Consider an equation $$\Delta u=-he^{u}$$ over a compact 2-manifold $M$, where $u\in C^{\infty}(M)$.
In paper "Curvature functions for Compact 2-Manifolds" by Kazdan&Warner it is said that integration of the above equation over the manifold $M$ leads to the identity:
$$\int_{M}he^udA=0,$$ where $dA$ is an area element. The paper is not the only place I find this conclusion, but I cannot trace from what kind of argument would this follow.
But to have this conclusions, means that $\int_M\Delta udA=0$. But why is that so?
Recall that a compact manifold has empty boundary. You can apply the divergence theorem, $$ \int_M \operatorname{div}{X} \, dV_g = \int_{\partial M} \langle X, N \rangle \, dV_{\bar{g}}, $$ where $g$ is the metric, $\bar{g}$ the metric induced on the boundary, and $N$ a unit normal vector field (this is basically the Hodge dual of the general fundamental theorem of exterior calculus, for example). In this case, $\partial M= \emptyset$ and $X=\operatorname{grad} u$, so $$ \int_M \Delta u \, dV_g = 0. $$