Volume of manifolds embedded in $\mathbb{R}^n$ Let $N$ be a closed, connected, oriented hypersurface of $\mathbb{R}^n$. Such a manifold inherits a volume form from the usual volume from on $\mathbb{R}^n$ and has an associated volume given by integration of the inherited volume form over $N$.
I would like to know if there is a proof of the following intuivitely obvious (I think) fact: let $X$ be a vector field on $\mathbb{R}^n$ everywhere transverse to the hypersurface. Let $\psi_\epsilon^X$ be the flow of such a vector field for a small time $\epsilon$ (we can choose $\epsilon$ small enough so that $\psi_\epsilon^X(N)$ is diffeomorphic to $N$). Then the volume of $\psi_\epsilon^X(N)$ is not equal to the volume of $N$. 
Does such a result exist? I assume, if so, it would not generalise to other manifolds/volumes other than the canonical one?
 A: I presume by the volume form on $N$ inherited from $\mathbb{R}^n$, you mean the induced hypersurface area measure. I'll write $N_{\epsilon} = \psi_{\epsilon}^X (N)$
The answer to your question is


*

*Yes if $N$ is mean convex,

*Maybe otherwise depending on $\int_{\partial N} \langle X, \nu \rangle H d\sigma$.


Here's the explanation:
The variation of volume (i.e. hypersurface area) is
$$
\partial_{\epsilon}|_{\epsilon = 0} V(N_{\epsilon}) = \int_{\partial N} \langle X, \nu \rangle H d\sigma
$$
where $d\sigma$ is the induced measure on the boundary $\partial N$ and $\nu$ is a choice of unit normal vector field and $H$ the mean curvature with respect to $\nu$.
Writing $\eta = \langle X, \nu \rangle$ we see that $\partial_{\epsilon}|_{\epsilon = 0} V(N_{\epsilon}) = 0$ if and only if
$$
\int_{\partial N} \eta H d\sigma = 0.
$$
The assumption that $X$ is everywhere transverse is that $\eta$ has a sign everywhere. Then in the case that $N$ is mean convex (i.e. $H$ has a sign everywhere) we get $H \eta$ has a sign everywhere and hence
$$
\int_{\partial N} \eta H d\sigma \ne 0
$$
so the volume is definitely different for small $\epsilon$.
Even in the case that $N$ is not mean convex, it could be that $\int_{\partial N} \eta H d\sigma \ne 0$ and the volume is different. If $\int_{\partial N} \eta H d\sigma = 0$, then the area is infinitesimally preserved - but it could still be that the volume is different if $\int_{\partial N_{\epsilon}} \eta Hd\sigma \ne 0$ for $\epsilon \ne 0$.
A: Yes if $N$ is convex - the flow will even increase or decrease distances. No if not convex, looking at $n=2$. The curve length of those parts of $N$ that are concave with respect to $X$ will decrease, and this may balance out the contribution from the convex parts.
