AN INTERRUPTED MEDIUM. 525 



sin k r 3 ,sinkr coskr, 



... / sin k r 3 /smkr coskr\\ 



From the latter c 2 = 2 ( __+ _ (__ __ ) ) 



\ k r* r 5 V k* r A J ' / 



A 2 3 A 2 / 1 1 \ 1 



67+ (120-24)} 



* 15 r 



It is clear that an approximation of this kind is illusory, inasmuch as 



is infinite. We are not justified, therefore, in arguing any thing relative to the 

 sign of m from this process. 

 To find the value of M. 



M = 2m(d> r + ^dx 2 } sine 3 xcosfd y + . -r 2 m $3" 8 y cos e d x sin f8 v 

 r sin <p r 



taken for Aa^the medium. 



Now m=gr 2 sin$ dd d<$> dr, 8 x=r cos 6, 8y=r sin 6 sin cf>. 



Instead of cosfdy write 1, and instead of cos e 8 x write 1, for the other 

 parts of these expressions respectively are very small compared with these, as we 

 have just proved. 



Now 2 m <pr sin e 8 x=g 1 1 1 r 2 r sin 6 sin (pr cos 6) dr d(f) d6 



PC v j. 1 cos e r , 

 = ?JJ r <t> r - -^T- drd( t> 



n Co j. 1 cos er , 



= 27TO I r 2 m r - d r 

 *J er 



= IT p / e r 3 (p r dr 

 2 m $-L dx 2 sin e d x= fff ^ -^-r 2 cos 2 d sin 6 sin (e r cos 6) dr d$ d 6 







= 2 TT p r 3 $' r - 3 sin p dpdr 



from j=0 to p er 



//^ (f)' T d T C ~\ 



*-s I 2 + 2coser + 2ersin ere* r s cos er i 

 e 3 ^ (. j 



TTpe /* 

 = ^ / r 1 r dr 



2 OT _L 5 a; 5y sin/5y=> III ^ 4 sin 2 Q cos sin ^> sin (Jr sin6 sin (f>) dr d6 



