This is a sequel to the previous post on some integral formulas for hypersurface in Euclidean space. As in the previous post, always denotes the outward normal of whenever this makes sense.
Proof: Let be the standard coordinate vectors, let be the region enclosed by , then by divergence theorem,
Since is arbitrary, .
For , the above theorem is trivial because and where is the anticlockwise rotation by . It turns out that there is another way to prove the above theorem when . The observation is the following
Theorem 2 (Hon Leung-Raymond’s theorem, 2011) For a closed surface with outward unit normal , suppose is a domain with smooth one-dimensional boundary (not necessarily connected). Suppose is the positive tangent vector of (i.e. if we arclength-parametrize such that always lies on the left hand side of , then ), and is the position vector, then
Proof: Note that
(as a vector-valued 2-form) for any positively oriented local coordinates . We will simply denote by , by , and so on. Note that on one hand, by product rule,
On the other hand, we have
Combining, we have
(In a coordinate-free language, the above is just , which is understood as for differential forms . Note that in this setting !) So by Stokes theorem,
But for positively oriented parametrized by arclength, , so
In particular, for a closed surface , since , we have recovered Theorem 1: .
I do not know the answer to the following
Question: Is there a higher dimensional analogue of the above result?
As remarked by Edward, the above result can be generalized to the higher dimension.
Theorem 3 (Hon Leung-Raymond-Edward Theorem, 2011) For a closed hypersurface , if is its unit outward normal and is a domain in with smooth -dimensional boundary with induced orientation, then
where is a positively oriented orthonormal basis for the tangent space of and is the position vector.
Proof: As in the previous proof, it is not hard to see that
For example for (),
Thus by Stokes theorem,
With a similar computation as before, it can be seen that
Therefore we have
Let me also state the discrete version of Theorem 1:
Here is the -dimensional volume of .
Proof: Clearly we only have to prove the identity for to be a –simplex (triangle for , tetrahedron for ), for, given a polytope we can triangulate into finitely many -simplices, since the result is true for these simplices, summing up these identities, and noting that the terms involving a pair of adjacent faces of two adjacent simplices cancel each others, we can get the result.
The simplex case is very easy. Let us take for notational simplicity. Without loss of generality the 4 vertices are where are linearly independent. Let us denote the normal of simply by . Then