Question

∫∫x^2dydz+y^2dzdx+z^2dxdy∑是抛物面z=x^2+y^2被平面z=1所截下的有限部分的下侧

怎么做啊

Answer 1

补上平面S：z = 1，取上侧
∫∫(Σ+S) x²dydz + y²dzdx + z²dxdy
= ∫∫∫Ω (2x + 2y + 2z) dV、<== ∫∫∫Ω (2x + 2y) dV = 0
= 2∫∫∫Ω z dV
= 2∫(0→2π) dθ ∫(0→1) r dr ∫(r²→1) z dz
= 2π/3
∫∫S x²dydz + y²dzdx + z²dxdy
= ∫∫S dxdy
= π
即∫∫Σ x²dydz + y²dzdx + z²dxdy = 2π/3 - π = - π/3