>>407

∫√(2a^2 -y^2)dy = (1/2)y√(2a^2 -y^2) + a^2･arcsin(y/(a√2))

∬[x=0,1][y=0,x] √(2x^2 -y^2) dydx
　= ∫[x=0,1] {(1/2) + (π/4)}x^2 dx
　= {(1/2) + (π/4)}∫[x=0,1] x^2 dx
　= {(1/2) + (π/4)}[ (1/3)x^3 ](x=0,1]
　= {(1/2) + (π/4)}(1/3),