Question

D={(x,y)l x^2+y^2<=1,x>=0} 求二重积分 I=∫∫D(1+xy)/(1+x^2+y^2)dxdy

D={(x,y)l x^2+y^2<=1,x>=0}
求二重积分 I=∫∫D(1+xy)/(1+x^2+y^2)dxdy

Answer 1

I = ∫∫ (1 + xy)/(1 + x² + y²) dxdy，D = { (x，y) | x² + y² ≤ 1，x ≥ 0 }
{ x = rcosθ，{ y = rsinθ
I = ∫(- π/2→π/2) dθ ∫(0→1) (1 + r²sinθcosθ)/(1 + r²) • rdr
= ∫(- π/2→π/2) dθ • ∫(0→1) [r/(1 + r²) + r³/(1 + r²) • sinθcosθ] dr
= ∫(- π/2→π/2) (1/2)ln(r² + 1) + sinθcosθ • [r²/2 - (1/2)ln(r² + 1)] |(0→1) dθ
= ∫(- π/2→π/2) (1/2)ln(2) + [1/2 - (1/2)ln(2)] • sinθcosθ dθ
= (1/2)ln(2) • (π/2 + π/2) + [1/2 - (1/2)ln(2)] • 0
= (1/2)ln(2) • π
= (π/2)ln(2)

Answer 2

用极坐标变换试试看，现在没时间做，对不起啊