Question

求由曲面z=根号下a^2-x^2-y^2及z=根号下x^2+y^2所围立体的体积

Answer 1

设x=rcosu,y=rsinu,则dxdy=rdrdu,
所求体积=∫<0,2π>du∫<0,a/√2>rdr∫<r,√(a^2-r^2)>dz
=2π∫<0,a/√2>r[√(a^2-r^2)-r]dr
=π[-(2/3)(a^2-r^2)^(3/2)-2r^3/3]|<0,a/√2>
=(2π/3)[1-2^(-3/2)-2^(-3/2)]a^3
=(π/3)[2-√2]a^3.
解2 两曲面围成的几何体被平面z=a/√2分成两个部分：上部是底面半径为a/√2，高为a-a/√2的球缺，下部是底面半径和高都是a/√2的圆锥，所以
所求体积=(π/6)(a-a/√2)[3(a/√2)^2+(a-a/√2)^2]+(π/3)(a/√2)^3
=(π/3)[2-√2]a^3.

Answer 2

简单计算一下即可，答案如图所示