Differentiate under the integral sign Let $\Sigma\subset \Bbb{R}^n$ be some compact hypersurface without boundary in $\Bbb{R}^n$,fixed $v = -H$ where $H$ is mean curvature vector, assume we have a family of hypersurface $$\Sigma_{s} = \{x+sv(x)\mid x\in \Sigma\}$$
Prove the differentiation under the integral sign formula
$$\frac{d}{dt}\int_{\Sigma_t} u = \int_{\Sigma_t} u_t -\int _{\Sigma_t}u|H|^2$$
where $H$ is the mean curvature vector.
My attempt using the differentiation under the integral sign formula in calculus we have $$\frac{d}{dt}\int_{\Sigma_t}u\ dV_g=  \int_{\Sigma_t}u_t  \ d V_g + \int_{\partial\Sigma_t} u \langle v,n\rangle\ dS$$
Correct where $v$ is the velocity direction of the change of $\Sigma _t$ and $n$ is the out normal direction of $\Sigma_s$.
However, as $\Sigma_t$ does not has boundary, the second term on the right hand side is zero?Sorry for the silly question, I can't see where goes wrong with my reasoning?
 A: The problem of the reasoning has been stated in the comment.
Let's see how to deduce it correctly (by the way, the similar argument has already given in Colding & Minicozzi's minimal surface book), I just mimic it.
Consider the embedding $F:\Sigma \times (-\epsilon, \epsilon) \to \Bbb{R}^{N}$  that maps $x\mapsto x- H(x)t\ $ with image $F(\Sigma , t) = \Sigma_t$, therefore the manifold $\Sigma _t$ isometric isomorphic to $(\Sigma,(F(\cdot,t))^*(\bar{g}))$ where $F(\cdot, t): \Sigma\to \Bbb{R}^N$ and $\bar{g}$ is standard Euclidean metric. therefore the new metric on $\Sigma$ is $$g_{i,j}(t) = \bar{g}(dF(\partial_i),dF(\partial_j))$$
therefore we have under the local coordinate $$\int_{\Sigma_t} u(x,t) \ dV_{t} = \int_{\Sigma} u \sqrt{\det{g_{ij}(t)} \det{g^{ij}(0)}}\sqrt{\det(g_{ij}(0))} dx^1\wedge...\wedge dx^n$$
denote $\nu(t) = \sqrt{\det{g_{ij}(t)} \det{g^{ij}(0)}}$Now use the fact in the book referenced above page 7, we have
$$\frac{d}{dt}\nu(t) = \text{div}_{\Sigma}(F_t) = \text{div}_{\Sigma}(-H)$$
therefore $$\frac{d}{dt}\int_{\Sigma_t}u  = \frac{d}{dt}\int_{\Sigma} u\nu dV_0 = \int_{\Sigma} u_t \nu + \frac{d}{dt}\nu u dV_0 = \int_{\Sigma_t} u\ dV_t - \int_{\Sigma}|H|^2 u$$
For the global case, just use the partition of unity which finish the proof.
