设柱面x2+y2=R2介于z=0，z=R之间部分的外侧，则z2/x2+y2+z2的曲线积分是多少？

Question

如图所示，求具体结果

Answer 1

取 z = 0+，z = R 之间部分的外侧， 用高斯公式
I = ∯<∑>[z^2/(x^2+y^2+z^2)]dxdy
= ∫∫∫ <Ω>[2z(x^2+y^2)/(x^2+y^2+z^2)^2]dv

= ∫<0, 2π>dt∫<0, R>rdr∫<0+, R>[2zr^2/(r^2+z^2)^2]dz
= ∫<0, 2π>dt∫<0, R>r^3dr∫<0+, R>[1/(r^2+z^2)^2]d(r^2+z^2)
= ∫<0, 2π>dt∫<0, R>r^3dr[-1/(r^2+z^2)]<0+, R>
= 2π∫<0, R>r^3[1/r^2-1/(r^2+R^2)]dr
= 2π[∫<0, R>rdr - ∫<0, R>[r^3/(r^2+R^2)]dr (后者令 r = Rtant)
= πR^2 - 2πR^2∫<0, π/4>(tant)^3dt

= πR^2 - 2πR^2∫<0, π/4>tant[(sect)^2-1]dt
= πR^2 - 2πR^2[∫<0, π/4>tantdtant-∫<0, π/4>tantdt]
= πR^2 - 2πR^2[(1/2)(tant)^2+lncost]<0, π/4>
= πR^2 - πR^2(1-ln2) = πR^2 ln2