EQUILIBRIUM OF ROTATING MASSES OF FLUID. 
385 
Thus the second line of (22) when transferred is 
(22—ii.) 
Then (22—i.) and (22—ii.) together constitute the potential now entirely referred to 
origin o. 
We want to choose the h’s and U s so that each spheroid may be a level surface, 
save as to the outstanding term of the first degree. 
In order that (17) may be a level surface, when we substitute for r its value (17) in 
(22), the whole potential must be constant. In effecting this substitution, w T e may put 
r — a in the small terms, but in -f 7 rct s /r we must give it the full value (17). 
The constancy of the potential is secured by making the coefficient of each harmonic 
term vanish separately—excepting the first harmonic, which remains outstanding by 
supposition. 
We may consider each harmonic term by itself. 
As far as concerns the term involving W/ c , we have, from (22 —i.) and (22-ii.), as the 
value of the potential, 
4t tA 3 
3c 
A* = 00 
/a\ 
V 
i -1 
•*w 
& = 0 
W 
W/c 
1 + 
<x\3 * = “ * = » Jc + i ! 
P 
i =2 k -0 
k\i\ i 
H; 
* 
U 
a k 
4vA 3 3 hk 
3c 2k-2 
aV w k 3 / a\ ?j (aV w k i - 150 k + i ! P~ 1 jr 
c) a* + 2 [cJ[cJ 0**2 k!i! i-1 ‘ 
and the value of r which must make this constant is 
n =1 + (^* g±Uf 
a \ a) 2k — 2 \c J r h 
but in the small terms inside [ ] we may put r = a simply. 
Make, therefore, the substitution, and equate the coefficient of W; c to zero. 
4:7 T 
dividing that coefficient by — 
On 
3c 
a 
- ) , we find 
2k + 1 7 , 3 h k i i i s / a \ ?J 1 -o'” h + i \ T l 1 TT 
ht + W -2 + 1 + 1 7 A TVu —1 H ‘ 
= 0. 
Therefore 
7 , o k + il P" 1 TT 
h=l+i -) t r.// , 
c) i -2 k ! i\ i — 1 
and, by symmetry, 
V,= 1+ #) 3i 2^4 ! P)A, 
\C 1 i = 2 r ! 1 ■ 1> — 1 
• (23) 
• (24) 
Multiplying both sides of (24) by the coefficient of H r in (23), we have 
k ! r ! 
r 
s /a\ 3 k + r ! P _1 , 3 v 2 /«\ 3 /M\ 3 + 4 ! r + k ! 7 ; 1 P' 1 7 
*\c) k\r\ r- l + \c) \c) rli\k\r\ 
3 D 
MDCCCLXXX VII.—A. 
