Mathematical Physics Vol 1

Chapter 4. Field theory

178

for any surface S passing through curve C . From here it follows that ∇ × A = 0 .

Exercise 120 Let ∆ S be a surface bounded by a simple closed curve c , P any point on ∆ S outside the curve c , n the unit vector of the normal to ∆ S at P . Show that at point P the following is true ( rot A ) · n = lim ∆ S → 0 H c A · d r ∆ S , where the boundary is taken in such a way that ∆ S " tightens" around P .

Solution According to Stokes’ theorem x ∆ S

( rot A ) · n d S = I c

A · d r .

(4.205)

And according to the integral mean value theorem x ∆ S

( rot A ) · n d S =( rot A ) · n ∆ S .

(4.206)

From (4.205) and (4.206) it follows that

( rot A ) · n = H c

A · d r ∆ S

,

that is

( rot A ) · n = ( rot A ) · n | P = lim ∆ S → 0 H c

A · d r ∆ S

lim ∆ S → 0

,

which is the required result.

Exercise 121 If rot A is defined as in Example 120, find the projection of rot A on the z axis.

Solution Let EFGH be a quadrangle parallel to the xy plane (its normal vector parallel to the z axis) with its center at point P ( x , y , z ) (Figure 4.32). Let A x and A y be the projections of vector A on the x and y axes, respectively.