TITLE: How to derive the divergence theorem from the General Stokes theorem?
QUESTION [15 upvotes]: So, given the generalized Stokes theorem:
$$\int_{\partial M} \omega = \int_M d\omega$$
where M is an n-dimensional surface and $\omega$ is a p-form on M (p < n). How can I derive the Divergence Theorem?
$$\iint_S {\bf F} \cdot d{\bf S} = \iiint_R \text{div}\;{\bf F}\; dV$$
I also have another related question. I'm learning that there are several theorems, like the divergence theorem, that are special cases of the generalized Stokes Theorem. For example, apparently, the Kelvin-Stokes Theorem is a special case of the General Stokes Theorem where n=2. So my 2nd question is, what if n=1 in the general stokes theorem? What does that imply or lead to?
Thank You.

REPLY [6 votes]: Rather late to the thread but since it seems that you are asking in particular about the 3-dimensional case, I shall provide an answer tailored to that case which is rather intuitive although a similar argument can be used for higher dimensions. 
First a preliminary. Note that the volume element on the manifold $dA$ is given by $$dA(v,w)=|v \times w|$$ where $v,w$ are vectors in $T_pM$.  Use this and the fact that $v \times w$ is aligned with $\vec{n}=n^x\hat{i}+n^y\hat{j}+n^z\hat{k}$ to derive the relations (see comments below for a proof if you don't see how to prove these). 
$$n^xdA=dy \wedge dz \tag{1}$$
$$n^ydA=dz \wedge dx \tag{2}$$
$$n^zdA=dx \wedge dy \tag{3}$$
Now, given the vector field $F=(F^x,F^y,F^z)$ , consider the 2-form $$\omega=F^xdy\wedge dz +F^ydz\wedge dx+F^zdx\wedge dy \tag{4}$$ on the boundary $\partial M$ of the 3-manifold M. 
Using the three relations above $$\omega=F^xn^xdA+F^yn^ydA+F^zn^zdA=F \cdot n dA \label{a}\tag{5}$$
Note that you can compute from (4) that $$d\omega=(F^x_x+F^y_y+F^z_z)dx\wedge dy \wedge dz$$ where the subscripts denote partial derivatives. In other words, $$d\omega=\operatorname{div}FdV\tag{6}$$ where $dV$ denotes the volume form. So, we are ready to use ($\ref{a}$) and (6) to write 
$$\int_M \operatorname{div}F dV= \int_M d\omega=\int_{\partial M}\omega= \int_{\partial M} F \cdot n dA$$