1886.] Equilibrium for a Rotating Mass of Fluid. 323 

 or 



E sec 2 *[l + (sin * tan ft cos 7 ) 2 ] - (2E-E) 



— sec 2 asin«tan^ cos 7(l-fsin 2 ^)=0. . . . (16) 



In order to adapt this for computation, we may introduce the 

 auxiliary angles defined by 



tan £=sin a tan ft cos 7, tan 8= sin ft, . . . . (17) 



and the equation becomes 



Esec 2 «sec 2 £-(2F-E)-sec 2 *sec 2 £tan£=0. . . (18) 



The second of (12) gives 



to 2 sin 2 * sin 3 ry 



4?r cos ft cos 7 



tan 2 * cos 2 7 (E — tan 7 cos ft) — (E — E) , 



_ co 2 E— E E cos 2 7 cos ft cos 7 



whence =-r~n — H — 2 9 • a 1 



Am cos ft cos 7 sm-a surfy sm°7 cos~« cos~« sm"7 



j- =cot /3 cosec /3 cot 7 (F— E) +cot 3 7 cos ft sec 2 *E— cos 2 /3 cot 2 7 sec 2 *. 



& • (19) 



Some of the subsequent computations were, however, actually 

 made from a formula deduced from the third of (12), which leads to 



— =cot ft cot 7 cosec 3 /3(l + cos 2 /3)(F— E) — cot 3 /3 cot 7 tan 3 * cosec /3E 



+ cot 2 /3cot 2 7sec 2 «. . . (20) 



By subtracting (20) from (19) we can deduce (16) ; hence it follows 

 that (19) and (20) lead to identical results. Most of the subsequent 

 results were computed from both (19) and (20), thus verifying the 

 solution of (18). 



The formulas (18) and (19) are suitable for finding the solution, 

 except when a is small or nearly 90°, when the elliptic integrals 

 become awkward to use. I have, therefore, found approximate 

 formulas for these cases, but as the algebraic process necessary to 

 establish them is somewhat tedious, the details are given in a note.* 



* Approximate Solutions of the Problem. 



From (7) we have 



J. 43 * 4 Jo 



. J_ E _ JL ( 2F - E ) _EEO££!X 



