Question

∫(0－1)dx(0-√(x-x²))(x²+y²)dy

Answer 1

∫<0,1>dx∫<0,√(x-x²)>(x²+y²)dy
=∫<0,1>dx[x^2*√(x-x^2)+(1/3)(x-x^2)√(x-x^2)]
=(1/3)∫<0,1>(x+2x^2)√(x-x^2)dx,
设x=(1/2)(1+cosu)，则√(x-x^2)=(1/2)sinu,
上式=(1/3)∫<0,π>(1/8)(1+cosu)(2+cosu)(sinu)^2du
=(1/48)∫<0,π>[2+3cosu+(cosu)^2](1-cos2u)du
=(1/48)∫<0,π>[5/2+3cosu+(1/2)cos2u](1-cos2u)du
=(1/48)∫<0,π>[5/2+3cosu-2cos2u-3cosucos2u-(1/2)(cos2u)^2]du
=(1/48)∫<0,π>[5/2+3cosu-2cos2u-(3/2)(cos3u+cosu)-(1/4)(1+cos4u)]du
=1/48*9π/4
=3π/64.

追问

非常感谢