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There are many examples where the notation of $\partial{U}$ is used to represent the surface boundary of some volume $U$. For example, the following representation of the Divergence Theorem: $$ \int_U{\nabla \cdot \mathbf{F}}\,dV = \oint_{\partial{U}}\,\mathbf{F}\cdot\mathbf{n}\,dS $$ What is the history and reasoning for the notation of using what appears to be a partial derivative of the volume for the surface. I am guessing that the derivative represents the tangent at the surface and thus refers to surface boundary but is there more then that?

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  • $\begingroup$ My guess would be it comes from the topological notation $\partial A$ for the boundary of a set. en.wikipedia.org/wiki/Boundary_(topology) $\endgroup$ – Gerald Edgar May 19 '16 at 21:48
  • $\begingroup$ @GeraldEdgar Or, closely related, from homology with its boundary operators. $\endgroup$ – Danu May 19 '16 at 22:27
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There seem to be two different rationales, one for how it first appeared, and the other for why it stuck, see MO thread. Francois Ziegler dug up the following remark in Michèle Audin's Henri Cartan & André Weil: du vingtième siècle et de la topologie

$d$ is for differential, no doubt, but what is $\partial$ for? Neither boundary nor frontier, in neither French nor German (nor even English), so? well, Rand, in German, written in gothic and abbreviated $\mathfrak{Rd}$, then $\mathfrak d$ (two letters, that was too much), then $\partial$!

He also found the corresponding use in §16 of Seifert and Threlfall's Lehrbuch der Topologie (1934):"Der Rand wird mit dem Symbole $\mathscr R\partial$ bezeichnet" [the boundary is denoted by the symbol $\mathscr R\partial$].

It seems unlikely that this convoluted justification would have impressed many in adopting the symbol. Fortunately, it has other things to recommend itself. E.g. it dovetails nicely with the dual use of $d$ in cohomology and calculus on manifolds, e.g. in the Stokes formula, $\int_M d\omega = \int_{\partial M} \omega$, of which all integral formulas of vector analysis, like the OP one, are particular cases. And it has some derivative like properties: $\partial(A\cap B)=(\partial A\cap B)\cup(A\cap\partial B)$ and $\partial(A\times B) =(\partial A\times B) \cup (A \times \partial B)$.

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  • $\begingroup$ Your answer justifies time to study but I can't get to that until this weekend. $\endgroup$ – K7PEH May 20 '16 at 4:37
  • $\begingroup$ I finally had a chance to study the answers and comments on the thread you linked (MO Thread) and find that the discussion there gives more sophisticated and better examples to help in answering. Therefore, I am marking your post as the answer due to this thread lead. Thanks for your efforts in looking this up. $\endgroup$ – K7PEH May 20 '16 at 15:33

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