一道利用直角坐标系计算三重积分的题

1个回答

  • h > 0 ==> z = (h/R)√(x² + y²)

    截面:x² + y² = R²,- √(R² - x²) ≤ y ≤ √(R² - x²)

    ∫∫∫ z dxdydz

    = ∫(- R→R) dx ∫(- √(R² - x²)→√(R² - x²)) dy ∫(0→h) z dz

    = (1/2)h²∫(- R→R) dx ∫(- √(R² - x²)→√(R² - x²)) dy

    = (1/2)h²∫(- R→R) 2√(R² - x²) dx

    = 2h²∫(0→R) √(R² - x²) dx,x = Rsinp,dx = Rcosp dp

    = 2h²∫(0→π/2) R²cos²p dp

    = h²R²∫(0→π/2) (1 + cos2p) dp

    = h²R² * [ p + (1/2)sin2p ] +(0→π/2)

    = h²R² * π/2

    = (1/2)πh²R²