324 Prof. D. N. Mallik on Elastic 



a sphere of radius a, we have to evaluate jX' r dx' dy' dz 

 over a sphere. For a point outside, this is 



sm ^ 

 r* It 



Jo ^ 



= 2ttX' #1 — ,-^-sin^^, 



where r x , r 2 are the distances of the point at which the 

 displacements are to be found from the surface, measured 

 in the direction 6, and R the distance of this point from the 

 centre. 



Now, since r 1? r 2 are the roots of r 2 + R 2 — 2rRcos# = a 2 , 

 the integral becomes, putting a 2 — R 2 sin 2 # = ®, 



-=SH ®{4© + 2(R 2 -a 2 )}tfe 



3B'r 5 /• 



.*. if R>a, taking p = l, 



1 ci 3 X' (\ + ^)a 3 3 / v , 3 



w= 



&&(*&+ +X»*H 



3/z. r 6/a(X + 2/a) 



if R<a, 



J r *fc» f <i/ t/z' over a sphere, 



sin 6 d6, 



-in 



where r 1? r 2 are the roots of r 2 — Rrcos0 + cos6 = a?. 



/R 4 9 \ 



This is = - W ~ 5 - - **a 2 R 2 -a 4 ) . 



4. If X', Y', Z' are derived from gravitational potential, 

 we may take 



X' = ■$*■*', &c, 

 since X'.?— = ^ , and we have to evaluate, 



fr£fc'<fy'<fe' [|^ J> -2x\rdx'dy'dz'. 



