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Suppose $\iint\mathbf{F}\cdot d\mathbf{a}=0$ for any closed surface. Prove that $\iint\mathbf{F}\cdot d\mathbf{a}$ is independent of surface, for any given boundary line.

My attempt:

I have the right idea but I'm having trouble with the signs.

Let $A_1$ and $A_2$ be two surfaces with the same boundary line.

$A_1\cup A_2$ is a closed surface. Or should it be $A_1\cup(-A_2)$?

The problem is that the sign of a surface integral is intrinsically ambiguous.

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if $\unicode{x222F} F\ dS = 0$ for all closed surfaces, then the divergence of $F =\nabla \cdot F= 0$

If F is a divergence free field then there exists some $G$ such that the curl of $G =\nabla \times G = F$

By Stokes theorem

$\oint G\cdot dr = \iint \nabla \times G \ dA = \iint F \ dA$

And all integrals with the same contour evaluate to the same thing.

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