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For a two-dimensional surface in $\mathbb{R}^3$, I thought that the total mean curvature was equal to the first variation of area:

$\frac{d}{dt}SA(t)\Big\vert_{t\to 0} = \int H dA,$

where $SA(t)$ is the surface area of the surface after flowing it along its normal vector field for time $t$.

But when I try this formula for a cylinder of unit height and radius $r$, I get that $SA(t) = 2\pi (r+t)$, $H=\frac{1}{2r}$, and

$2\pi \stackrel{?}{=} 2\pi r \frac{1}{2r} = \pi.$

Where have I gone wrong? Am I missing a factor of two in the first variation of area formula?

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    You could write that up as an answer and accept it so the question doesn't remain unanswered.2012-06-12

1 Answers 1

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As Giusepppe mentions in the comments, the right hand side in the formula is indeed incorrect. It should be

$\int 2 H dA.$

Sometimes in the literature mean curvature is defined as $H = k_1 + k_2$ instead of $(k_1+k_2)/2$ and this discrepancy was the source of my error.