Proof of equality of the integral and differential form of Maxwell's equation Just curious, can anyone show how the integral and differential form of Maxwell's equation is equivalent? (While it is conceptually obvious, I am thinking rigorous mathematical proof may be useful in some occasions..)
 A: Well, as the people said in the comments, the Theorems of Green, Stokes and Gauss will do the job, and are about as mathematically rigorous as you could hope for here!
The two different sets of formula follow directly.
I don't want to write all four of them out, you should be able to do them yourself, but for example, let's consider the Gauss Law.
Starting with the integral form, we have (ignoring physical constants)
$$ \int_{\partial \Omega} \vec{E} . d\vec{S} = \int_{\Omega} \rho\space dV$$
Then by Gauss, we have
$$ \int_{V} \mbox{div} \vec{F} \space dV = \int_{S} \vec{F} .d \vec{S} $$
Hence, we can replace 
$$ \int_{\partial \Omega} \vec{E} . d\vec{S} \rightarrow \int_{\Omega} \mbox{div} \vec{E} \space dV  $$
to give
$$ \int_{\Omega} \mbox{div} \vec{E} \space dV =  \int_{\Omega} \rho\space dV  $$
or dropping the integrals,
$$ \mbox{div} \vec{E} =   \rho\space   $$
which is the differential form.
You should try to derive the other three. This may be helpful in showing you where to start, and where you want to get to.
As for proofs of Green's, Stoke's and Guass Theorems, I recall learning them for some maths exams some years ago, but I wouldn't know where to begin now! Look at any differential geometry course or book and they should be somewhere early on. I can assure you though that the mathematicians have rigirous proofs for them, so we do not need to be shy in using the results of the theorems!
