210 Prof. A. Gray on Electric 



Keep i/r fixed and consider 



i 



rfdf. 



J ° 

 By the figure 



6 2 = r 2 +/ 2 -2>/cos^, 

 which gives 



r ■= /cos i/r+ ylr—f* sin- l/r. 



We take the upper sign and integrate for the semicircle- 

 above OP in the diagram. Substituting, we obtain 



j>=j 



(/- cos ^ +/ v/6 2 -/ 2 sin 2 -f) df 







6 3 



= i ( a 8 cos ^ +-_ -* (^_ a 2 sin 2 ^ } | 1 

 3 (. sm 2 -v/r sin 2 \jr K r J 



=-j« , o«++i a ^ll-(l-« , «in , ^ i i, • (30) 



where A- 2 is written for a 2 /b 2 . 



We have now to integrate this with respect to -ty. First, 

 it is clear that ^a 3 cosi/r contributes nothing to the definite 

 integral between and ir as limits. Next we get 



6 sf^ T= _J3 cott 



J sin- y\r T 



and this is infinite at each limit. It will be found, however, 

 to be cancelled by the same integral, arising from the term 

 which remains to be integrated and affected by the — sign. 

 Integrating by parts we obtain 



j' 



d ^ (1 - P sin 2 i/r)i = - (cot yfr + *» (1 - P sin 2 tyf 



sin 2 yjr 



-h C ot++Pir) k2s[n + bos ta*. 



(L-Psin 2 -f) 

 But 



i 



" . , k 2 sin -vlr cos yjr 7 , 



cot vr 4 dxr 



( T (1-Psin 2 ^ Y 



J (1-Fsin 2 t/ Jo (l-#sin»yr)* 

 = —2PF-2E + 2F, (31) 



where F and E are Legendre's complete elliptic integrals of 

 the first and second kinds to modulus Jc = a/b. 



