from a Cylindrical Wall. 363 



In the first part put y 2 — a 2 — 6 2 =2a& sin 6, and we get 



, f 2 2x a 2 + a& sin . . . 



2 L T"^ + a 2 + ^ + 2a6sin^ ■ 



*/ - 



2 



which gives after integration 



irx C a 2 — x 2 — b 2 1 



~ ^\^~(^+a 2 + b 2 ) 2 -4:aW 2 + J ' 



Usin o- the same substitution in the 2nd part, i. e. 



_ , r-" *(f+a*-vy , 



we get 



x_ r~ _ 4a 4 + 8a 3 6 sin 6 + Aa 2 b 2 sin 2 0^ ™ 



2a J ff (a 2 + & 2 + 2a6sin0)(# 2 + a 2 + & 2 + 2a&sin0) 



(3) 



2aJ_ v \ cr 



+ ^^iU 



+ /> 2 + 2a/<siu<9 x 2 + a 2 + b 2 + 2ab sin 



where B = v - = - and CJ=— ^ -, 



which gives after integration 



™\ x ■ B , C 1 



2a L « 2 -^ 2 v / (^+a 2 +6 2 ) 2 ~4a 2 6 2 J' 



The whole integral (1) 

 47raN 



a" — b" — ar \ 



~V^+a 2 + b 2 ^£aW + j 



+ 2a \ + a 2 -6 2 + "V(^ + « 2 + & 2 ) 2 -4^ 2 / J ' ' 

 Now */Jx 2 + a 2 -b 2 ) 2 +Wx 2 « V (^+i 2 + 6 2 ) 2 - 4a 2 rV r . 



Using this and substituting the values for B and C the 

 integral reduces to 



J*! \<s/(x 2 + a 2 + b 2 ) 2 -U 2 b 2 i 



