20/06/24 21:21:24.00 NoItLLRp.net
>>601
・∫[0,q] dt √(1-t^2) = ( arcsin(q) + q√(1-q^2) )/2
・∫[0,1]dz z^{3/2}/√(1-z) = B(5/2,1/2) = Γ(5/2)Γ(1/2)/Γ(3) = 3π/8
・∫[0,1] z*arcsin(√z) = 1/2*arcsin(1) - 1/4*∫[0,1]dz z^{3/2}/√(1-z) = π/4 - 1/4*3π/8 = 5π/32
・∫[0,1]dz z^{3/2}*√(1-z) = B(5/2,3/2) = Γ(5/2)Γ(3/2)/Γ(4) = π/16
V=∫[0,1]dz { πz/2 + 2∫[0,z]dt √(z-t^2) }
= ∫[0,1]dz { πz/2 + 2z∫[0,√z]dt √(1-t^2) }
= π/4 + ∫[0,1]dz { z*arcsin(√z) + z^{3/2}*√(1-z) }
= π/4 + 5π/32 + π/16
= 15π/32
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