How do they rigorously define a "curve bounding a surface" in Stokes' Theorem? Can more than one curve be the bound for any given surface with the integral remaining the same? And why is the integral the same no matter the surface?
2026-04-14 03:32:22.1776137542
Stokes' Theorem Details
335 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
Here's a stab at a start of an answer.
One way to formalize the notion of a surface and its boundary is to say that you have a (sufficiently smooth) map from some 2-dimensional manifold with boundary into your 3-dimensional space. This basically says that every point of the surface has a neighborhood in which the surface can be smoothly deformed into the shape of a standard "edge", like a neighborhood of a point that's either in the interior or on the edge of an ordinary square. It turns out that no matter which such neighborhood of a point you look at, they will always agree whether the point is on the edge or not; if it is, we say it's in the boundary. The boundary always forms a 1-dimensional manifold, meaning that every point has a neighborhood which can be continuously deformed to look like an open interval.
The independence of which surface you choose to close the boundary is analogous to the fact that the line integral of a gradient is independent of the path. There is a formalism where they are instances of the same thing, called differential forms. The wikipedia article talks a bit about Stokes' theorem for differential forms, and a relatively short ( less than 100 pages) elementary introduction can be found in Spivak's Calculus on Manifolds or any number of other places. The transition from vector calculus to differential forms is an important rite of passage in any mathematician's education: see some discussion and further references here.
I really should say something about how, geometrically, this works, and maybe I or somebody else will.