设∑是锥面z=√(x²+y²)(0≤z≤1)取下侧,求∫∫∑ xzdydz -ydzdx + zdxdy求解

Question

如题所述

Answer 1

补面Σ1：z = 1取上侧
由高斯公式：
∫∫(Σ+Σ1) xzdydz - ydzdx + zdxdy
= ∫∫∫Ω [∂/∂x (xz) + ∂/∂y (- y) + ∂/∂z (z)] dV
= ∫∫∫Ω (z - 1 + 1) dV
= ∫(0→1) z dz ∫∫Dz dxdy：x² + y² = z →Dz：πz
= ∫(0→1) πz² dz
= (1/3)πz³：(0→1)
= π/3
∫∫Σ1 xzdydz - ydzdx + zdxdy
= ∫∫Σ1 dxdy
= ∫∫D dxdy：Dxy：x² + y² = 1
= π
即∫∫Σ xzdydz - ydzdx + zdxdy
= - 2π/3

Answer 2

∑是锥面z=√(x²+y²)(0≤z≤1)取下...