244 Mr. J. J. Thomson on the Electric and Magnetic 



the surface of a sphere vanishes, we may substitute in the 



(/- 1 d 2 1 

 integral -r-s for -j-^-i^ then, transforming to polars, the 

 da? r da? r' & f } 



integral 



= flrj (I 4Q2-J sin 6d<j>dvdr 



_ lQircr 

 for values of r < R, 



I I + ^.rQ?. 



v' R -1 " R 2 ^ R 3 i "••• ' 



Now r n Q n is a solid harmonic of the nth order ; hence 



d 2 

 ~T~2 /l Q») * s a son( i harmonic of the (n — 2)th order ; and in 



d 2 

 particular -y-^ (? ,4 Q4) is a solid harmonic of the second order ; 



and, by the same reasoning as before, we may substitute in the 



integral i- 5 *" (^Q 4 ) for *, * . Now 



4n Sox*-S0x 2 (x 2 +y 2 + z 2 ) + 3(x 2 + y 2 + z 2 y . 

 r 4 Q 4 = * 8 ' 



d 2 



.-. ^(r^) = l2x 2 -Q(y 2 + z 2 ) = l2r 2 Q 2 . 



So for values of r < R the integral becomes 



Si { i ( 24Qlr 3 sin<9#d0cfr 



• Jo Jo Jo 



R 



*/ u v o < 



24 



5R 



Adding this to the part of the integral for r > R, we get for 



the coefficient of uvf , —rz- 



vanish by inspection. 

 The coefficient of vv r 



-iff 



r 2 -, — - n ; — ; -7 dx dy dz. 



dx dy r dx d r 



Now when r > R we may, by the same reasoning as before, 



d 2 1 d 2 1 



substitute -j — j- - for 3—3- —, in the integral, and it becomes 

 dx dy r dx dy r 



-m 



x 2 y 2 



^f- dx dy dz, 



