# How to make a 3D plot using polar coordinates?

From the beginning: I have function $u(r)$ and radial symmetry in system. Also I've got results as data array ${u_i(r_i)}$. And I want to plot it as $u(x,y)$. Due to radial symmetry its gonna be like $x=r \cos(\varphi), y=r \sin(\varphi)$ In other words I have function profile and want to "integrate" it over $2\pi\,\mathrm d\varphi$. Like this for Gaussian

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• Use RevolutionPlot3D[] instead. – J. M.'s ennui May 19 '15 at 19:03
• Have you looked at ParametricPlot3D? I believe it can be applied. – Mr.Wizard May 19 '15 at 19:04
• All this works if I have a rule for my U(r) -> U(x,y). But I have to assign a numerical value from list of already obtained values to new rule for (x,y) to build concentric circles. Maybe I'm missing something.I'm novice in mathematica, all my knowledges based on F1 help – Dima Sakovich May 19 '15 at 19:28

One option would be to use RevolutionPlot3D.

u = Table[Sin[2 \[Pi]*r], {r, 0, 1, 0.1}]; (*u is a dummy u[r]*)
f = ListInterpolation[u, {0, 1}]; (*Create an interpolating function over the range {0,1}*)
(*Plot it over the domain.*)
RevolutionPlot3D[f[r], {r, #1, #2}] & @@@ f["Domain"]


You could also generate the points yourself and use ListPointPlot3D

u = Table[{r, Sin[2 \[Pi]*r]}, {r, 0, 1, 0.1}];(*table of {r,u[r]}*)
xyz = Flatten[Table[{#1*Cos[\[Theta]], #1*Sin[\[Theta]], #2} & @@@ u, {\[Theta],0, 2 \[Pi], 2 \[Pi]/100} ], 1];
ListPointPlot3D[
xyz
, Filling -> Axis
]