∫∫∫(G)(x^2+y^2)dv,其中G为旋转抛物面z=1/2(x^2+y^2)与平面z=3所围成求三重积分 详细过程

1个回答

  • { z = 3、在上方

    { 2z = x² + y²、在下方

    柱坐标(投影法):2z = x² + y² --> 2z = r²、x² + y² = 2(3) = 6 --> r² = 6 --> 0 ≤ r ≤ √6

    ∫∫∫(G) (x² + y²) dV

    = ∫∫(Dxy) dxdy ∫(r²/2~3) r² dz

    = ∫(0~2π) dθ ∫(0~√6) r dr ∫(r²/2~3) r² dz

    = 2π • ∫(0~√6) r³ • (3 - r²/2) dr

    = π • ∫(0~√6) (6r³ - r⁵) dr

    = π • [ (6/4)r⁴ - (1/6)r⁶ ] |(0~√6)

    = π • [ (3/2)(√6)⁴ - (1/6)(√6)⁶ ]

    = 18π

    柱坐标(切片法):x² + y² = 2z --> x² + y² = (√2√z)² --> 0 ≤ r ≤ √(2z)

    ∫∫∫(G) (x² + y²) dV

    = ∫(0~3) dz ∫∫(Dz) (x² + y²) dxdy

    = ∫(0~3) dz ∫(0~2π) dθ ∫(0~√(2z)) r² • r dr

    = ∫(0~3) dz • 2π • (1/4)[ r⁴ ] |(0~√(2z))

    = (π/2)∫(0~3) 4z² dz

    = 2π • (1/3)[ z³ ] |(0~3)

    = 2π • (1/3)(27)

    = 18π