# Closed surface integral of the surface's normal vector

Is it true that the surface integral over any closed surface (we are in $$\mathbb R^3$$) of the normal vector $$\hat n$$ of that surface, say $$\hat n$$ is pointing outward, is zero? In other words, is it true that

$$\iint_S \hat n \ dS = 0$$

for any closed surface $$S$$?

I recently ran across such an integral and while intuitively it would seem to have to be zero (it's quite obvious if the closed surface is a cube or a sphere), is it true for any closed surface and what would be the best way of proving this statement?

Yes, the integral is always $0$ for a closed surface. To see this, write the unit normal in $x,y,z$ components $\hat n = (n_x,n_y,n_z)$. Then we wish to show that the following surface integrals satisfy $$\iint_S n_x dS = \iint_S n_y dS = \iint_S n_z dS = 0.$$
Let $V$ denote the solid enclosed by $S$. Denote $\hat i = (1,0,0)$. We have via the divergence theorem $$\iint_S n_x dS = \iint_S (\hat i \cdot \hat n) dS = \iiint_V (\nabla \cdot \hat i) dV = 0,$$ since $\hat i$ is constant. The other two integrals are computed analogously.