202 
SIR G. II. DARWIN ON THE 
It is possible to express this equation in terms of elliptic integrals and to use 
Legendre’s tables for finding the solution, but the method is very tedious, and after 
finding a few. solutions in that way I abandoned it. It may, however, be worth 
while to mention that 
$h-A/ = 
K 
sin 2 y 
-X 2 smy COS y cos $ + F „ 2k '~ E 
« /C“ 
2 /2 
KK 
wher 
= ^hP^ ~ 72 sin y cos y cos ft - r °^ F + — °^ s - E 
^ 1,7(1 £, = i„V( 
When the forms of the ellipsoids have been determined, the radius vector becomes 
determinable from either of the equations ( 34 ). 
The conditions that the internal potential of an ellipsoid satisfies Poisson’s equation 
and that xjj is homogeneous in a, b, c of degree - 1 , give the two following equations 
Ai 1 'Hi 1 k n 
*2 7,2 ' “ ~ZTZ — P 
a 
cibc 
Ax 1 + iih 1 + — Tjrkxfl = 0 . 
Our two equations for 1 /r 3 may be written 
Ai 1 -^! 1 + ~(a 2 +b 2 \ + c 2 e) = 0 , 
3Xr 3 v ’ 
Ai 1 -^i+3^T3[(3 + X)c 2 + « 2 +c 2 7 7 ] = 0. 
These four equations afford a determinant by which Afi, & 1 , 3 h may be eliminated. 
On reduction we find 
w- 
S/abc 
3 X?" 
l/a 2 +l/b 2 +l/c 2 
3 c 2 -a 2 + \(b 2 +c 2 ) + c 2 U + .)- G ( 1+X + jp + ic 2 e/6 2 ) 
’ W ; l/a 2 +l/b 2 +l/c 2 
Therefore Roche’s second integral is equal to — , 2 cos ' g - Ad), and his third integral is equal to 
k snv y cos y ° 1 
cos 3 (3 
•03d-Ad). 
kk 2 sin 3 y cos y 
Using these transformations of the second and third of (A), and dropping redundant factors, we get 
(cos 2 y + A cos 2 0) (Si - Ad) = (3 + A + cos 2 y) («d - Ad). 
This agrees with the result in the text when e and ?/ are neglected. 
