1016 Dr. E. H. Kennard on a Simplified Proof for 

 Now by (1) 



.But cdt I I (\/ 2 <j))r 2 dw is, in the limit when dt=0, equal to 



the volume integral o£ V 2 <P throughout the shell bounded by 

 the two limiting positions of the sphere and by the ring s 

 which these cut out of S, and this volume integral in turn, 

 by the divergence theorem (or " Green's theorem "), equals 

 the outward normal flux of the vector V<£ over the surface 

 of the shell. Hence we can write 



477tJJ^ ^ ~~ kirr'drj} dr 1 ^ 47rrJJ \ ~dr ) 



a g 



s 



where n = distance along inward normal to S. 



Substituting from this equation in (4) and then from (4) 

 back into (3), we find that many terms cancel out and 



Let us now write instead of dco in the second integral, 

 ■c?S cos #/r 2 where 6 — angle between r drawn toward P and 

 the inward normal to S ; and let us write in place of the 

 first term N\(p/r)dr taken throughout the shell bounded by 

 s and the two limiting positions of the sphere. We can then 

 integrate and obtain I. At time t=— go, 1 = 0; at time 

 t, l = (f>, as noted above. Hence 



c c 



.... (7) 



Here the first integral extends throughout all the space 

 enclosed by 8 while the second extends over all of S, and the 

 subscript reminds us that values of quantities inside the 

 brackets are to be taken at a time t—r/^ 



