Let $F$ be a vector field in $\mathbb{R}^3$. If $F$ is divergence free, we may deform the surface. Why?

Question

In working through a solution, I can across the following generalization about vector fields and the Divergence Theorem. Can someone furnish a standard proof of this or at least its intuition?

Let $F$ be a divergence-free vector field in $\mathbb{R}^3$. This means that we can deform our piecewise smooth, compact surface in $\mathbb{R}^3$ without changing the value of the surface integral of $F \cdot \vec{n}$.

Answer 1

One can see the claim intuitively as an application of the Divergence Theorem: Suppose first that two surfaces $D, D'$ intersect only along their common boundary, so that they together bound a solid region $R$. If we choose orientations of $D, D'$ so that (1) the orientation of $D'$ is outward-pointing (w.r.t. $R$) and (2) the deformation takes the orientation of $D$ to the orientation of $D'$, then the oriented boundary of $E$ is $$\partial E = D' \cup -D ,$$ where $-D$ denotes the surface $D$ with the reverse of the specified orientation.

Then, by the Divergence Theorem the difference of the integrals over the surfaces is $$\iint_{D'} {\bf F} \cdot d{\bf S} - \iint_{D} {\bf F} \cdot d{\bf S} = \iint_{D'} {\bf F} \cdot d{\bf S} + \iint_{-D} {\bf F} \cdot d{\bf S} = \iiint_{\partial E} \text{div } {\bf F} \,dV = \iiint 0 \, dV = 0 .$$

Rearranging gives that $$\color{#bf0000}{\boxed{\iint_{D'} {\bf F} \cdot d{\bf S} = \iint_{D} {\bf F} \cdot d{\bf S}}} ,$$ that is, that the surface integrals are the same.

More generally, the union of two surfaces $D, D'$ with the same boundary is the boundary of multiple solid regions with disjoint interiors, and roughly we can apply the above argument to each region, add the resulting equalities, and conclude the same independence of the surface integral.

Remark This is, by the way, the surface analogue of the independence of path of line integrals of irrotational vector fields under deformation of curves, whose proof follows from Stokes' Theorem:

Let $\bf F$ be a $C^1$ vector field on an open set $U \subseteq \Bbb R^3$ such that $\text{curl } {\bf F} = {\bf 0}$, and let $\gamma, \gamma'$ be homotopic curves in $U$ with the same endpoints. Then, $$\int_{\gamma} {\bf F} \cdot d{\bf s} = \int_{\gamma'} {\bf F} \cdot d{\bf s} .$$

These are both special cases of a general result that follows from the (Generalized) Stokes' Theorem.

Answer 2

If the vector field is divergence-free everywhere, then the divergence theorem reads

$$\oint_{\partial M} F \cdot \hat n \, dS = \int_M \nabla \cdot F \, dV = 0$$

for any, all, and every possible region $M$ that satisfies the conditions of the general theorem.

The given statement merely concerns deforming a given region $M$ into another region, and so for the corresponding boundaries.