Integral form of Gauss's law for magnetism from Stokes' theorem? How can the integral form of Gauss's law for magnetism be described as a version of general Stokes' theorem? How does it follow?
 A: Maxwell's equations in curved spacetime are written in the form
$$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with $F$ the Faraday two-form, $J^a$ the current four-vector, $\nabla$ the covariant derivative and $[]$ denotes antisymmetrization of the indices. In terms of exterior calculus they become:
$$ \begin{split} d\star F &= 4\pi \star J\\ dF&=0,\end{split}$$ with $\star$ the Hodge dual, which sends p-forms to $4-p$-forms in dimension 4. If we integrate the left side of the first equation over a space-like hypersurface of dimension 3, $\Sigma$, with normal time-like vector $t^a$, then Stokes' theorem yields
$$\int_\Sigma d\star F = \int_S \star F,$$ with $S$ the boundary of $\Sigma$ with normal $n_a$. Since $\Sigma$ is space-like and $(\star F)_{cd} = \frac{1}{2} F^{ab} \epsilon_{abcd}$, $S$ is also space-like and one component of $F$ must be time-like, therefore $\star F = F^{ab} t_a n_b dS$. This is easy to see also if one takes the restriction of the dual on $S$ in local coordinates, all the 2-forms are space-like. But we know that $E_a = F_{ab} t^b$, hence 
$$\int_S \star F = -\int_S F^{ba} t_a n_b d S = -\int_S E_b n^b d S.$$
Now we integrate the right side,
$$\int_\Sigma \star J = \int_\Sigma J^a \epsilon_{abcd}=\int_\Sigma J^a t_a d \Sigma = -q,$$ and after combining this and the previous one, we obtain:
$$\int_S E_a n^a d S = 4\pi q,$$ Gauss's law applies also in curved spacetime. Note that $\epsilon_{abcd}$ is the Levi-Civita (volume) tensor, not the symbol. In local coordinates its components are the product of the symbol with $\sqrt{|\det{g_{\mu\nu}}|}$.
For the case of the magnetic field, $B_a = - \frac{1}{2} \epsilon_{abcd} F^{cd} t^b$, only the space-like components of $F_{ab}$ are used, and the magnetic field part of the Faraday tensor is
$$F = F_{12} dx^1\wedge dx^2 + F_{23} dx^2\wedge dx^3 + F_{31} dx^3\wedge dx^1,$$ and the components of the field are $B^1 = - |\det g_{\mu\nu}|^{-1/2} F_{23}$ etc, therefore the integrals become
$$0=\int_S F = -\int_S \sqrt{|\det g_{\mu\nu}|} (B^1 dx^2\wedge dx^3 + B^2 dx^3\wedge dx^1 + B^3 dx^1\wedge dx^2) = - \int_S B^a n_a dS.$$
A: So on in three-dimensional Euclidean space we have an isomorphism between vectors and 1-forms, the usual way $$\eta_\mu = g_{\mu\nu} \eta^\mu.$$
We also have an isomorphism between 1-forms and 2-forms, given by $\star : dz\mapsto dx\wedge dy$ and cyclically. This isomorphism has a fancy name, the Hodge dual, if you want to know about it in general. Then if $B^\mu$ is the magnetic field, we can make a 3-form -- something that can be integrated over a volume -- out of it by (i) lowering the index to get a 1-form (ii) taking the Hodge dual to get a 2-form (iii) using $d$ to get a 3-form. More explicitly, $$B_\mu = B_x dx + B_y dy + B_z dz$$ 
$$(\star B)_{\mu\nu} = B_x dy\wedge dz + B_y dz\wedge dx + B_z dx\wedge dy$$
$$(d\star B)_{\mu\nu\rho} = \frac{\partial B_x}{\partial x} dx\wedge dy \wedge dz + \frac{\partial B_y}{\partial y} dy\wedge dz\wedge dx + \frac{\partial B_z}{\partial z} dz\wedge dx\wedge dy$$
But this is $$(d\star B)_{\mu\nu\rho} = \left(\frac{\partial B_x}{\partial x}  + \frac{\partial B_y}{\partial y}  + \frac{\partial B_z}{\partial z} \right) dx\wedge dy\wedge dz$$
which is what those poor souls who don't know about differential forms call $\nabla \cdot \mathbf B$. Now you can just apply Stokes's wonderful theorem!
This is what they should teach you in multivariable calculus but don't!
A: Without specifying any particular scenario, and ignoring any proportionality constants, simply consider some general differential form $\omega$, and let this represent the electric flux through a closed surface which bounds some volume V. In classical electromagnetism, the Gauss law tells us that the flux through a closed surface is proportional to the amount of charge enclosed within that surface; in other words, visualise the flux as "flux tubes" which terminate inside the volume V. If $\omega$ represents the flux tubes, then $d\omega$ represents their end points, and we can write the Gauss law simply and intuitively as 
$\displaystyle{\int_{\partial V}\omega =\int_{V}d\omega}$
This is just precisely the generalised Stokes Theorem - the amount of flux tubes ending inside the volume equals the amount of flux tubes crossing the surface. 
