补面S:z = 0的下侧
∫∫(Σ+S) dydz + dzdx + dxdy
= 0
∫∫S dydz + dzdx + dxdy
= ∫∫S dxdy
= - ∫∫D dxdy
= - π * 1²
= - π
综上得∫∫Σ dydz + dzdx + dxdy = 0 - (- π) = π
补面S:z = 0的下侧
∫∫(Σ+S) dydz + dzdx + dxdy
= 0
∫∫S dydz + dzdx + dxdy
= ∫∫S dxdy
= - ∫∫D dxdy
= - π * 1²
= - π
综上得∫∫Σ dydz + dzdx + dxdy = 0 - (- π) = π