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.
Best Answer
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: $$\int_{\Omega}\operatorname{div}X~\operatorname{vol}_{\Omega}=\int_{\partial\Omega}\langle X,\mathbf{n}\rangle \operatorname{vol}_{\partial\Omega}.$$