求由曲面z=x2+2y2及z=6-2x2-y2所围的立体的体积．
由题意知：v=∫∫x2+y2=2 dxdy∫x2+2y26-2x2-y2dz =∫∫x2+y2=2(6-3x2-3y2)dxdy= ∫02πdθ∫02√2r(6-3r2)dr =2π∫0√2(6r-3r3)dr=2π[3r3-(3/4)r4]∣0√2 =2π(6-3)=6π
求由曲面z=x2+2y2及z=6-2x2-y2所围的立体的体积．
由题意知：v=∫∫x2+y2=2 dxdy∫x2+2y26-2x2-y2dz =∫∫x2+y2=2(6-3x2-3y2)dxdy= ∫02πdθ∫02√2r(6-3r2)dr =2π∫0√2(6r-3r3)dr=2π[3r3-(3/4)r4]∣0√2 =2π(6-3)=6π
