Is there a closed surface in 3D bounding a domain of non-zero volume such that $\int_{surface}\vec{n}dS \neq 0$?
Is there a closed surface in 3D such that $\int_{surface}\vec{n}dS \neq 0$?
0
$\begingroup$
vector-analysis
-
0No. That integral is always zero. – 2017-01-05
-
0That seems like a condition required for closed surfaces though. Is there a specific definition of closed surface? (needed for the statement to be provable) – 2017-01-05
-
0How about the Klein bottle? – 2017-01-05
-
0@TheMathsGeek There is no Klein bottle in 3D. If you have self intersections, the integral is not well defined. – 2017-01-05
-
0Ah ok - just a thought – 2017-01-05
-
0@gebra A closed surface (embedded in 3D space) is an orientable surface without boundary. Compact ones have been classified as the $n$-tori, and for non-compact ones the integral $\int \vec{n} dS$ is not well defined. – 2017-01-05
1 Answers
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.