∫(D)∫ln(1+x^2+y^2)dxdy
D:x^2+y^2=1与 两坐标所围成的位于第一象限内的闭区
ρ=1,θ从0,到π/2
dS=ρdθdρ
∫(D)∫ln(1+x^2+y^2)dxdy
=∫[0,1]∫[0,π/2]ln(1+ρ^2) ρdθdρ
=∫[0,1]ln(1+ρ^2) ρdρ∫[0,π/2]dθ
=(π/4)∫[0,1]ln(1+ρ^2)d(1+ρ^2) ∫lnxdx=xlnx-x+C
=(π/4)(2ln2-1)
换极坐标
积分变为∫θ∫r ln(1+r^2)rdrdθ
0<θ<π/2 0
=π/4∫ lntdt 1
=π/4*(2ln2-1)