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.
Asked
Active
Viewed 3,723 times
0
m_goldberg
- 107,779
- 16
- 103
- 257
rsri
- 1
- 1
- 1
-
1related – bobthechemist May 15 '14 at 15:03
1 Answers
4
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]
m_goldberg
- 107,779
- 16
- 103
- 257
-
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.. – george2079 May 15 '14 at 18:43