计算三重积分∫∫∫zdv,其中Ω由z=-√(x^2+y^2)与z=-1围成的闭区域

1个回答

  • { z = - √(x² + y²)

    { z = - 1

    - 1 = - √(x² + y²)

    x² + y² = 1 --> r = 1

    切片法:

    ∫∫∫ z dV

    = ∫(- 1→0) z dz ∫∫Dz dxdy

    = ∫(- 1→0) z * πz² dz

    = ∫(- 1→0) πz³ dz

    = (π/4)(z⁴):[- 1→0]

    = π/4 * 0 - π/4 * (- 1)⁴

    = - π/4

    投影法:

    ∫∫∫ z dV

    = ∫(0→2π) ∫(0→1) ∫(- 1→r) rz dzdrdθ

    = 2π∫(0→1) r * z²/2:[- 1→r] dr

    = 2π∫(0→1) r/2 * (r² - 1) dr

    = π∫(0→1) (r³ - r) dr

    = π(r⁴/4 - r²/2):[0→1]

    = - π/4