0970132人目の素数さん
2019/02/24(日) 07:58:34.45ID:tbaAoo1o= 2∫ dx ∫ d(y/√(4ax)) 4ax √( 1 - (y/√(4ax))^2 )
= 4a ∫ dx x { asin(s) + s √(1-ss) } (∵ @A)
= 64 a^3 ∫ [s:0,1/√2]ds (s-2s^3){ asin(s) + s√(1-ss) }
= 32 a^3 ∫ d{ (ss-s^4)(asin(s) + s√(1-ss)) } - ∫ ds 2(ss-s^4)√(1-ss)
= 32 a^3 { (π/16 + 1/8) - (1/8)(π/4 + 1/3) } (∵ D)
= a^3 (π + 8/3)