based upon Electromagnetic llieory. f)ll 



the various powers of! S 2 and C 2 , and the problem of inte- 

 grating (42) with respect to 6 2 from zero to 2tt is reduced 

 to the process of obtaining the definite integrals of the 

 powers of the sine and cosine of this angle. One example 

 will suffice, namely, to find the integral of the cosine. 

 We have V) find 



( 2 *C 2l W 2= \cJl-J c ^f dh , . . (43) 



JO J \ C CW 2> 



■between the limits and 2tt for 2 , or between — Rg> 2 /c and 

 2-7T — Ro) 2 /V' for <£ 2 . This involves finding a value for dR/t/0 2 , 

 since this contains JS2 and C 2 implicitly. By (13) we have 



R=t(1 + u/2-u 2 /8...,.)- • • • (44) 



Hence rf R/^= J* - J.**. . . .). . . . (45) 



Only the first term need be retained, because the next intro- 

 duces higher orders in r. Also only those terms in u in 

 (15) need be retained which contain ,r, y, or z. Recalling 

 the definition of f in (33), we have 



> 



?[-a 1 XS 1 -a 1 YC 1 + a 2 (5S 2 +YG 2 )]. . (46) 



Since the first two terms are independent of # 2 , we have 



ZR ?* ^w / n VQ \ / oj 2 dR 



</6> 2 2 dtf 2 



= « 2 (*C 2 -YS 2 )(l-^).. (47, 



Whence ^R/^ 2 = a 2 B/(l +/i 2 B), . * . . . . (48) 

 where B = fC 2 -YS 2 (49) 



Hence — ^ = -r^Vlj, (50 ) 



c ad 2 1 + P2& 



, a> 2 ^R 1 /ci\ 



and 1 --77= =- — pr-^i, . ( 0l ) 



c </# 2 1 + /3 2 B' 



an d A_^^" 1 = i + / 8 2 B = l-f &(gC,-YS 8 ), . (52) 



\ c da 2 / 



The integral of the cosine (43) required is. therefore. 



( 2ir C 2 ^ 2 =fr0 3 + y3 o f( ) 2 2 -/9 2 YS 2 C 2 ] <ty*= 2tt( fe0rf). (53) 



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