TITLE: How do I compute mean curvature in cylindrical coordinates?
QUESTION [5 upvotes]: If I have a surface defined by $ z=f(r, \theta) $, does anyone know the expression for the mean curvature? There is a previous post dealing with Gaussian instead of mean curvature, the answer I'm looking for is similar to that given by J.M. on that post.
The mentioned post:
How do I compute Gaussian curvature in cylindrical coordinates?
Many thanks in advance for your help,

REPLY [4 votes]: Unfortunately, Vittorio gave the mean curvature with respect to the parameters $\theta$ and $z$ in his answer instead of the parameters $r$ and $\theta$ that OP needed. To get the mean curvature of $z=f(r,\theta)$, we start with the parametrization
$$\begin{align*}x&=r\cos\,\theta\\y&=r\sin\,\theta\\z&=f(r,\theta)\end{align*}$$
Using the usual formula for mean curvature (equation 7 here) and simplifying, we obtain the expression
$$\small \frac{\frac{\partial f}{\partial r}\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+2\frac{\partial f}{\partial \theta}\left(\frac{\partial f}{\partial \theta}-r\frac{\partial f}{\partial r\partial \theta}\right)\right)+r\left(\left(\frac{\partial f}{\partial \theta}\right)^2+r^2\right)\frac{\partial^2 f}{\partial r^2}+r\frac{\partial^2 f}{\partial \theta^2}\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)}{2\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+\left(\frac{\partial f}{\partial \theta}\right)^2\right)^{3/2}}$$