求曲面x^2+y^2=2ax(a>0),z=αx,z=βx(α,β为常数,且α>β>0)所围成立体的体积.

Question

求曲面x^2+y^2=2ax(a>0),z=αx,z=βx(α,β为常数,且α>β>0)所围成立体的体积.

Answer 1

所围成立体的体积=∫dθ∫(β-α)r²cosθdr (作柱面坐标变换)
=(β-α)∫cosθdθ∫r²dr
=(β-α)∫[(2acosθ)³/3]cosθdθ
=[(8a³/3)(β-α)]∫cos³θdθ
=[(8a³/3)(β-α)](sinθ-sin³θ/3)│
=[(8a³/3)(β-α)](1-1/3+1-1/3)
=32a³(β-α)/9.