Is there a closed surface in 3D such that $\int_{surface}\vec{n}dS \neq 0$?

Question

Is there a closed surface in 3D bounding a domain of non-zero volume such that $\int_{surface}\vec{n}dS \neq 0$?

Answer 1

Provided that $S$, is orientable, this is easy: Let $f$ be a constant vector field. Then $\operatorname{div}{f}=0$, and the Divergence theorem shows $$ \int_S n \cdot f \, dS = \int_V \operatorname{div}{f}=0. $$ One can choose two more orthogonal constant vector fields to show that the rest of $n$ is zero.