Question

7、求由曲面z=x^2+2y^2 以及 z=6-2x^2-y^2 所围成立体的体积

Answer 1

∵解方程组z=x²+2y²与z=6-2x²-y²,得x²+y²=2

∴所求立体在xoy面上投影区域为D={(x,y)lx²+y²≤2}

故 所求立体体积=∫∫[(6-2x²-y²)-(x²+2y²)]dxdy

=∫∫[6-3(x²+y²)]dxdy

=∫dθ∫(6-3r²)rdr (应用极坐标变换)

=2π∫(6r²-3r³)dr

=2π(2r³-3r^4/4)│

=2π(4√2-3)