Question

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

Answer 1

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)