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,


3 Answers 3


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

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

  • $\begingroup$ Your fraction renders strangely for me. Seems related to this meta thread, but unfortuntately your LaTeX is too complicated for me to parse. $\endgroup$
    – t.b.
    Commented Jul 14, 2012 at 8:55
  • $\begingroup$ Oh dear. I haven't seen that meta thread before; thanks! I'll have to think about how to reformat... $\endgroup$ Commented Jul 14, 2012 at 8:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Jul 14, 2012 at 16:42
  • $\begingroup$ The mean curvature $H$ given above for a Monge patch parameterized in the polar coordinates needs to be corrected. The denominator is correct, but the numerator is not. $\endgroup$ Commented Sep 20, 2022 at 13:16

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}

  • 1
    $\begingroup$ Lemma 13.34 in Gray et al. For a Monge patch $(u, v)\mapsto (u, v, h(u,v))$ we have the mean curvature $H=[ (h_{uu}(1+h_v^2) -2h_u h_v h_{uv} +(1+h_u^2)h_{vv}]/[2(1+h_u^2 +h_v^2)^{3/2}]$. $\endgroup$ Commented Sep 20, 2022 at 11:25
  • $\begingroup$ One can easily verify the correctness of curvatures provided above in the post by calculating $H$ and $K$ for well-known surfaces, such as a paraboloid $(u\cos v, u\sin v, u^2)$ in polar coordinates and compare the result from Lemma 13.34 in Gray et al. for the same paraboloid given by $(x, y, x^2+y^2)$ in Cartesian coordinates . $\endgroup$ Commented Sep 20, 2022 at 11:44

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 $f_{xy}$.

  • $\begingroup$ 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. $\endgroup$
    – Rodrigo
    Commented Jun 4, 2012 at 19:14
  • $\begingroup$ 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. $\endgroup$
    – Rodrigo
    Commented Jun 4, 2012 at 22:30
  • $\begingroup$ @Vittorio Patriarca I am not sure your answer is right or not. Especially if we are dealing with a Monge patch, $\rho$ is the radial coordinate, which should be independent. I have worked out the mean and Gaussian curvatures if a patch is expressed in $(u\cos v, u\sin v, h(u,v))$ . Any comments and corrections are welcome. $\endgroup$ Commented Sep 20, 2022 at 2:30

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