CONTEMPORARY ADVANCES IN PHYSICS 305 



in which the integrand on the right is integrated over any surface 

 completely enclosing the origin. For any such surface, then, the 

 integral on the right of (63) must be equal to 47r cos nt. 



Fig. 1 



This we proceed to test for the simplest of such surfaces, a sphere 

 centred at the origin. 



Denote by R the radius of the sphere ; by d, the angle between the 

 positive direction of the x-axis and the line drawn from the origin out- 

 ward through any point P of the sphere. The inward-pointing normal 

 at P is directed oppositely to this line, and hence cos («, r) = — \ and 

 cos {n, x) = — cos d. The function U and its derivatives are these: 



U = cos w ( / I — mx = cos {nt — mr — mx), 



dU/dr = m sin (nt — mr — mx), 



dU/dn = (dU/dx) cos («, x) = (dU/dx) cos (tt - 6) (64) 



= — m cos 6 sin (nt — mr — mx). 



In each of these we are to set r = R and x = R cos 6 in preparation 

 for integrating over the sphere. The element of area is 



dS = 2xi?2 sin edd (65) 



and the limits of integration are and tt. Therefore the integral is 

 this: 



I = 2r r dd sin Sfcos (p — mR(l — cos Q) sin (pi; 



(p = nt — mR(l + cos 6), 



which is easy to evaluate, since on putting z for (1 + cos 6) and —dz 

 for sin Odd it becomes 



I = - 2ir j dzlcos (nt — mRz) — mR(2 — z) sin (nt — mRz)2, (67) 



