inside a Hollow Unlimited Boundary. 237 



Differentiating this and dividing by sin 0, we should get 



- 5 P 2 + 2 (2n + 1) a n P„ = 1 - 2 cos 0. 

 4 4 



But this is true, for it is the expansion of cos or P x between the 

 limits and 7r/2 in terms of Legendre Coefficients of even order, 

 in fact the same series that occurs in (44) remembering that 



„ =2( _ iy+w/2x a-5---(rc-3) 



*•-*{ L) X 2A...(n + 2)' 



If using the notation of Arts. 1 — 7 we call a, the Z AOx (Fig. 1) 

 we can get its value by putting r = a either in (59) or (60). 



12. The supposition b = \i gives an extremely simple solution 

 of the problem. We thus get a tubular surface closed at one end 

 whose equation is from (59) and (60) 



Fr 2 sin 2 = '2fia (1 + cos 0) (62). 



The general shape is given by Fig. 4. We see that in this case the 

 equation to the boundary is the same inside and outside the circle 

 r = a. 



13. The next simplest case is got by supposing V=0 and 

 the /.AOx (Fig. 1) to be tt/2. We may do this because the ex- 

 pansion for P 1 which we have made use of in (44) is true even at 

 both limits and tt/2. Moreover this expansion has never been 

 differentiated. Accordingly putting r = a and = tt/2 in (59) or 

 (60) we shall get since P n ' vanishes when = 7r/2 if n be even 



6 = 4/x (63), 



and this satisfies the condition derived from (56). If Y be the 

 serai-angle of the asymptotic cone, we get from (60) X = irj'd. 



From (55) we get for determining OB (Fig. 1), putting OBja = p 



2__2-^+&c. = (64) 



P A 



from which we get OBja = *5726 nearly. 



From (60) we get for the approximate form of AD (Fig. 1) at 

 infinity the equation 



r_ sin 2 ,p 5 s 



a ~ 1 - 2 cos {:>0) 



from which we see, comparing it with a hyperbola, that the 

 boundary crosses and ultimately lies within its asymptotic cone. 



