From a bank of past master's exams:
Let $u(x,y)$ satisfy $ -\left( \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right) = \lambda u $ in a bounded region $\mathcal{D} \subset \mathbb{R}^2$ with smooth boundary $\mathcal B$. Assume $u=0$ on $\mathcal B$ (but $u \not \equiv 0$ on $\mathcal D$). Show that $ \lambda = \frac{\iint_{\mathcal D} |\nabla u|^2dxdy}{\iint_{\mathcal D} u^2dxdy}. $ (Suggestion: Compute the divergence $\operatorname{div}(u\nabla u)$.)
So taking the suggestion, I get the following: $ \begin{align} \nabla \cdot (u \nabla u) &= \nabla \cdot \left( u \frac{\partial u}{\partial x}, u \frac{\partial u}{\partial y} \right) \\ &= u \left( \frac{ \partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) + \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 \\ &= -\lambda u^2 + |\nabla u|^2 \end{align} $
If I get $ \iint_D \nabla \cdot (u \nabla u) dA = 0,$ then I'll have $\begin{align} \iint_D (|\nabla u|^2 - \lambda u^2) dA &= 0 \\ \iint_D |\nabla u|^2 dA = \iint_D \lambda u^2 dA &= \lambda \iint_D u^2 dA \\ \frac{\iint_D |\nabla u|^2 dA}{\iint_D u^2 dA} &= \lambda \end{align}$
But I don't know how to show that (I was thinking Stokes's Theorem, but I don't think the integrand is the curl of a vector field). It's probably something very simple (my vector calculus is quite rusty) -- what am I missing here?