376 Rev. B. Bronwin on certain Definite Multiple Integrals. 



dxdydz = d R sin u du dv, 

 because (R) decreases while (x) increases. Therefore 



U = III R d R sin u du dv = — / /R 2 sin u du do. 



Putting the values of .r, y, and z in 



^LlISI z ! - 1 

 a 2 + & + c2 - j > 



and making 



1 . . 1 1 



A = -o sin 2 « sin 2 fl + to sin 2 « cos 2 v + -s 

 a" 1 tr c 2 



= -^ sin u sin » + 75 sin ^* cos u + -a cos m, 

 a b c 



o- 2 A 2 yfc 2 



55+j5 + ? => W 



we find 



2 B /*/»B 2 



R = -spi and U = 2 / / -^ sin w c?« afr;. 



Between # = and u = tt, R will be positive, and then negative 

 up to u = 2ir. We must therefore integrate from «=0, tt=0 to 

 u — TTf v=7r. If we leave out terms containing the first power 

 of cos u, cos v, as these would give nothing in the value of the 

 integral, we may make 



p-2 ^2 £2 



B 2 _ s sin 2 u sin 2 v + 7-3 sin 2 a cos 2 u + -r cos 2 u. 

 cr tr <r 



sm^ u = w, 



Put 



/'o-2 ^2\ g /t 2 /# 2 ff 2 \ 



(l + ^si^a-f-^-cos 2 ^^ (g- 1) 



(-S + iy sin 9 M + ^cos 2 M =p, (1 - ij sin 2 w = gj 



and we have 



B 2 = — (m-f- wcos2w), A = •— -(p + ^cos2u). 



We may obviously integrate from 



« = 0, w = to a = — ; u = — , 

 2 2 



multiplying the integral by 4. Therefore 



tt i r PC m + n cos 2 v • 77 o /*»».P — nq. . 



U = 1 6 / / t — ■ — rssin u dudv=8v I — f — — , smudu 



J J (p + $/cos2») 2 »/ q>2_ g«\| 



