6 Lord Rayleigh on the Correction to the 



cylindrical wall contributes nothing, since V vanishes along 

 it. At 2 = 



V = 2(A + B)J (£r), -dY/dz = $k(k + B)J (kr) ; 

 and 



(15)=^ 2 S/t(A + B) 2 J 1 2 (^). . . . (16) 



On the left the complete value of V includes (1) and (8). 

 There are here two cylindrical surfaces, but r = b contributes 

 nothing for the same reason as before. On r = a we have 

 V = land 



dY 1 



- ~~ = 1 \, -2 hR6'(ha) e hz ; 

 dr a log o a 



so that this part of the surface, extending to a great distance 

 z = — I, contributes to (15) 



-^Hf(k) (17) 



2 log b/ a 2 



There remains to be considered the annular area at 2 = 0. 

 Over this 



v= log|/r +2H _ _ _ 



log b/a T v 



dV/dz = $hK(j>(hr). .:.... (19) 

 The integrals required are 



f <f>(hr)rdr=-h- l {b<l>Xhh)-a(f>'(ha)}, . . (20) 



I a \ogr<p(hr) rdr=-li~ 1 {b \ogb$(lib)-a\oga<f> f (ha)},(2l) 



Jo 



P {^(hr)yrdr=^b 2 {^(hb)} 2 -ia 2 {4>Xha)} 2 ; . . (22) 



%) a 



and we get for this part of the surface 



iatB.\cj>'(ha)i-ithR 2 [b 2 {(t>\hb)}^-a 2 {(l> / (ha)yj. (23) 

 Thus for the whole surface on the left 



(15) = I^bfi +it *&(*+"<&) -<* S *'W], (24) 



the simplification arising from the fact that (1) is practically 

 a member of the series <f). 



The calculated capacity, an overestimate unless all the 

 coefficients H are correctly assigned, is given by addition o 



