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用极坐标计算二重积分 ∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdyD:x^2+y^2≤1,x≥0,y≥0

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用极坐标计算二重积分
∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdy
D:x^2+y^2≤1,x≥0,y≥0
极坐标下D:x^2+y^2≤1,x≥0,y≥0可表示为
0≤r≤1,0≤θ≤π/2
∫∫√(1-x^2-y^2)/(1+x^2+y^2)dxdy
=∫(0,π/2)dθ∫(0,1)[(1-r^2)/(1+r^2)]rdr
=π/2∫(0,1)[(1-r^2)/(1+r^2)]rdr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)r^3/(1+r^2)dr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)[(r^3+r)-r]/(1+r^2)dr
=2*(π/2)∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)rdr
=(π/2)ln(1+r^2)|(0,1)-(π/2)*(1/2)r^2|(0,1)
=(π/2)*(ln2-1/2)
再问: 只有D求对了 后面就太离谱了
再答: 如图: