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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.

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    $\begingroup$ To your last question, it's the fundamental theorem of calculus. $\endgroup$ – spaceisdarkgreen Nov 18 '17 at 1:51
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    $\begingroup$ You need to identify vector fields on $\mathbb{R}^3$ with differential forms. By the way, by "surface" we usually mean something $2$-dimensional. Also, we only integrate top-dimensional forms with the standard definition of the integral $\endgroup$ – leibnewtz Nov 18 '17 at 3:27
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Let $\Omega$ be an open subset of $\mathbb{R}^n$ with $\partial\Omega$ of class $\mathscr{C}^\infty$, and let $X$ be a smooth vector field on $\Omega$. Now we compute \begin{align} d(i_X\operatorname{vol}_{\Omega}) & = d\left(i_X\left(dx^1\wedge\cdots\wedge dx^n\right)\right) \\ & = d\left(\sum_{i=1}^{n}(-1)^{i+1}X_i~dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n\right) \\ & = \sum_{i=1}^{n}\frac{\partial X_i}{\partial x^i}~dx^1\wedge\cdots\wedge dx^n \\ & = (\operatorname{div}X)\operatorname{vol}_{\Omega}, \end{align} where $\operatorname{vol}_\Omega$ denotes the volume form on $\Omega$, and $i_X$ denotes the interior product with $X$. From Stokes' theorem we obtain $$\int_{\Omega}\operatorname{div}X~\operatorname{vol}_{\Omega}=\int_{\Omega}d(i_X\operatorname{vol}_{\Omega})=\int_{\partial\Omega}i_X\operatorname{vol}_{\Omega}.$$ Now decompose $X$ into it's tangential and normal components on $\partial\Omega$, i.e. $X=X^\top+X^\bot$. Then one easily computes $$i_X\operatorname{vol}_{\Omega}=\operatorname{vol}_{\Omega}(X^\top+X^\bot,\cdots)=\operatorname{vol}_{\Omega}(\langle{X,\mathbf{n}}\rangle\mathbf{n},\cdots)=\langle X,\mathbf{n}\rangle \operatorname{vol}_{\partial\Omega},$$ where in the above $\mathbf{n}$ is the outward facing unit normal vector on $\partial\Omega$. Now the usual divergence theorem follows immediately: $$\int_{\Omega}\operatorname{div}X~\operatorname{vol}_{\Omega}=\int_{\partial\Omega}\langle X,\mathbf{n}\rangle \operatorname{vol}_{\partial\Omega}.$$

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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 $$\int_M \operatorname{div}F dV= \int_M d\omega=\int_{\partial M}\omega= \int_{\partial M} F \cdot n dA$$

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    $\begingroup$ Proof of (1): let $\vec{c}$ denote the cross product $v \times w$. Since it is in the direction of the unit normal, $\hat{n}C=\vec{C}$. Examine the first component of either side of this vector equation. The LHS is $n^xC=n^xdA(v,w)$. The RHS is$ (v^yw^z-v^zw^y)=dy \wedge dz(v,w)$. That proves the relation. Likewise, use the other two components of the vector equation to prove (2) and (3) $\endgroup$ – Mathemagical Jan 22 '18 at 12:48

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