设z为锥面z=(x^2+y^2)^1⼀2在柱体x^2+y^2<=2x的内部,则曲面积分zds

设z为锥面z=(x^2+y^2)^1/2在柱体x^2+y^2&lt;=2x的内部,则曲面积分zds
2025-05-17 09:23:49
推荐回答(2个)
回答1:

z=x^2+y^2为选择抛物面 S曲面在z=0平面的投影为平面区域D(如下图阴影部分) 则积分区域D换为极坐标为: x=rcosθ,y=rsinθ,z=r^2;0≤r≤1,π/4≤θ≤5π/4 ∴∫∫S [y^2+xyf(z)]dS =∫∫D [y^2+xyf(z)]dxdy =∫∫D [(rsinθ)^2+rcosθ*rsinθf(r^2)]*rdrdθ =∫<0,1>r^3dr*∫<π/4,5π/4> (sinθ)^2dθ+∫<0,1>r^3f(r^2)dr*∫<π/4,5π/4>sinθcosθdθ =[<0,1>r^4/4]*[<π/4,5π/4> θ/2-(sin2θ)/4]+∫<0,1>r^3f(r^2)dr*[<π/4,5π/4>(sinθ)^2/2] =π/8+0*∫<0,1>r^3f(r^2)dr =π/8

回答2:

简单分析一下,详情如图所示