 Solution

The divergence of $\mathbf{F}=\nabla z$ is the LaplacianIts area integral over the square isFor the boundary flux, ${\int }_0^{2}A(s)ds=-4$. On $x=0$, $\mathbf{F}\cdot \mathbf{n}=4A(y)$, giving $-16$; on $x=2$ it is $8A(y)$, giving $-32$. The corresponding horizontal sides give $-16$ and $-32$. The total is again $-96$, verifying the divergence theorem with outward normals. With a counterclockwise boundary, this is the same orientation as $Mdy-Ndx$.