What is the "boundary" in Stoke's theorem? The classical Stokes' theorem "relates the surface integral of the curl of a vector field over a surface Σ in Euclidean three-space to the line integral of the vector field over its boundary". As far as I can tell, this definition assumes that the boundary of Σ is a closed path.
However, the topological boundary of a surface in R^3 is not always a path -- it's typically the entire surface. For example, every point of the unit disk in R^3 is a boundary point (in the standard topology of R^3).
How exactly is the "boundary of Σ" defined? Is it the topological boundary of Σ in the subspace topology of Σ?
 A: You need some regularity requirements on $\Sigma$ for the boundary to be nice enough for Stoke's theorem to apply, or for the boundary integral to even make sense. It certainly won't be the topological boundary in the subspace topology of $\Sigma$, since this is always empty ($\Sigma$ is open by definition of the subspace topology!).
The simplest case is when $\Sigma$ is a (sub-)manifold with boundary; in which case $\partial \Sigma$ is defined by the boundary points of the "half-charts". This generalises to rectifiable subsets with $C^1$ boundary (which essentially means "piecewise submanifold with boundary"), and even further to the ideas of currents and varifolds in geometric measure theory.
For the case of surfaces in $\mathbb{R}^3$, the definition of a $C^1$ submanifold with boundary goes something like this: for every $p \in \Sigma$ either:


*

*there is a neighbourhood $U$ of $p$ such that $U \cap \Sigma$ is the image of an open disc under a $C^1$ immersion; or

*there is a neighbourhood $U$ of $p$ such that $U \cap \Sigma$ is the image of the half-plane $\{(x,y) \in \mathbb{R}^2 : x \ge 0\}$ under a $C^1$ immersion.


The submanifold boundary is then the set of points that satisfy only the second condition. The identification of pieces of the boundary with the vertical axis $\{y=0\}$ under the various $C^1$ maps means it is a collection of $C^1$ curves, so the boundary integral makes sense.
