I have a computational problem that I am solving that ultimately relies on either the areas (for $\mathbb{R}^2$) or the volumes (for $\mathbb{R}^3$ and above) for a pair of cropped objects. We can think of these cropped objects as closed, concave (but sometimes convex), orientable polytopes with finite surface area and no self-intersections that may have holes in their interior. In the case where the polytopes are in $\mathbb{R}^2$ or $\mathbb{R}^3$, one can use Green's theorem and Gauss' divergence theorem, respectively, to arrive at a closed-form solution to the contour integral. For $\mathbb{R}^d$, it is my guess that Stokes' theorem could be applied. Is this correct? For instance, when $d = 4$ and assuming no holes, would I write $\int\int\int\int_V (\nabla \cdot {\mathbf F})dV = \oint\oint\oint_{\partial V} ({\mathbf F} \cdot {\mathbf n})dS$ where $\mathbf{F}$ is a suitably chosen, continuously differentiable vector field defined on $V$ and $\mathbf{n}$ is the normal for the closed surface boundary $\partial V$?
(Closed, Concave, Orientable) Polytope Volume
3
$\begingroup$
calculus
geometry
1 Answers
2
Your intuition is correct.
This is best phrased in the language of differential forms (see the Wikipedia page), where we have an $(n-1)$-form $\omega$ defined on a smooth $n$-manifold with boundary ($n=4$ in you example). There are two useful things to do with this form:
- Restrict it to the boundary: $\omega|_{\partial M}$
- Take the exterior derivative to make an $n$-form: $d\omega$
Then, Stokes' Theorem says
$\displaystyle \int_M d\omega = \int_{\partial M} \omega$
All forms of Green, Gauss, Stokes theorems are higher-dimensional analogs of the Fundamental Theorem of Calculus.
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0I just wanted to point out that a general n-dimensional polytope is NOT a smooth n-manifold with boundary. An example to show that is a square on the plane, which is NOT a smooth 2-manifold with boundary. Anyway, I have read some proofs of Stokes Theorem for smooth n-manifolds with corners, which is a bigger class of regions than manifolds with boundary (Introduction to Smooth Manifolds by John M. Lee.). But still, there are certain polytopes which are not smooth manifolds with corners. I agree with the intuition, but I haven't found any proof of that, even less speaking in dimension 4. – 2011-08-14