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In one of my calculus exercises, I am asked the following question:

Use the Gauss theorem to show that if B is a vector field defined on a open set U in $R^{3}$ that has $div(B)=0$ then the flux of B through a oriented surface $\Sigma$ contained in U and with border $\gamma$ doesn't depend on $\Sigma$ but only on $\gamma$.

I don't have a clue as to how to show this said independence, should I assume that maybe $\Sigma$ is the border of another surface, and then apply the divergence theorem? Still I wouldn't be able to get any useful information out of that.

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Yes you can take two surfaces with a common border, apply the divergence theorem in the three dimensional region between them and then split your surface integral into two pieces, corresponding to the original surfaces with common boundary $\gamma$. Observe that the two surfaces will have opposite orientation w.r.t. the curve $\gamma$. By changing one of them you obtain two equal surface integrals. Therefore that common value depends on $\gamma$ only. Here are the details . Let $S_1$ and $S_2$ be two surfaces with common border $\gamma$, the orientation of $S_1$ being in agreement with that of $\gamma$ (Think of $S_1$ as being the northern hemisphere of a ball and $\gamma$ being The equator oriented in such a way that the corresponding normal vector at the North Pole is pointing up and $S_2$ some other surface with the same equator as its border but located below the equator). Let the region in between be $D$. By the divergence theorem you have $$ 0=\int_D div\,F\,dV=\int_{S_1\cup S_2} F\cdot dS $$ $$ =\int_{S_1}F\cdot dS+\int_{S_2}F\cdot dS $$ Where both pieces $S_1$ and $S_2$ are oriented with normals pointing out of $D$. Therefore $$ \int_{S_1} F\cdot dS=-\int_{S_2} F\cdot dS. $$ The minus sign in the second integral can be removed if we change the orientation of $S_2$ to the opposite. By doing that, the orientation of both surfaces are now in agreement with the orientation of $\gamma$ and the equality is thus established.

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  • $\begingroup$ How exactly do you split the surface integral, and what do you mean change one of them to obtain two equal surface integrals ? $\endgroup$ – Guilherme takata Jun 26 at 23:41
  • $\begingroup$ See my edited answer $\endgroup$ – GReyes Jun 27 at 0:28

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