Question

∫∫xdydz+ydzdx+(z^2-2z)dxdy 其中∑为锥面 z=根号x^2+y^2 被平面z=0 和z=1所截得的外侧，求答案 详解

Answer 1

Gauss公式。
∂P/∂x + ∂Q/∂y + ∂R/∂z = 1 + 1 + 2z - 2 = 2z
∫∫Σ xdydz + ydzdx + (z² - 2z)dxdy
= ∫∫∫Ω 2z dxdydz
= 2∫(0→1) z dz ∫∫Dz dxdy
= 2∫(0→1) z * πz² dz
= 2π * (1/4)[ z⁴ ]|(0→1)
= 2π * (1/4)
= π/2

普通方法。Σ₁：z = √(x² + y²)下侧、Σ₂：z = 1上侧
∫∫Σ xdydz + ydzdx + (z² - 2z)dxdy
= ∫∫Σ₁ xdydz + ydzdx + (z² - 2z)dxdy + ∫∫Σ₂ xdydz + ydzdx + (z² - 2z)dxdy
= - ∫∫D (- P * ∂z/∂x - Q * ∂z/∂y + R) dxdy + ∫∫D (1 - 2) dxdy
= - ∫∫D [- x * x/√(x² + y²) - y * y/√(x² + y²) + (z² - 2z)] dxdy - ∫∫D dxdy
= - ∫∫D [- x²/√(x² + y²) - y²/√(x² + y²) + (x² + y²) - 2√(x² + y²)] dxdy - π(1)²
= - ∫∫D [x² + y² - 3√(x² + y²)] dxdy - π
= - ∫(0→2π) dθ ∫(0→1) (r² - 3r)r dr - π
= - 2π * [1/4 * r⁴ - r³]|(0→1) - π
= - 2π * (1/4 - 1) - π
= π/2