设曲面为抛物面z=1-x^2-y^2(0

Question

设曲面为抛物面z=1-x^2-y^2(0

Answer 1

取Σ：x^2 + y^2 = 1 - z.(0 ≤ z ≤ 1)抛物面曲顶向上.
补面Σ1：z = 0.取下侧
∫∫(Σ+Σ1) 2x^3dydz + 2y^3dzdx + 2dxdy
= ∫∫∫Ω (6x^2 + 6y^2 + 0) dxdydz.Gauss定理
= 6∫∫∫Ω (x^2 + y^2) dxdydz
= 6∫(0,2π) dθ ∫(0,1) r dr ∫(0,1 - r^2) r^2 dz
= 6 * 2π * ∫(0,1) r^3 * (1 - r^2) dr
= 12π * ∫(0,1) (r^3 - r^5) dr
= 12π * (1/4 * r^4 - 1/6 * r^6)：(0,1)
= 12π * 1/12
= π
∫∫Σ1 2x^3dydz + 2y^3dzdx + 2dxdy
= ∫∫Σ1 0 + 0 + 2dxdy
= 2∫∫D dxdy.x^2 + y^2 ≤ 1
= 2 * π * 1^2
= 2π
于是∫∫Σ + ∫∫Σ1 = ∫∫(Σ+Σ1) = π
∫∫Σ 2x^3dydz + 2y^3dzdx + 2dxdy = π - 2π = - π