrix} \cos{\theta} \\ \sin{\theta} \\ f'(r) \\ \end{pmatrix} \; \; \; \; \vec\theta = \begin{pmatrix} -r\sin{\theta} \\ r \cos{\theta} \\ 0 \\ \end{pmatrix}$$ $$ d\vec{S} = ||\vec r \times\vec\theta|| \; dr d\theta = \sqrt { r^2 + f'(r)^2} \; dr d\theta \\ $$ $$ S = \iint_{\Omega} {\sqrt { r^2 + f'(r)^2} \; dr d\theta} $$ Using Euler-Lagrange Equations to find the extrema of the functional: $$ \frac{\partial \mathcal L}{\partial f} - \frac{d}{d r}\frac{\partial \mathcal L}{\partial f'} = 0 $$ We have: $$ f'(r) = r \\ rf''(r) = f'(r) $$ Solving this two equations gives: $$ f_1(r) = \frac{1}{2}r^2 + C \\ f_2(r) = C_1 r^2 + C_2 $$ Does this means that the cylindrical surfaces (without a $\theta$ dependence) which maximizes its area are paraboloids? Is my line of reasoning right? Did I made any mistakes?