Question

计算∭1/√(x^2+y^2+(z-2)^2 )dv，其中Ω={(x,y,z)|x^2+y^2+z^2<=1}

Answer 1

解：由Ω={(x,y,z)|x²+y²+z²≤1}
令x=rsinφcosθ，y=rsinφsinθ，z=rcosφ，其中0≤r≤1，0≤θ≤2π，0≤φ≤π
则∭1/√[x²+y²+(z-2)²]dv
=∭r²sinφ/√[(rsinφcosθ)²+(rsinφsinθ)²+(rcosφ-2)²] drdφdθ
=∭r²sinφ/√(r²+4-4rcosφ) drdφdθ
=∫dθ ∫dr ∫[r²sinφ/√(r²+4-4rcosφ)]dφ
=-2π∫dr∫[r/√(r²+4-4rcosφ)]d(rcosφ)
=2π∫r²dr
=2π/3