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Re: v.6 RevolutionPlot3D

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  • Subject: [mg75939] Re: [mg75852] v.6 RevolutionPlot3D
  • From: Andrzej Kozlowski <akoz at>
  • Date: Sun, 13 May 2007 06:00:46 -0400 (EDT)
  • References: <>

On 12 May 2007, at 16:00, Helen Read wrote:

> We just got 6.0 on our site license, and I installed on my computer(s)
> yesterday.
> I see that SurfaceOfRevolution (which was in an add on package) has =

> been
> replaced by RevolutionPlot3D. Sounds great, except that 
> RevolutionPlot3D
> only revolves around the vertical axis. We (my calculus students) used
> the old SurfaceOfRevolution all the time for visualizing surfaces of
> revolution for computing volume and surface area, and we need to be =

> able
> to revolve around both the vertical axis and the horizontal axis. This
> was a simple matter of setting RevolutionAxis->{0,0,1} or
> RevolutionAxis->{1,0,0} respectively. Now, I can write a function for
> revolving around the horizontal axis and provide it to the 
> students, but
> I can already foresee the confusion it will cause when they can use a
> built-in function for revolving in one direction, and have to do it a
> different way to revolve in the other direction.
> --
> Helen Read
> University of Vermont

First of all, you can still use SurfaceOfRevolution in Mathematica 6:

<< Graphics`SurfaceOfRevolution`

(ignore the compatibility message)

SurfaceOfRevolution[x^2, {x, 0, 1},
     RevolutionAxis -> {1, 1, 1}]

(example taken form te Documentation for 5.2).

You get the same picture as before but now you can do all the new 
things that Mathematica 6 graphics allow you to do.

Another possibility is simply to forget about RevolutionPlot3D and 
use ParametricPlot3D:

ParametricPlot3D[Evaluate[{X, Y, Z} /. Solve[RotationMatrix[{{1, 1, 
1}, {0, 0, 1}}].{X, Y, Z} == {t Cos[=CE=B8],  t Sin[=CE=B8], t^2 }, =
{X, Y, 
Z}]], {t, 0, 1}, {=CE=B8, 0, 2 =CF=80}]

Andrzej Kozlowski=

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