Question

如何求由曲面z=√x^2+y^2,x^2+y^2=2ax与平面z=0围成的立体的体积，谢谢

Answer 1

它是由圆锥面、圆柱面和XOY平面围成。
用极坐标做较方便。
z=√x^2+y^2变成z=ρ，
,x^2+y^2=2ax变成ρ=2acosθ,
积分区域D：0<=ρ<=2acosθ,
-π/2<=θ<=π/2,
V=∫[-π/2，π/2]dθ∫[0,2acosθ] ρ*ρdρ
=∫[-π/2，π/2]dθ[0,2acosθ]ρ^3/3
=2∫[0,π/2](8a^3)/3*(cosθ)^3dθ
=16a^3/3∫[0,π/2][1-(sinθ)^2dsinθ
=16a^3/3[sinθ-(sinθ)^3/3] [0,π/2][
=16a^3/3(1-1/3)
=32a^3/9.

Answer 2

它是由圆锥面、圆柱面和XOY平面围成。
用极坐标做较方便。
z=√x^2+y^2变成z=ρ，
,x^2+y^2=2ax变成ρ=2acosθ,
积分区域D：0<=ρ<=2acosθ,
-π/2<=θ<=π/2,
V=∫[-π/2，π/2]dθ∫[0,2acosθ]
ρ*ρdρ
=∫[-π/2，π/2]dθ[0,2acosθ]ρ^3/3
=2∫[0,π/2](8a^3)/3*(cosθ)^3dθ
=16a^3/3∫[0,π/2][1-(sinθ)^2dsinθ
=16a^3/3[sinθ-(sinθ)^3/3]
[0,π/2][
=16a^3/3(1-1/3)
=32a^3/9.