# Independent Colorfunction in RevolutionPlot3D

probably a trivial questions, but I couldn't find a solution for over 2 hours:

I want to produce a RevolutionPlot3D of a function, f1, with a separate colorfunction, f2. I thought it might be possible to specify something like:

RevolutionPlot3D[r^2*Sin[2*\[Theta]], {r, 0, 1}, {\[Theta], 0, 2*Pi},
ColorFunction -> Function[{x, y, z, r1, \[Theta]1}, ColorData["Rainbow"][Abs[z]]]]


where

f1 = r^2*Sin[2*\[Theta]]


and

f2 = Abs[r^2*Sin[2*\[Theta]]]


instead of the expected outcome that should look like this: Thank you very much for your help.

• Bottom plot is not rainbow see this mathematica.stackexchange.com/questions/101268/… Apr 8 '18 at 15:22
• Based on @Okkes's comment: RevolutionPlot3D[r^2 Sin[2 θ], {r, 0, 1}, {θ, 0, 2 π}, ColorFunction -> Function[{x, y, z, r, θ}, Hue[2 (1 - Abs[r^2 Sin[2 θ]])/3]], ColorFunctionScaling -> False, Mesh -> False] Apr 8 '18 at 16:00
• Just add ColorFunctionScaling -> False to your original plot.
– user484
Apr 8 '18 at 16:19
• @J.M. - recommend that you also increase the PlotPoints Apr 8 '18 at 16:20
• Thanks a lot, Rahul! It finally clicked and I understand the comment now. Thanks J.M. & Dulgerci - I wouldn't have spotted it. Problem solved (in case you want to post it as an answer) Apr 8 '18 at 17:13

RevolutionPlot3D[r^2 Sin[2 θ], {r, 0, 1}, {θ, 0, 2 π}, 