$$(\nabla \times \mathbf{F}) \cdot \mathbf{\hat{n}} \ \overset{\underset{\mathrm{def}}{}}{=} \lim_{A \to 0}\left( \frac{1}{|A|}\oint_{C} \mathbf{F} \cdot d\mathbf{r}\right)$$

This is how the curl operator is usually defined, and I want to know the proof that the left hand and the right hand are equivalent.

  • $\begingroup$ I think you should use Stock's formula here. $\endgroup$ – Nikita Evseev Oct 5 '12 at 8:49
  • $\begingroup$ oh right...... thanks. $\endgroup$ – dervi Oct 5 '12 at 8:53
  • $\begingroup$ @nikita2 Stock=Stokes? $\endgroup$ – Siminore Oct 5 '12 at 9:57
  • $\begingroup$ @Siminore Definitely, I meant it. en.wikipedia.org/wiki/Stokes%27_theorem $\endgroup$ – Nikita Evseev Oct 5 '12 at 10:06

If A is an element of surface area in bounded by a simple closed curve C, let P be an interior point in C and $\mathbf{n}$ a unit normal at P.

By Stokes' theorem, $\iint(\nabla\times \mathbf{F})\cdot \mathbf{n}~dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r}.$

Using the mean value theorem for integrals* we can write this as

mean[$(\nabla \times \mathbf{F} )\cdot \mathbf{n}$] =$ \frac{\oint_{C} \mathbf{F} \cdot d\mathbf{r}}{\Delta A},$ and the result follows from taking the limit as $\Delta A \to 0. $

In words, the expression for (curl$\cdot\mathbf{n}$) reaches a limiting value as the area A shrinks around the point P.

*The MVT for integrals is:

If a function f is continuous on [a,b] there exists a point c on [a,b] such that $$f(c) = \frac{1}{b-a}\int_a^b f(x)dx.$$


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