有些看不懂你的式子,我只能理解为∫∫z√(x^2+y^2)dv,y=√(2x-x^2),z=0,z=a围成做变换x=rcosθ,y=rsinθ,y=√(2x-x^2)可得y>0,y^2+(x-1)^2=1推出(要结合图才好看出,但不上传啊)θ∈[0,π/2],r∈[0,2cosθ],z∈[0,a]∫∫z√(x^2+y^2)dv=∫dz?√(x^2+y^2)dxdy=∫dz?r*rdrdθ=∫[0,a]dz∫[0,π/2]dθ∫[0,2cosθ]r*rdr=a∫[0,π/2] 8/3*(cosθ)^3dθ=8a/3*∫[0,π/2] (1-(sinθ)^2)dsinθ=16a/9... |