1个回答
展开全部
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
∫∫(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
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询
广告 您可能关注的内容 |