Revolution of two curves around the y-axis

Question

I've looked through the other posts on rotating a function around the x or y axis and haven't had any luck.

my question is this how do i rotate the area between the functions $f(x)=x^2+1$, and $f(x)=2x^2-2$, whereby $x>0$ and $y>0$ and the region is bounded by the y and x axis.

This is a plot of the area I'm talking about.

Revolving the area around the y-axis is suppose to produce a vase, although I picture something of a shallow bowl.

Any help would be much appreciated. As I'm still quite new to Mathematica, an explanation of any code posted would be helpful.

Dr. belisarius · Answer

Show[RevolutionPlot3D[#, {x, 0, Sqrt[3]}, MeshStyle -> None, 
                      PlotStyle -> Opacity[.5]] & /@ {2 x^2 - 2,  x^2 + 1}]

m_goldberg · Answer

Not an answer but an extended comment on belisarius' answer.

I think the plot will look more like what I believe the OP is expecting if a few options are added.

Show[RevolutionPlot3D[#, {x, 0, Sqrt[3]},
  PlotRange -> {Automatic, Automatic, {0., Automatic}},
  MeshStyle -> None,
  Axes -> None,
  Boxed -> False,
  PlotStyle -> Opacity[.5]] & /@ 
    {2 x^2 - 2, x^2 + 1, Piecewise[{{0, 0 < x < 1}}, Null]}]

plot

Also note that I have put a bottom on the bowl.

ubpdqn · Answer

Just for something different: pp = ParametricPlot3D[{Sqrt[3] Cos[t], Sqrt[3] Sin[t], 4}, {t, 0, 2 Pi}, PlotStyle -> {Red, Thickness[0.04]}, Boxed -> False]; rp = RegionPlot3D[ 0 < x^2 + y^2 < 3 && Max[0, 2 x^2 + 2 y^2 - 2] < z < x^2 + y^2 + 1, {x, -2, 2}, {y, -2, 2}, {z, 0, 5}, Mesh -> None, PlotStyle -> Red, PlotPoints -> 200, PerformanceGoal -> "Quality", Boxed -> False]; Show[pp, rp, Axes -> False, Background -> Black] Top ring just to deal with boundary of intersection...

Revolution of two curves around the y-axis

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