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

1个回答

  • 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