<s>[INST] How do I compute mean curvature in cylindrical coordinates? 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,
 [/INST] It's similar to the Gaussian Curvature. J.M. actually hide his/her calculation.
You start with the parametrization:
\begin{align}
x &= \rho(\vartheta, z)\cos\vartheta \\
y &= \rho(\vartheta, z)\sin\vartheta \\
z &= z \\
\end{align}
you need to find the values of $L$, $M$ and $N$ of the Second fundamental form and then to find the trace of the matrix:
\begin{pmatrix}
L & M \\
M & N
\end{pmatrix}
and divide it by 2. Practically you only need $L$ and $N$.
Let $\mathbf{r}\colon [0, 2\pi)\times \mathbb{R}^2 \to \mathbb{R}^3$ defined by the parametrization, then:
\begin{align}\mathbf{n} &= \frac{\mathbf{r}_{\vartheta} \times \mathbf{r}_z}{\lvert\mathbf{r}_{\vartheta} \times \mathbf{r}_z\rvert} \\
L &= \mathbf{r}_{\vartheta\vartheta}\cdot \mathbf{n} \\
N &= \mathbf{r}_{zz}\cdot \mathbf{n} \\
\end{align}
You have:
\begin{align}\mathbf{r}_{\vartheta} &= \bigl(\rho_{\vartheta} \cos\vartheta - \rho \sin\vartheta, \rho_{\vartheta}\sin\vartheta + \rho\cos\vartheta, 0\bigr) \\
\mathbf{r}_{z} &= \bigl(\rho_{z} \cos\vartheta, \rho_{z}\sin\vartheta, 1\bigr) \\
\mathbf{r}_{\vartheta\vartheta} &= \bigl(\rho_{\vartheta\vartheta} \cos\vartheta - \rho_{\vartheta} \sin\vartheta - \rho_{\vartheta} \sin\vartheta - \rho \cos\vartheta, \rho_{\vartheta}\sin\vartheta + \rho\cos\vartheta, 0\bigr) \\
 &= \bigl(\rho_{\vartheta\vartheta} \cos\vartheta - 2\rho_{\vartheta} \sin\vartheta - \rho \cos\vartheta, \rho_{\vartheta\vartheta}\sin\vartheta + \rho_{\vartheta}\cos\vartheta + \rho_{\vartheta}\cos\vartheta - \rho\sin\vartheta, 0\bigr) \\
\mathbf{r}_{zz} &= \bigl(\rho_{zz} \cos\vartheta, \rho_{zz}\sin\vartheta, 0\bigr) \\
&= \rho_{zz}\bigl(\cos\vartheta, \sin\vartheta, 0\bigr)
\end{align}
If I define $\mathbf{k} = (0,0,1)$, $\mathbf{u} = (\cos\vartheta,\sin\vartheta,0)$ and $\mathbf{v} = \mathbf{u}_{\vartheta} = (-\sin\vartheta,\cos\vartheta,0)$, I can rewrite them as:
\begin{align}\mathbf{r}_{\vartheta} &= \rho_{\vartheta}\mathbf{u} + \rho\mathbf{v} \\
\mathbf{r}_{z} &= \rho_{z}\mathbf{u} + \mathbf{k} \\
\mathbf{r}_{\vartheta\vartheta} &= \rho_{\vartheta\vartheta}\mathbf{u} + 2\rho_{\vartheta}\mathbf{v} - \rho\mathbf{u} \\
&= (\rho_{\vartheta\vartheta} - \rho)\mathbf{u} + 2\rho_{\vartheta}\mathbf{v} \\
\mathbf{r}_{zz} &= \rho_{zz}\mathbf{u}
\end{align}
\begin{align}\mathbf{u} \times \mathbf{v} &= (0,0, cos^2\vartheta + \sin^2\vartheta) =  \mathbf{k}\\
\mathbf{u} \times \mathbf{k} &= (\sin\vartheta, -\cos\vartheta, 0) = -\mathbf{v}\\
\mathbf{v} \times \mathbf{k} &= (\cos\vartheta, \sin\vartheta, 0) = \mathbf{u}\\
\mathbf{u} \cdot \mathbf{u} &= 1\\
\mathbf{v} \cdot \mathbf{v} &= 1\\
\mathbf{k} \cdot \mathbf{k} &= 1\\
\mathbf{u} \cdot \mathbf{v} &= -\cos\theta\sin\theta + \sin\theta\cos\theta =  0\\
\mathbf{u} \cdot \mathbf{k} &= 0\\
\mathbf{v} \cdot \mathbf{k} &= 0\\
\end{align}
Now I can calculate $\mathbb{n}$, $L$ and $N$.
\begin{align} \mathbf{r}_{\vartheta} \times \mathbf{r}_z &= \bigl(\rho_{\vartheta}\mathbf{u} + \rho\mathbf{v}\bigr)\times \bigl(\rho_{z}\mathbf{u} + \mathbf{k}\bigr) \\
&= \rho_{\vartheta}\mathbf{u}\times\rho_{z}\mathbf{u} + \rho\mathbf{v}\times\rho_{z}\mathbf{u} + \rho_{\vartheta}\mathbf{u}\times\mathbf{k} + \rho\mathbf{v}\times\mathbf{k} \\
&= \rho_{\vartheta}\rho_{z}\mathbf{u}\times\mathbf{u} + \rho\rho_{z}\mathbf{v}\times\mathbf{u} + \rho_{\vartheta}\mathbf{u}\times\mathbf{k} + \rho\mathbf{v}\times\mathbf{k} \\
&= -\rho\rho_{z}\mathbf{k} - \rho_{\vartheta}\mathbf{v} + \rho\mathbf{u} \\
\lvert\mathbf{r}_{\vartheta} \times \mathbf{r}_z \rvert &= \rho^2 + \rho_{\vartheta}^2 + \rho^2\rho_{z}^2 = \rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2\\
\mathbf{n} &= \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
\end{align}
\begin{align}
L &= \bigl[(\rho_{\vartheta\vartheta} - \rho)\mathbf{u} + 2\rho_{\vartheta}\mathbf{v}\bigr]\cdot \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
&= \frac{(\rho_{\vartheta\vartheta} - \rho)\rho + 2\rho_{\vartheta}^2}{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
N &= \rho_{zz}\mathbf{u}\cdot \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
&= \frac{\rho\rho_{zz}}{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2}
\end{align}
Now, the mean curvature is: 
\begin{align}
H &= \frac{L + N}{2} \\
&= \frac{(\rho_{\vartheta\vartheta} - \rho)\rho + 2\rho_{\vartheta}^2 + \rho\rho_{zz}}{2\rho^2\bigl(1 + \rho_{z}^2\bigr) + 2\rho_{\vartheta}^2}
\end{align}
I suggest you to check my calculations. Anyway this is the general methods to find the mean curvature when you have a parametrization of a suface (in $\mathbf{R}^3$).
P.S. I abbreviate $\frac{\partial f}{\partial x}$ with $f_x$ and $\frac{\partial f}{\partial x\partial y}$ with $f_{xy}$.
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