To color surfaces using z values with ContourPlot3D

*To*: mathgroup at smc.vnet.net*Subject*: [mg23881] To color surfaces using z values with ContourPlot3D*From*: "Valeria Olivati" <olivaleNNN at tin.it>*Date*: Thu, 15 Jun 2000 00:51:18 -0400 (EDT)*Organization*: TIN*Sender*: owner-wri-mathgroup at wolfram.com

Hallo eveybody, I'm a student of Mathematics at University of Pavia, Italy. I start using Mathematica 4 two months ago, and I need to plot a cubic function and color it without using LightSources. I use ContourPlot3D to plot the function but I don't know how to color the surface using the z values, as you can do using Plot3D and ColorFunction->Hue. I cannot use Plot3D because the function is quite complicated: eq = 81*(x^3 + y^3 + z^3) - 189*(x^2*y + x^2*z + y^2*x + y^2*z + z^2*x + z^2*y) + 54*x*y*z + 126*(x*y + x*z + y*z) - 9*(x^2 + y^2 + z^2) - 9*(x+y+z) + 1 Now I plot this funtion in this way: sup = ContourPlot3D[eq, {x, -1, 3/2}, {y, -1, 3/2}, {z, -1, 3/2}, PlotPoints->{5,7}, Axes->True, ViewPoint->{0.056, -3.382, -0.076}, ContourStyle->{Hue[1], EdgeForm[GrayLevel[0.6]]}, Lighitng->False] but with only one color It's not a clear shape. If I use the following LightSources instead of ConturStyle LightSources->{{{0,0,0.7}, RGBColor[0,1,0]}, {{1,0,1}, RGBColor[1,1,0]}, {{-1,0,1}, RGBColor[0,0,1]}, {{-1,0,0}, GrayLevel[1]}} I don't like the colors I obtain, they are not linked to the surface but they are the best LightSources I found after a big number of changes. Of course my problem with LightSources is that I haven't understood how they work.... I hope someone has suggestions for me Thanks a lot Best Regards Olivati Valeria