How to prove this gradient theorem in vector calculus I encountered a theorem in some online notes labeled the 'gradient theorem':
$$\int_\Omega \nabla u = \int_\Gamma u \mathbf n$$
I've never seen this theorem before, and searches for 'gradient theorem' yield the fundamental theorem of calculus for line integrals which appears to be a different theorem. I've also never seen an explicit integral of a vector - only integrals of scalar quantities derived from vector quantities, like in Stoke's theorem or the divergence theorem.
How do I prove this theorem, and how do I even interpret a vector integrand?
 A: Take a constant vector field $\mathbf a$. Then by Divergence Theorem
$$
\mathbf a \cdot \int_\Omega \nabla u 
= \int_\Omega \mathbf a \cdot \nabla u 
= \int_\Omega \nabla \cdot (\mathbf a u) 
= \int_\Gamma \mathbf a u \cdot \mathbf n 
= \mathbf a \cdot \int_\Gamma u \mathbf n
$$
Since this is valid for all $\mathbf a$ we have
$$\int_\Omega \nabla u = \int_\Gamma u \mathbf n$$
A: That's only a notation to write the formula 
$$\tag{1}
\int_\Omega \frac{ \partial u}{\partial x_j}\, dV = \int_{\Gamma} u n_j\, dS,\quad j=1\ldots n.$$
Here $\Omega\subset \mathbb R^n$ is an open set with smooth enough boundary $\Gamma$ and $n_j$ are the components of the normal unit vector $\mathbf n$. The connection with the formula in the OP is given by
$$
\nabla u = \sum_{j=1}^n \partial_{x_j} u\, \mathbf e_j, $$ 
(here $\mathbf e_j$ is an orthonormal basis of $\mathbb R^n$). This shows that the formula in this post is equivalent to the one in the OP by linearity of the integral.
EDIT I will add here some hints to prove (1).
Without loss of generality assume that $\Omega\subset \mathbb R^2$. Also, assume that 
$$
\Omega=\{ (x_1, x_2)\ : -\infty<x_1<\infty,\ \phi(x_1)\le x_2\},$$
for a smooth function $\phi\colon \mathbb R\to \mathbb R$. With this assumption, $\mathbf n(x_1, \phi(x_1))=\frac{\phi'(x_1)\mathbf e_1 -\mathbf e_2}{\sqrt{\phi'(x_1)^2+1}}$. And finally, assume that $u\colon \Omega\to\mathbb R$ is an integrable function such that 
$$u(x_1, x_2)=0\qquad \text{for all sufficiently big }x_1, x_2.$$
These assumptions are motivated by the machinery of the partition of unity, which is used to remove them. 
The proof now goes as follows: 
$$\iint_\Omega \frac{\partial u}{\partial x_1}\, dx_1dx_2= \int_{-\infty}^\infty \left( \int_{\phi(x_1)}^\infty 
\frac{\partial u}{\partial x_1}\, dx_2\right)\, dx_1=\int_{-\infty}^\infty \left[\partial_{x_1}\left(\int_{\phi(x_1)}^\infty u(x_1, x_2)\, dx_2\right) +u(x_1, \phi(x_1))\phi'(x_1)\right]\, dx_1.$$
Here we have used Leibniz integral rule to take $\partial_{x_1}$ out of the integral. Now the first summand vanishes by the fundamental theorem of calculus, so we are left with 
$$\int_{-\infty}^\infty u(x_1, \phi(x_1))\phi'(x_1)\, dx_1$$
and you can check that this is equal to $\int_{\Gamma} u\, n_1\, dS.$
