Question

∫∫∫√（x^2+y^2+z^2)dV 其中Ω由z=1 z=√(x^2+y^2) 围成

用球坐标系计算怎么写？

Answer 1

球面坐标：
{ x = rsinφcosθ
{ y = rsinφsinθ
{ z = rcosφ
圆锥面z = √(x² + y²) ==> 0 ≤ φ ≤ π/4
顶z = 1 ==> rcosφ = 1 ==> r = secφ ==> 0 ≤ r ≤ secφ
dV = r²sinφdrdφdθ
∫∫∫Ω √(x² + y² + z²) dV
= ∫(0→2π) dθ ∫(0→π/4) sinφdφ ∫(0→secφ) √(r²) r²dr
= 2π∫(0→π/4) sinφdφ * (1/4)[ r⁴ ] |(0→secφ)
= (π/2)∫(0→π/4) sinφsec⁴φ dφ
= (π/2)∫(0→π/4) tanφsec³φ dφ
= (π/2)∫(0→π/4) sec²φ d(secφ)
= (π/2)(1/3)[ sec³φ ] |(0→π/4)
= (π/6)(2√2 - 1)
= (1/6)(2√2 - 1)π

还有以下两种柱坐标方法，供参考：
切片法：
Dz：x² + y² = z²
∫∫∫Ω √(x² + y² + z²) dV
∫(0→1) dz ∫∫Dz √(x² + y² + z²) dxdy
= ∫(0→1) dz ∫(0→2π) dθ ∫(0→z) √(r² + z²) rdr
= (1/6)(2√2 - 1)π
投影法：
∫∫∫Ω √(x² + y² + z²) dV
= ∫∫Dxy dxdy ∫(r→1) √(x² + y² + z²) dz
= ∫(0→2π) dθ ∫(0→1) rdr ∫(r→1) √(r² + z²) dz
= (1/6)(2√2 - 1)π