Field of an Electrical Current. 209 



vanishes, / being any rational symbol of functionality, while 



_ w /(cos%) .^%=2J o f(cosx)dx- 



Now, the values of x being understood to range from to 

 ft), and being equal to <£— %, 



\ndS = 2{7r-^)A-^^\\'Lnhm.cosxdx 



- T5T [|JJ R8dR - cos 2^/ % - JJlR^R. cos 2 X d X 

 + 4m j LR 2 dR . cos X d X ~ ± m U R2 ^ R • cos X d x\ ' ( 33 ) 



If we first perform the integration in each case with respect 

 to R, taking x constant, we shall have, for example, 



JJR 3 iR.cos2 X rf X =ij(R»-KVo S 2;^ X , • (34) 



where W = Vk! and R = P& (fig. 2) ; but since in integrating 

 throughout the semicircle pk'^tkO the values of % are repeated 

 in the revolution of the radius-vector, Yk', the values of dx 

 being negative as the radius-vector revolves from the position 

 of the tangent from P, it is clear that both terms in the 

 integral at right side of (34) are included in the single 

 expression 



l$Wco S 2 x d x , 



R now being a radius-vector from P to a point on the circum- 

 ference of the circle. 

 Similarly 



JJ LR 3 <ffi . cos 2 x d X =S(iIM i +i 5 R l ) cos 2 x d x . 



To calculate the double integrals involved in (33), we shall 

 take as the independent variable the angle JJDp, or ^ ; and 

 we have, if c is the radius of the cross-section, 



R2 = m 2 + 2 C mcos^ + c 2 , (35) 



R 2 6^ = c(c + mcos i|r)^, (36) 



R 2 cos 2%=2R 2 cos 2 %-R 2 = ??i 2 + 2cm cos ^ + c 2 cos 2^. (37) 



Hence 

 $ R 3 tf R . cos 2 X d X = iJR 4 cos 2 X d X 



c f' r 



= t 1 (c + )ncosi/')(m 2 + 2fl»icos-i|r + t: 2 cos2-</<')rfi|c = JA»i 2 , 



(38) 



