Question

求曲面积分∑∫∫2x^2zdxdz+y(z^2+1)dzdx+(9-z^3)dxdy,∑为曲面z=x^2+y^2+1(1<=z<=2)下侧

期末考试题，谢谢
∑∫∫2x^2*zdxdz+y(z^2+1)dzdx+(9-z^3)dxdy,∑为曲面z=x^2+y^2+1(1<=z<=2)下侧

Answer 1

Σ为x² + y² = z - 1
1 ≤ z ≤ 2、下侧
补面S：z = 2、上侧
∫∫S 2x²zdydz + y(z² + 1)dzdx + (9 - z³)dxdy
= ∫∫S dxdy
= ∫∫D dxdy、x² + y² ≤ 1
= π
∫∫(Σ+S) 2x²zdydz + y(z² + 1)dzdx + (9 - z³)dxdy
= ∫∫∫Ω (4xz + z² + 1 - 3z²) dV
= ∫∫∫Ω (1 + 4xz - 2z²) dV
= ∫(0→2π) dθ ∫(0→1) r dr ∫(r² + 1→2) (1 + 4rz cosθ - 2z²) dz
= - 7π/3
即∫∫Σ 2x²zdydz + y(z² + 1)dzdx + (9 - z³)dxdy = - 7π/3 - π = - 10π/3

Answer 2

题目可能有错误，大概是：
∑∫∫2x^2zdydz+y(z^2+1)dzdx+(9-z^3)dxdy,∑为曲面z=x^2+y^2+1(1<=z<=2)下侧
添加z=2上侧为Σ1，封闭图形所谓空间为Ω,利用高斯公式，并用柱坐标计算：
原式=Ω∫∫∫（4xz+z^2+1-3z^2)dxdydz-∑1∫∫2x^2zdydz+y(z^2+1)dzdx+(9-z^3)dxdy
=∫[0,2π]dθ∫[0,1]ρdρ∫[ρ^2+1,2](4ρcosθz-2z^2+1)dz-∫∫[x^2+y^2<=1]dxdy
=∫[0,2π]dθ∫[0,1]ρ[2ρcosθz^2-2/3z^3+z][ρ^2+1,2]dρ-π
=∫[0,2π]dθ∫[0,1]ρ[8ρcosθ-2ρcosθ(ρ^2+1)^2-16/3+2/3(ρ^2+1)^3+2-(ρ^2+1)dρ-π
=下面自己算