Parametric Region from Surface of Revolution

I would like to create a region defined by the volume inside (or outside) a surface of revolution, and am struggling with understanding how the ParametricRegion function works. The help documentation only gives 2D examples unfortunately.

For example I would like to take the surface defined by:

ParametricPlot3D[{Cos[s] (1 + t^2), Sin[s] (1 + t^2),t}, {t, -2, 2}, {s, 0, 2 \[Pi]}]

and then us this to define one of my region boundaries.

If I define a 3D region like:

reg = ParametricRegion[{Cos[s] (1 + t^2), Sin[s] (1 + t^2), t}, {{t, -2, 2}, {s, 0, 2 \[Pi]}}]

and then try to plot this using RegionPlot[reg]

I get errors related that say such a region cannot be automatically discretised and that this is not a valid region to plot.

As far as I understand this code should give me the surface ? But does anyone know how one can define the volume inside, and then visualise this using RegionPlot or DiscretizeRegion ?

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It’s not a closed surface. What you mean by “inside”?
– Michael E2
Oct 17 at 22:40

  

 

@MichaelE2. Sorry, perhaps this was unclear. By inside I mean the region close to the z-axis. I would also like to cap the region eventually, to make a closed shape (i.e. constraining to -a 3, PlotPoints -> 35,
PlotRange -> {{-5, 5}, {-5, 5}, {-2, 2}}]

  

 

Thanks for pointing that out. Exactly what I was looking for.
– Dunlop
Oct 18 at 5:21

I am not exactly sure what region is required.

region1 =
ImplicitRegion[
0 < x^2 + y^2 < (z^2 + 1)^2, {{x, -5, 5}, {y, -5, 5}, {z, -2, 2}}]; region2 = ImplicitRegion[ x^2 + y^2 > (z^2 + 1)^2, {{x, -5, 5}, {y, -5, 5}, {z, -2, 2}}];
opts = Sequence[Background -> Black, Boxed -> False, ImageSize -> 400]
Column[{RegionPlot3D[region1, PlotPoints -> 40, PlotStyle -> LightRed,
Evaluate@opts],
RegionPlot3D[region2, PlotPoints -> 40,
PlotStyle -> {LightBlue, Opacity[0.5]}, Evaluate@opts],
RegionPlot3D[{region1, region2}, PlotPoints -> 40,
PlotStyle -> {LightRed, {LightBlue, Opacity[0.5]}}, Evaluate@opts]

}]

  

 

The First image was the one I was looking for, although in fact either of the surfaces is usable for my calculations. Many thanks!
– Dunlop
Oct 18 at 10:52