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?
Best Answer
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.