1886.] Equilibrium for a Rotating Mass of Fluid. 325 



Approximate Solutions of the Problem — continued. 



I Q2 1 [~ 3 3 3 



pYj—dy=pq*-p*qKy + -k 2 ^pq 5 --pq 5 + -pq3--ypy I + . .. 



(1 3 3 \ 

 - -£P<f + ^p<i 3 —^p\ 2 \ + • • • (h) 



The equation (b) is p 2 q 2 ^dy=A^-^ 3 , whence, equating (g) and (h), 



+ . . . = 0. 



If Jc be zero, we hare ^<Z 3 P + j^M — 7 ^ + P 2 9^j — & 



This easily reduces to sm 27 sin 47 ^ 



1 — i cos 47 .... V / 



of which the solution is 7=54° 21' 2*7", as stated in the text. 

 Now 



^7 [j2 3 2> + ~ 7 (| + 2> y)] = - ^-[1 -2 cos 2 7 -f cos 47 + 7 sin 4y], 



and with the above value of 7 is equal to — O1355014. 



Also, with this value of 7 the coefficient of ¥■ is 0-0160432 ; so that 



0-0160432 sin (y-54° 21' 27") =0-1355014P, 



or, sin 2 a = 10' 9266528 sin (y - 5 i° 21' 27"), 



which is the equation (21) of the text. 



Again j^.«^ + ^ + l^(-^3 p _| 2p + |^, 



P^dy^qp - (1- 2 2 ) 7 + \ 7c 2 (<fp + 1^(1 - 2 2 ) -| 7 (1 - ^ 



l/l 3 3 3 \ 

 = TP ~ 7 + 2*7 + 2 F \ 2 ^ + 2 qV ~2 7 + 2 ^ )' 



Therefore 



^dy-pj^dy~ -\qp 4- (| - 2 2 ) 7 + \& \j^P~1P + ■ 1 



Now A = l-|^V-.. f 



