计算∫∫yzdydz+(x²+z²)dzdx+xydxdy,其中∑为曲面4-y=z²+x²在平面xoz右侧部分的外侧

Question

计算∫∫yzdydz+(x²+z²)dzdx+xydxdy,其中∑为曲面4-y=z²+x²在平面xoz右侧部分的外侧

Answer 1

补Σ1y = 0、dy = 0的左侧
∫∫(Σ+Σ1) yzdydz + (x² + z²)dzdx + xydxdy
= ∫∫∫Ω (0 + 0 + 0) dxdydz = 0
∫∫Σ1 (x² + z²)dzdx
= - ∫∫Dzx (x² + z²)dzdx、z² + x² ≤ 4
= - ∫(0→2π) dθ ∫(0→2) r³ dr
= - 2π * (1/4) * 16
= - 8π
于是∫∫Σ yzdydz + (x² + z²)dzdx + xydxdy = 0 - (- 8π) = 8π