Let $\textbf{F}: \mathbb{R}^2\rightarrow \mathbb{R}^2$ be a continuously differentiable vector field. Write $\textbf{F}(x,y) = (f(x,y), g(x,y))$ and define the divergence of $F$ as $div\: \textbf{F} = f_x(x,y)+g_y(x,y)$. For a bounded piecewise smooth domain $\Omega$ in $\mathbb{R}^2$, let $\partial\Omega$ denote its boundary.
If $\Omega$ is the unit disk in $\mathbb{R}^2$, prove that $$\iint_{\Omega}div\:\textbf{F}\: dxdy = \int_{\partial \Omega} \textbf{F}\cdot\nu\:dS(x,y)$$ where $\nu$ is the exterior normal to the boundary and $dS$ denotes the area element.
Given an $n\times n$ matrix $M$, use the Divergence Theorem in general dimensions to prove that for any $R>0$, $$\frac{1}{\left|B_{R}(0)\right|}\int_{B_{R}(0)}My\cdot y\:dy=\frac{R^2}{n+2} \text{trace}(M),$$ where trace$(M)=M_{11}+M_{22}+...+M_{nn}$ is the trace of the square matrix $M$.
Any ideas how to solve this will be appreciated.