Question

积分问题求解！

会几道写几道就可以了！十分感谢！

Answer 1

（1）x^2+y^2<=2x
r^2<=2rcosθ
r<=2cosθ
原式=∫(-π/2,π/2)dθ∫(0,2cosθ)r^2dr
=∫(-π/2,π/2)dθ*(1/3)*r^3|(0,2cosθ)
=∫(-π/2,π/2) (8/3)*cos^3θdθ
=(16/3)*∫(0,π/2) (1-sin^2θ)d(sinθ)
=(16/3)*[sinθ-(1/3)*sin^3θ]|(0,π/2)
=(16/3)*(1-1/3)
=32/9
（2）x^2+y^2<=1，r^2<=1，r<=1
x<=x^2+y^2，rcosθ<=r^2，r>=cosθ
原式=∫(0,π/2)dθ∫(cosθ,1)r^2dr
=∫(0,π/2)dθ*(1/3)*r^3|(cosθ,1)
=(1/3)*∫(0,π/2) (1-cos^3θ)dθ
=(1/3)*[θ-sinθ+(1/3)*sin^3θ]|(0,π/2)
=(1/3)*(π/2-1+1/3)
=π/6-2/9
（3）√x+√(y/2)<=1
√(2x)+√y<=√2
√(2rcosθ)+√(rsinθ)<=√2
√r*[√(2cosθ)+√(sinθ)]<=√2
√r<=√2/[√(2cosθ)+√(sinθ)]
r<=2/[2cosθ+sinθ+2√(sin2θ)]
原式=∫(0,π/2)dθ∫(0,2/[2cosθ+sinθ+2√(sin2θ)])r^3*cosθsinθdr
=(1/4)*∫(0,π/2)cosθsinθdθ*r^4|(0,2/[2cosθ+sinθ+2√(sin2θ)])
=4*∫(0,π/2) cosθsinθ/[2cosθ+sinθ+2√(sin2θ)]^4dθ
=4*∫(0,π/2) cosθsinθ/[3cos^2θ+1+6sin2θ+8cosθ√(sin2θ)+4sinθ√(sin2θ)]^2dθ
=