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I'm trying to rotate the area between the graphs of $y=3$ and $y=x^2$ around $y=1$. I'm not attempting to just find the area as that would just be π Integral[9 - x^4], but rather I'm trying to create a 3D plot that I can print with a 3D printer.

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I can show you how to generate a RevolutionPlot3D of your curve, but having no experience with 3D-printing, I don't know if the results will be in the form you need for 3D-printing.

First, rewrite your function so that it can be revolved about y = 0, because RevolutionPlot3D is designed to revolve curves about an axis the passes trough the origin.

y[x_] := x^2 - 1

RevolutionPlot3D also assumes the curve lies in the xz-plane, so the revolution should be made about the x-axis, not the default z-axis.

RevolutionPlot3D[y[t], {t, -Sqrt[3], Sqrt[3]},
  RevolutionAxis -> "x", Boxed -> False, Axes -> False]

plot

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  • $\begingroup$ Suspect he meant for the plot range to be {1,Sqrt[3]} (at least that gives something physically sensible) -- then you need to close the plot to make a solid.. $\endgroup$ – george2079 May 15 '14 at 18:43

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