例2-1求由下列曲面围成的立体的体积 (1)x=6-x2-y2,z=√x2+y 〖解〗 FEFF: g1=Plot3D(6-x2-y 2+3, ( x, 2, 2),y,-2, 2), Shading->Falsel g2=Plot3D/2+y/+3,, -2, 23,3,-2, 2), Shading- gfapetametricPlot3D[ r* Coslul, r SinU), 0), r,0, 2), u, 0, 2Pi), Shading-> False ShowIgl, g2, g3, BoxRatios->(1, 1, 1.5, Shading->Falsel 由于透影区域为Dn:x2+y2≤4 故选用柱坐标积分: v-1101m=2n1--n=37 00 3 0 teA: Integrate r, v,0, 2Pi), r,0, 2),z,r, 6-r/2)=32 Pi/3例2-1 求由下列曲面围成的立体的体积 g2=Plot3D[Sqrt[x^2+y^2+3],{x,-2,2},{y,-2,2},Shading- >False] 〖解〗 程序：g1=Plot3D[6-x^2-y^2+3,{x,-2,2},{y,-2,2},Shading->False] g3=ParametricPlot3D[{r*Cos[u],r Sin[u],0},{r,0,2},{u,0,2Pi}, Shading->False] 故选用柱坐标积分: 由于透影区域为 Show[g1,g2,g3,BoxRatios->{1,1,1.5},Shading->False] 2 2 2 2 (1) z = 6 − x − y , z = x + y : 4 2 2 Dxy x + y    V = dv    − =   2 0 2 0 6 2 r r d rdr dz 3 32 2 (6 ) 2 0 2  =  − − =  r r r dr 语句: Integrate[r,{v,0,2Pi},{r,0,2},{z,r,6-r^2}=32 Pi/3