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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,

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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


Using the usual formula for mean curvature (equation 7 here) and simplifying, we obtain the expression

$$\small \frac{\frac{\partial f}{\partial r}\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+2\frac{\partial f}{\partial \theta}\left(\frac{\partial f}{\partial \theta}-r\frac{\partial f}{\partial r\partial \theta}\right)\right)+r\left(\left(\frac{\partial f}{\partial \theta}\right)^2+r^2\right)\frac{\partial^2 f}{\partial r^2}+r\frac{\partial^2 f}{\partial \theta^2}\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)}{2\left(r^2\left(\left(\frac{\partial f}{\partial r}\right)^2+1\right)+\left(\frac{\partial f}{\partial \theta}\right)^2\right)^{3/2}}$$

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Your fraction renders strangely for me. Seems related to this meta thread, but unfortuntately your LaTeX is too complicated for me to parse. – t.b. Jul 14 '12 at 8:55
Oh dear. I haven't seen that meta thread before; thanks! I'll have to think about how to reformat... – J. M. Jul 14 '12 at 8:58
Now it looks fine. I'll leave my comment in order to make clear that my report on meta addressed an earlier version of your answer. – t.b. Jul 14 '12 at 16:42

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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Thank you for the comprehensive response! I asked for quite a different thing but anyway what you did definitively help me to obtain the correct expression. – Rodrigo Jun 4 '12 at 19:14
Sorry, yet another question. Since $ \textbf{n} $ is dimensionless (though you forgot a root square, I think!), $ \bf{r_{\theta \theta}} $ and $ \bf{r_{rr}} $ should have the same units. But in your calculations they haven't - perhaps a $ 1/ \rho^{2} $ factor is missed somewhere? Thank you once more. R. – Rodrigo Jun 4 '12 at 22:30

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