integral of Laplacian of a positive function

I've encountered the following, rather elementary, problem:

$K$ is a compact subset of some 2-dimensional oriented manifold with smooth boundary, $f$ is a positive smooth function on $K$ that vanishes on the border $\partial K$. I want to show that the integral of the Laplacian of $f$ is non-positive, where Laplacian is given by $\Delta f = (f_{xx} + f_{yy}) dx \wedge dy$ in local coordinates.

It seems that one can use Stokes to compute: $$\int_K \Delta f = \int_{\partial K} \nabla f \cdot \mathbf{n} \ dl \leq 0$$ where $\mathbf{n}$ is the normal vector and the inequality follows from the fact that since "$f$ decreases in the direction of the boundary", on the boundary we have $\nabla f \cdot \mathbf{n} \leq 0$.

I was wondering if there is some more elegant reason for the inequality in question. Also: is the given argument correct (my analysis is a bit rusty ;) ) ?

-
It's the other way around; the directional derivative along the outward normal is non-positive, so the integral is non-positive. Also, in the displayed inequality you allow for the possibility that it's zero, but in the text you don't, which is wrong, since the gradient could vanish on the boundary, in which case the integral of the Laplacian would vanish. –  joriki Mar 2 '12 at 14:15
@joriki: Thank you! I was a bit careless with the formulation, and I got the sign wrong. I'm just about to correct it. –  Feanor Mar 2 '12 at 14:46