The integral of the mean curvature vector over a closed immersed surface Suppose we have a closed, orientable, smooth surface $\Sigma$ immersed smoothly in $\mathbb R^n$ via $f:\Sigma \rightarrow \mathbb R^n$.  Impose a Riemannian structure on $\Sigma$ by taking $g_{ij} = \partial_if\cdot\partial_jf$, the metric induced on $\Sigma$ by the immersion $f$.  The inner product here is just the usual inner product from $\mathbb R^n$.
The mean curvature vector is
$$
\vec H = \Delta f,
$$
where $\Delta$ is the Laplace-Beltrami operator on $(\Sigma,g)$.
Consider the integral of the mean curvature vector over the surface $\Sigma$:
$$
\int_\Sigma \vec H\ d\mu.
$$
It seems rather plausible that this ought to be zero in the case where $\Sigma$ is closed, embedded, and has only one codimension.  Is this known?  Is it easy to prove?
If it is not zero in the generality above, as a surface immersed in $\mathbb R^n$, is it equal to some expression involving topological information of $\Sigma$?

For immersions into $R^3$ (rather than arbitrary codimension) there is basically a one-line proof:
$$2\int_\Sigma HN dA = \int_\Sigma df \wedge dN = \int_\Sigma d(fdN) = \int_{\partial\Sigma = \emptyset} f dN = 0.$$
Here $N$ is the Gauss map associated with the immersion $f: M \to \mathbb{R}^3$. The relationship $df \wedge dN = 2HNdA$ is also not hard: let $\kappa_i$ be the curvatures associated with principal directions $X_i$, so that $dN(X_i)=\kappa_i df(X_i)$.  Then
$$df \wedge dN(X_1,X_2) = df(X_1) \times dN(X_2) - df(X_2) \times dN(X_1) = (\kappa_1 + \kappa_2) df(X_1) \times df(X_2) = 2HNdA(X_1,X_2).$$
(Note that here the wedge product uses the cross product $\times$ to define an algebra on $\mathbb{R}^3$, i.e., $\alpha \wedge \beta(X,Y) := \alpha(X) \times \beta(Y) - \alpha(Y) \times \beta(X)$.)
