236 Mr Sharpe, Liquid Motion from a Single Source 
same quantity to be discharged in the same time through the 
asymptotic cylinder, we shall get 
C 2 — b 0 a = — 2 pa (57). 
Or we may suppose V—0, then from (52) it is easy to shew that 
the quantity of fluid discharged through the asymptotic cone at 
infinity in unit of time = 27 rb 0 a (1 — cos 0 X ). 0 X being the semi- 
angle of the asymptotic cone, therefore we must have 
27 rb 0 a (1 — cos 0 X ) = ^pair (58), 
and remembering from (53) that cos 0 X = C 2 /b 0 a we get for 0 2 the 
same value as is given by (57). From (58) we see that as 0 X 
cannot vanish without p vanishing, which we do not suppose, the 
boundary cannot be parabolic at infinity. 
11. Looking at (45) and (47) let us put for brevity a n for 
a n /(p — b 0 ). Then (51) may be written, being the equation to 
BA (Fig. 1), 
0 = — pa — b 0 a cos 6 + ~ Vr 2 sin 2 0 - (p — b 0 ) a cos 0 — 
(p — b 0 ) sin 6 
-ta P ‘ + HiF' +S 
„n+i 
P / 
l [a n (n + 1) n 
.(59). 
(53) may be written, being the equation to AD (Fig. 1), 
0 = — pa — b 0 a cos 6 + ^ kV 2 sin 2 6 — (p — b 0 ) a 
— (p— b 0 ) sin 0 
P '_s 
Sr 1 8 r ! 2 7 
2 (a»a n+1 p , 
V nr n n 
... (60). 
As the curves BA, AD meet in the same point A on the circle 
AG A' it is evident that when r — a the two equations (59) and (60) 
should give us the same value for 0, therefore the coefficients of 
in both equations must be equal. We must therefore 
establish the truth of the following equation, 
sin 0 
5 n , " 2n + 1 p , 
24 4 n (n + 1) 
= 1 — cos 0 — sin 2 0 ... (61). 
This can be done thus. From (50) 
sin 0 . Pn [ 0 jj • qjq 
—7 tT =_ Pn^0d0. 
7l(?2 + l) Jo 
Applying this to (61) it becomes 
f 
J o 
sin 0 
— 7 P 2 T S (2 n + 1) a n P r 
* 4 
d0 = 1 — cos 0 — sin 2 0. 
