∫∫ln(1+x2+y2)dxdy=∫∫ln(1+r2)rdrdθ,x=rcosθ,y=rsinθ
0≤r≤1,0≤θ≤π/2
∴∫∫ln(1+x2+y2)dxdy=∫∫ln(1+r2)rdrdθ
=∫ln(1+r2)rdr∫dθ
=π/2*∫ln(1+r2)rdr(0～1)
=π/4*∫ln(1+r2)dr2
=π/4*[ln(1+r2)*r2-∫r2dln(1+r2)]
=π/4*[ln(1+r2)*r2-∫r2/(1+r2)dr2]
=π/4*[ln2-∫(1-a)/ada]
其中,r自0至1,故ln(1+r2)*r2=2；
a=1+r2,故a自1至2,∫(1-a)/ada=∫1da-∫1/ada=1-ln2
再帶回去,就得到：∴∫∫ln(1+x2+y2)dxdy=π/4*[2ln2-1]
注意,2ln2=ln4；r2表示r的平方