I'm trying to prove the following: $$\iint_R(\phi_1\nabla^2\phi_2-\phi_2\nabla^2\phi_1)\ dR = 0$$
For $ 0 < x < L, 0 < y < H$. Given $\phi_i(x,y) $ satisfying the boundary conditions $$ \phi_i(x,0) = 0 = \phi_i(x,H) \ \ ||| \ \frac{\partial \phi_i}{\partial x}(0,y) = 0 = \frac{\partial \phi_i}{\partial x}(L,y) $$
where the $\phi_i(x,y) $ also satisfy $$\nabla^2 \phi_i + \lambda_i \phi_i = 0 $$
This is what I've done so far:
$$\iint_R(\phi_1\nabla^2\phi_2-\phi_2\nabla^2\phi_1)\ dR = \oint_{\partial R}(\phi_1\nabla\phi_2-\phi_2\nabla\phi_1) \cdot \textbf{n} \ ds$$ which is deduced from Green's Second Theorem. My issue is how to evaluate this last line integral given that there's the gradient and I don't know the normal vector.
$\oint_{\partial R} (\phi_1\nabla\phi_2-\phi_2\nabla\phi_1)\cdot \mathbf{n}$ = $$\int_{A} (\phi_1\nabla\phi_2-\phi_2\nabla\phi_1)\cdot \mathbf{n} \ d\sigma + \int_{C} (\phi_1\nabla\phi_2-\phi_2\nabla\phi_1)\cdot \mathbf{n} \ d\sigma$$
Where A and C are the vertical borders of the rectangle (our domain). Here, we use boundary conditions, that's why horizontal borders disappear. But the two remaining integrals have null integrands : $\phi_1\nabla\phi_2-\phi_2\nabla\phi_1$ is a $y$ coordinate vector, and the normal vector is a $x$ coordinate vector (since A and C are the vertical borders), so the dot product is null.