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?