I'm tasked with showing that $$\iint_D \nabla^2 u dA = \oint_C \frac {\partial u}{\partial n} ds$$ where $\frac {\partial u}{\partial n}$ is the outerward normal derivative.
I'm not sure if I can necessarily do things like let $dA=dxdy$ because this for any general $2D$ region. The only way I can think to show this trivially with the divergence theorem:
$$\oint_C \frac {\partial u}{\partial n} ds = \oint_C \nabla u \cdot \hat n ds = \iint_D (\nabla \cdot \nabla u)dA = \iint_D \nabla^2 u dA$$
But it seems a little underhanded to prove Green's theorem with the divergence theorem. Is there another way I can do this?
Provided that $u_{xx}$ and $u_{yy}$ are continuous on the region containing $C$ and its interior, we have:
$$\oint_C \frac {\partial u}{\partial n} ds = \oint_C \nabla u \cdot n ds$$ $$= \int_a^b u_x \frac{dy}{dt} - u_y \frac{dx}{dt} dt = \oint_C u_x dy - u_y dx$$ $$= \iint_D u_{xx} + u_{yy} dA$$
In the above, $t$ is the original parametrization that led to the arc-length parametrization $s = s(x) = \int_a^x |\alpha'(t)| dt$.