237 
inside a Hollow Unlimited Boundary. 
Differentiating this and dividing by sin 6, we should get 
- j P 2 + % (2 n + 1) <x n P n = 1 - 2 cos 6. 
^ 4 
But this is true, for it is the expansion of cos 6 or P 1 between the 
limits 0 and 7t/2 in terms of Legendre Coefficients of even order, 
in fact the same series that occurs in (44) remembering that 
= 2 (- 1)'+^ x 1^1 
(»~ 3 ) 
0 + 2 )’ 
If using the notation of Arts. 1 — 7 we call a x the Z A Ox (Fig. 1) 
we can get its value by putting r = a either in (59) or (60). 
12. The supposition b 0 = /jl 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) 
Vr 2 sin 2 0 = 2 pa (1 + cos 6) (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 Z AOx (Fig. 1) to be 7r/2. We may do this because the ex- 
pansion for P x which we have made use of in (44) is true even at 
both limits 0 and 7 t/ 2. Moreover this expansion has never been 
differentiated. Accordingly putting r = a and 6 = 7t/ 2 in (59) or 
(60) we shall get since P n ' vanishes when 0 = 7t/ 2 if n be even 
b 0 = 4/-t (63), 
and this satisfies the condition derived from (56). If 0 X be the 
semi-angle of the asymptotic cone, we get from (60) 0 1 = 7t/3. 
From (55) we get for determining 0B (Fig. 1), putting 0B/a = p 
+ & c. = 0 
(64) 
from which we get 0B /a = '5726 nearly. 
From (60) we get for the approximate form of AD (Fig. 1) at 
infinity the equation 
r _ sin 2 0 
a 1 — 2 cos 0 
(65) 
from which we see, comparing it with a hyperbola, that the 
boundary crosses and ultimately lies within its asymptotic cone. 
