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I've noticed that the boundary of a region $\Sigma$ as used in Stoke's theorem is always denoted by $\partial \Sigma$, yet I've always been told that a quantity of this form has no mathematical meaning as applied to a function in the sense that if $df = \frac{df}{dx} dx$ there is no analog with $\partial f = \frac{\partial f}{\partial x} \partial x$. I can't think of why it would be called anything else but on the other hand I can't quite work out why it's called $\partial \Sigma$.

$ \iint_{\Sigma} \nabla \times \mathbf{F} \cdot \mathrm{d}\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot \mathrm{d} \mathbf{r}. $

bjem
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  • RHS is a line integral – janmarqz Mar 01 '14 at 00:52
  • The thing that you were told has no mathematical meaning is $df$ where $f$ is a function, not a region, and this is wrong; it just doesn't have a simple meaning. See http://en.wikipedia.org/wiki/Differential_form for details. – Qiaochu Yuan Mar 01 '14 at 01:53

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The $\partial$ notation is not meant to indicate a "partial derivative" of the boundary or anything fanciful like that. Rather, it's just the particular symbol that we've chosen to represent the boundary of a region. The choice of $\partial \Sigma$ is perhaps more evocative when Stokes' theorem is written in the language of differential forms. Here, the theorem becomes $$ \int_{\Sigma} d\omega = \int_{\partial \Sigma} \omega, $$ so that the boundary operator $\partial$ is "adjoint" to the exterior differential $d$.

NKS
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    Well, or is it? Note that if $M, N$ are manifolds with boundary then $M \times N$ is a manifold with corners, and $\partial (M \times N) = (\partial M \times N) \cup (M \times \partial N)$. – Qiaochu Yuan Mar 01 '14 at 01:55