设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
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![设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy](/uploads/image/z/1534720-40-0.jpg?t=%E8%AE%BE%E5%8C%BA%E5%9F%9FD%3D%7B%28x%2Cy%29%7Cx%26%23178%3B%2By%26%23178%3B%E2%89%A41%2Cx%E2%89%A50%7D%2C%E8%AE%A1%E7%AE%97%E4%BA%8C%E9%87%8D%E7%A7%AF%E5%88%86I%3D%E2%88%AB%E2%88%AB%281%2Bxy%29%2F%281%2Bx%26%23178%3B%2By%26%23178%3B%29dxdy)
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
原式=∫(-π/2,π/2)dθ∫(0,1)[(1+r²sinθcosθ)/(1+r²)]rdr (极坐标变换)
=1/2∫(-π/2,π/2)dθ∫(0,1)[(1+rsinθcosθ)/(1+r)]dr (用r代换r²)
=1/2∫(-π/2,π/2)dθ∫(0,1){1/(1+r)+[1-1/(1+r)]sinθcosθ}dr
=1/2∫(-π/2,π/2){[ln(1+r)+(r-ln(1+r))sinθcosθ]│(0,1)}dθ
=1/2∫(-π/2,π/2)[ln2+(1-ln2)sinθcosθ]dθ
=1/2∫(-π/2,π/2)[ln2+(1-ln2)sin(2θ)/2]dθ
=1/2[θln2-(1-ln2)cos(2θ)/4]│(-π/2,π/2)
=1/2[(π/2)ln2-(1-ln2)cos(π)/4-(-π/2)ln2+(1-ln2)cos(-π)/4]
=(πln2)/2.