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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.