Zero Mean Curvature

I heard somewhere or another the claim that soap film forms a "surface of zero mean curvature". I wasn't sure exactly what that meant until I read this book which gave me the tools to understand what that means, and prove it. It turns out to be a simple consequence of Gauss's Divergence theorem. In tensor notation Gauss's Divergence theorem states \int_{\Omega}\nabla_iT^idV = \int_{\partial{\Omega}}T_iN^idA where \Omega is some volume of space \nabla_i is the covariant derivative along the ith coordinate z^i \delta{\Omega} is the surface of the volume T^i is any single index tensor defined over the whole volume N^i is the unit vector normal to the surface in contravariant form This works in any number of dimensions, so if you take surface embedded in 3 dimensional euclidean space \vec{z} = \vec{z}(s^1,s^2) and cut it, then the theorem tells us $$ \int_{S}\nabla_{\alpha}T^{\alpha}dA = \int_{\partial{S}}T_{\alpha}n^{\alph...