Re: v.6 RevolutionPlot3D

*To*: mathgroup at smc.vnet.net*Subject*: [mg75939] Re: [mg75852] v.6 RevolutionPlot3D*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 13 May 2007 06:00:46 -0400 (EDT)*References*: <200705120700.DAA23589@smc.vnet.net>

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=

**References**:**v.6 RevolutionPlot3D***From:*Helen Read <read@math.uvm.edu>