4 THE SATURNIAN SYSTEM. [§ 5 



But the value of/ 1 in § 1 gives 



/ D P /= Q—rcoscp. 

 Hence 



11, = J ^ log (/, - + b) -f p ifAfi _ z + by 



But, again, an easy reduction gives 



r g 2 (g — rcosqc) f g'(g — rcos qp) (/ 2 +g— 6) _ f g' 2 (g — r P0S 9-) I/2 + g — ^ 



J p 2/ 2 (/ 2 -z + 6) ~ J p 2/ 2 [/I— (z — 6)?] J p 2/ s (r*+e» — aprcosg.) 



-6) 



19) 



? 2 (g — r cos (y) (/ 2 -|-z — &) 

 /p 2/2/0(0) 



Put, for the moment, 



77 = — r cos (p, 



and we find 



/! = f + H sin 2 if -f (0 — £) 2 , 

 ^ 2 = jp 2 -f- 2j» r cos 9 -j- r cos 2 q>, 

 <» 2 (o — r cos </> ) = jo 3 — |— 2jr r cos 9 -\-p t 2 cos 2 9 



= (jr -\- r 2 sin 2 9) ( p -]- 2 r cos <j ) -|- /> r 2 cos 2 9 — r 3 sin cp sin 2 9), 



F + ' sm 9> = V + ' — 2 9 r cos 9" =/S(o)» 



r Q-(i> — rcosy) , C , in rr , n - m \ I C ( z — b ) (/> + 2 r cos g.) , f > r 2 cos 2 qp — r 3 sin qp sin -2 q p 



Jp»/.(/.-«+rt~ U^" 1 " 9;_t "i 2/ 2 -T-J, 2/J w) 



, f (z — b) (p r 2 cos 2 gj — ; -3 sin g sin 2 g>) 



~r i, " 2/ 2 /§ ( „r" ~~- 



The terms of this last integral, which do not involve — b can be omitted, because 

 the same terms must recur in the expression of 12", and must consequently disappear 

 from the value of 12. 



We find also, by the integral tables, 



if, =/■ 



£j = log (/=+/>), 



Xzk=?<^) l0 ='.fe^=T^ I °s«- s +'')-^j |0 ^ 



'0(0) 



zri lo fe r /o(o) — rzij lo g (A + * — *), 



f-i- 



tan [ 



-,?M) 



r sin qp (2 — It) r sin gj/ 



Hence 



Q.I = i (? 2 + r 2 cos 2 9) log (/„ + — 3) — i (0 - J)/ 2 



— (0 — i) ;• cos (p log ( /o -f- 9 — r cos 9) 

 -{- i r sm 2 w tan 1 ] - — ~ -, 



' j. 2 r sin qp 



and 



In o — o" Q" o" _i_ o " 



i -^0 - 1 ' "^22 • i '-12 "21 T M 11- 



(266) 



