Why does the curl of a vector field converge?

[guitarphysics]

This is more of an intuitive question than anything else: the curl of a vector field $\mathbf{F}$, $\nabla \times \mathbf{F}$ is defined by
$$(\nabla \times \mathbf{F})\cdot \mathbf{\hat{n}} = \lim_{a \to 0} \frac{\int_{C} \mathbf{F}\cdot d\mathbf{s}}{a}$$

Where the integral is taken around a closed curve $C$, $\mathbf{\hat{n}}$ is the normal unit vector to that curve, and $a$ is the area of the curve.

Now, my question stems from the following: roughly speaking (if we have, for example, a constant vector field and a "flat" curve), the area of the curve decreases as the square of the perimeter, as we make the curve smaller. On the other hand, the line integral of the vector field along the curve decreases proportionally to the perimeter. So how can the ratio of the line integral to the area converge, if the area decreases more rapidly than the integral?

(A similar question could be asked of the divergence, of course.)

 

[Orodruin]

If you have a constant field, the circulation integral and hence also the curl is zero. The first order correction to this is proportional to the area and some derivatives of the components of the vector field. You. Should be able to find the derivation in any introductory text on vector analysis.

 

[guitarphysics]

Oh, right- the constant field was a bad choice.
Anyway, I just checked the argument given by Purcell (unfortunately don't have any vector analysis books)- it was that you taylor expand your vector field, and the first order terms in your expansion are proportional to the perimeter of the curve; when multiplied by the $d\mathbf{s}$ term, you get a factor proportional to the area :) (Maybe this is basically what you had explained, just more verbose.)

 

[wrobel]

Folland G. Real analysis.. modern techniques and their applications (2ed., PAM, Wiley, 1999)(ISBN 0471317160)(600dpi)(T)(402s)_MCat_