AN INTERRUPTED MEDIUM. 521 



By combining the two portions, it may be written thus : 



f r sin k r\ /r 3 sin k r r cos k r 



r 2 



sin k r r cos 



If we write/ 1 (r) for r </> (r), the force at distance r, we get 



r 2 sin k r , 



(ft), 



which is identical with M. Cauchy's equation (15), Nouveaux Exercises (Prague), 

 p. 187, and leads immediately to his equation (28), Exercises d" Analyse, &.C., 

 p. 299. The facility with which I have deduced this equation, is a proof of the 

 utility of the method of proceeding which I adopted in the Memoirs from which 

 the original value of c 2 is extracted. 



It is worthy of remark, that, by converting sums into integrals at once, we 

 obtain the same result as by the process we have followed. We shall prove this 

 as follows : 



Let r be the distance of a particle from the origin, Q the angle between r and 

 the axis of x, <$> the angle between the planes of x and r and x and z : 

 then the mass of an element is p r 2 sin 6 d 6 d (j> d r 



.. 2m (j) r cos k d x = g I I I dr dQ d$> r 2 (fir sin 6 cos k 8x 

 d xr cos 6, 8y = r sin 6 sin <f> 



r cos kx= 



) / / d r d Q r 2 $ r sin Q cos (k r cos 6) 



/, , sin kr cos 6 

 *rr0r(-- -j + C) 



/, sin A r 

 i*rr$r ^~ 



<',. r/"' A 23 " </>' 



and 2m Oy 2 = 1 1 I drdQdd)^- r 4 sin 3 5 sin 2 d) 



y A/o i/o r 



=-7T p ff dr d Q <$>' r . r 3 sin 3 



= <7T p / rf r (. cos Q H 5 + C) 0' r r 3 



4 T? / 



O *^ 



2 w ^ y* cos (&5a;) = 7rp// dr dd (f)' r r 3 sin 3 cos (A r cos 0) 



dr dv 



