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,
 A: 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
$$\small \frac{\frac{\partial f}{\partial r}\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+2\frac{\partial f}{\partial \theta}\left(\frac{\partial f}{\partial \theta}-r\frac{\partial f}{\partial r\partial \theta}\right)\right)+r\left(\left(\frac{\partial f}{\partial \theta}\right)^2+r^2\right)\frac{\partial^2 f}{\partial r^2}+r\frac{\partial^2 f}{\partial \theta^2}\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)}{2\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+\left(\frac{\partial f}{\partial \theta}\right)^2\right)^{3/2}}$$
A: 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 $f_{xy}$.
A: In Modern Differential Geometry of Curves and Surfaces with Mathematica, 3rd Edition, by Alfred Gray et al. (Chapman and Hall/CRC, 2006), Theorem 13.25 states that the mean and Gaussian curvatures for a regular patch $\mathbf{x}$ are given by the formulas
\begin{align}
    H &= \frac{eG -2fF + gE}{2(EG-F^2)},\\
    K &= \frac{eg -f^2}{EG-F^2},
\end{align}
where $e,f, g$ are the coefficients of the second fundamental form of patch $\mathbf{x}$,
and $E,F,G$ are the coefficients of the first fundamental form.
Patches which can be parameterized by two variables are called the Monge patches.  Gray et al. provides the $H$ and $K$ curvatures for a Monge patch $\mathbf{x}\mapsto (u, v, h(u,v))$ in Lemma 13.34.
Utilizing the framework in Gray et al., we provide the necessary details on how to find the curvatures of a regular patch parameterized in Polar coordinates:
\begin{equation}
(u,v)\mapsto (u\cos v, u\sin v, h(u,v)),
\end{equation}
where $u$ is the radial coordinate and $v$ the angular coordinate, respectively. For a regular patch $\mathbf{x}$ parameterized in $(u,v)$, $\mathbf{x}_u$ and $\mathbf{x}_v$ are the local surface tangent vectors at  point $\mathbf{p}$. The $E,F,G$ functions of $\mathbf{x}$ are defined by
$$ E=\lVert\mathbf{x}_u\rVert^2,\quad 
F=\mathbf{x}_u\cdot \mathbf{x}_v,\quad
G=\lVert\mathbf{x}_v\rVert^2.$$
See Definitions 12.1 and 13.15 in Gray et al.
The normalized normal vector at $\mathbf{p}$ can be readily found by the cross product of the two tangent vectors:
$$  \mathbf{U}=\frac{\mathbf{x}_u\times \mathbf{x}_v}{\left\lVert \mathbf{x}_u\times \mathbf{x}_v \right\rVert}.$$
For surface $\mathbf{x}=(u\cos v, u\sin v, h(u,v))$, we have the following immediately:
\begin{align}
    &\mathbf{x}_u=(\cos v, \sin v, h_u),\\
    &\mathbf{x}_v=(-u\sin v, u\cos v, h_v),\\
    &\mathbf{x}_{uu}=(0, 0, h_{uu}),\\
    &\mathbf{x}_{uv}=(-\sin v, \cos v, h_{uv}),\\
    &\mathbf{x}_{vv}=(-u\cos v, u\sin v, h_{vv}).
\end{align}
and
\begin{align}
    &\mathbf{U}=\frac{(h_v\sin v-u\, h_u\cos v,
    -h_v\cos v - u\, h_u\sin v, u(\cos^2v+ \sin^2v))}{\sqrt{u^2(1+h_u^2) + h_v^2}},\\
    &E=\lVert\mathbf{x}_u\rVert^2=1+h_u^2,\\
    &G=\lVert\mathbf{x}_v\rVert^2=u^2+h_v^2,\\
    &F=\mathbf{x}_u\cdot\mathbf{x}_v=h_u \,h_v,\\
    &e=\mathbf{x}_{uu}\cdot\mathbf{U}=\frac
    {u h_{uu}}{\sqrt{u^2(1+h_u^2) + h_v^2}},\\
    &f=\mathbf{x}_{uv}\cdot\mathbf{U}= \frac{-h_v+h_u h_{uv}}{\sqrt{u^2(1+h_u^2) + h_v^2}},\\
    &g=\mathbf{x}_{vv}\cdot\mathbf{U}=\frac{u^2 h_u + u h_{vv}}{\sqrt{u^2(1+h_u^2) + h_v^2}}.
\end{align}
From Theorem 13.25, the mean and Gaussian curvatures of a Monge patch in
Polar coordinates $(u\cos v, u\sin v, h(u,v))$ are:
\begin{align}
    H &= \frac{ uh_{uu} (u^2+h_v^2) -2 h_u^2(h_{uv}-1)h_v + (u^2h_u +v h_{vv})(1+h_u^2)}{2 \left[u^2(1+h_u^2)+h_v^2 \right]^{3/2}},\\
     \\\nonumber
    K &= \frac{u^2 h_{uu} (uh_u+h_{vv}) - (-h_v+h_u h_{uv})^2}{[u^2(1+h_u^2)+h_v^2]^2}.
\end{align}
Note that differentiations with respect to $u, v$ are
\begin{align}
    \frac{\partial}{\partial u}=\frac{\partial}{\partial r},\\
    \frac{\partial}{\partial v}=\frac{\partial}{\partial\theta}.
\end{align}
