# Gauss/Divergence Theorem special case

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.

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.