296 Lord Rayleigh on the 



potential at B due to the fluid which lies above the plane B F. 

 Thus 



PB = ^{f)dxdydz, 



where the integrations are to be extended through the region 

 above the plane B F which is external to the sphere BDCE. 

 On the introduction of polar coordinates the integral divides 

 itself into two parts. In the first from /=0 to/=2r the 

 spherical shells (e. g. D H) are incomplete hemispheres, while 

 in the second part from/=2r to/=c© the whole hemisphere 

 (e. g. IGF) is operative. The spherical area D H, divided 



by/ 2 , 





sin 6d0=27r cos 6 = irfjr. 



The area GF=2tt/ 2 . 



Thus, dropping the suffix B, we get the unexpectedly simple 

 expression 



P=^y(f)f4f+^£w)fdf- • • (27) 



If 2r exceed the range of the force, the second integral 

 vanishes and the first may be supposed to extend to infinity. 

 Accordingly 



in accordance with the value (12) already given for T . We 

 see then that, if the curvature be not too great, the pressure 

 in the cavity can be calculated as if it were due to a constant 

 tension tending to contract the surface. In the other extreme 

 case where r tends to vanish, we have ultimately 



V 



= 27r^U(f)fdf=K . 



In these extreme cases the results are of course well known; 

 but we may apply (27) to calculate the pressure in the cavity 

 when its diameter is of the order of the range. To illustrate 

 this we may take a case already suggested, in which <£(/) = e~P f , 

 Il(f) = l3~ 1 e~K Using these, we obtain on reduction, 



V 



=27r/3_4 {i-^ r ( 2 ^ +4+ J)}- • • ( 29 > 



From (29) we may fall back upon particular cases already 

 considered. Thus, if r be very great, 



2 

 p= - X37T/3- 5 ; 



