Question

用三重积分 求由曲面Z=X2+2Y2及Z=6-2X2-Y2所围成的立体的体积.

Answer 1

Ω由z = x² + 2y² 及 2x² + y² = 6 - z围成.
消掉z得投影域D：
x² + 2y² = 6 - 2x² - y²
==> x² + y² ≤ 2
体积 = ∫∫∫Ω dV
= ∫(- √2→√2) dx ∫(- √(2 - x²)→√(2 - x²)) dy ∫(x² + 2y²→6 - 2x² - y²) dz
= 4∫(0→√2) dx ∫(0→√(2 - x²)) [(6 - 2x² - y²) - (x² + 2y²)] dy
= 12∫(0→√2) dx ∫(0→√(2 - x²)) (2 - x² - y²) dy
= 12∫(0→π/2) dθ ∫(0→√2) (2 - r²)r dr
= 12 * π/2 * ∫(0→√2) (2r - r³) dr
= 6π * (r² - r⁴/4)：0→√2
= 6π * (2 - 4/4)
= 6π