I was wondering about a simple detail. As you probably already know, there are two definitions curl and divergence can be defined in the following way (in $\mathbb{R^{2}}$ but the question is also valid for higher dimensions) : $$ \text{rot(f)(u)} = \lim_{\epsilon \to \,0}\frac{1}{\pi r^{2}}\int_{\partial B_{\epsilon}(u)}f\; \cdot \; \text{d}s $$ and: $$ \text{div}(f)(u) = \lim_{\epsilon \to \,0} \frac{1}{4\epsilon^{2}}\int_{u + B_{ \infty}( \epsilon )} f \; \cdot \text{d}n$$

So here is my question. Is there a reason beyond semantics, why the divergence is defined over a square, as opposed to a disk ? Intuitively, you should be able to define div with a disk...
What do i mean by semantics ? I have seen divergence being introduced first in the calculus class on a square, before generalizing it to functions with a smooth boundary. So therefore it makes sense to introduce it over squares to begin with.

  • $\begingroup$ ? You can use either discs/balls or squares/cubes. It's a matter of taste.... $\endgroup$ – Lord Shark the Unknown Jul 19 '18 at 17:40
  • $\begingroup$ All right, that was the question, so the main reason is semantics. $\endgroup$ – Pastudent Jul 20 '18 at 16:45

Your formulas are true, and in a way convey an intuitive understanding of ${\rm curl}$ and ${\rm div}$; but they should not be considered as "definitions" of these concepts. The correct definitions are either in terms of orthogonal coordinates $x$, $y$, $z$, or in terms of exterior algebra. Using these definitions one then proves Stokes' theorem and the divergence theorem. These theorems immediately show that such formulas are valid for bodies of arbitrary (reasonable) shape and shrinking to a point by scaling.


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