Is it possible to create polar CountourPlot/ListCountourPlot/DensityPlot in Mathematica?

I am looking to plot something like the whispering gallery modes -- a 2D cylindrically symmetric plot in polar coordinates. Something like this:

I found the following code snippet in Trott's symbolics guidebook. Tried running it on a very small data set; it ate 4 GB of memory and hosed my kernel:

(* add points to get smooth curves *)
addPoints[lp_][points_, \[Delta]\[CurlyEpsilon]_] := 
Module[{n, l}, Join @@ (Function[pair,
       If[(* additional points needed? *)
          (l = Sqrt[#. #]&[Subtract @@ pair]) < \[Delta]\[CurlyEpsilon], pair, 
          n = Floor[l/\[Delta]\[CurlyEpsilon]] + 1; 
          Table[# + i/n (#2 - #1), {i, 0, n - 1}]& @@ pair]] /@ 
          Partition[If[lp === Polygon, 
                       Append[#, First[#]], #]&[points], 2, 1])]

(* Make the plot circular *)
With[{\[Delta]\[CurlyEpsilon] = 0.1, R = 10}, 
 Show[{gr /. (lp : (Polygon | Line))[l_] :> 
     lp[{#2 Cos[#1], #2 Sin[#1]} & @@@(* add points *)
       addPoints[lp][l, \[Delta]\[CurlyEpsilon]]], 
   Graphics[{Thickness[0.01], GrayLevel[0], Circle[{0, 0}, R]}]}, 
  DisplayFunction -> $DisplayFunction, Frame -> False]]

Here, gr is a rectangular 2D ListContourPlot, generated using something like this (for example):

data = With[{eth = 2, er = 2, wc = 1, m = 4}, 
   Table[Re[
     BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[
       I m phi]], {r, 0, 10, .2}, {phi, 0, 2 Pi, 0.1}]];
gr = ListContourPlot[data, Contours -> 50, ContourLines -> False, 
  DataRange -> {{0, 2 Pi}, {0, 10}}, DisplayFunction -> Identity, 
  ContourStyle -> {Thickness[0.002]}, PlotRange -> All, 
  ColorFunctio开发者_开发技巧nScaling -> False]

Is there a straightforward way to do cylindrical plots like this?.. I find it hard to believe that I would have to turn to Matlab for my curvilinear coordinate needs :)

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Previous snippets deleted, since this is clearly the best answer I came up with:

With[{eth = 2, er = 2, wc = 1, m = 4}, 
 ContourPlot[
  Re[BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[I phi m]]/. 
                                         {r ->Norm[{x, y}], phi ->ArcTan[x, y]}, 
  {x, -10, 10}, {y, -10, 10}, 
  Contours -> 50, ContourLines -> False, 
  RegionFunction -> (#1^2 + #2^2 < 100 &), 
  ColorFunction -> "SunsetColors"
 ]
]

Edit

Replacing ContourPlot by Plot3D and removing the unsupported options you get:

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This is a relatively straightforward problem. The key is that if you can parametrize it, you can plot it. According to the documentation both ListContourPlot and ListDensityPlot accept data in two forms: an array of height values or a list of coordinates plus function value ({{x, y, f} ..}). The second form is easier to deal with, such that even if your data is in the first form, we'll transform it into the second form.

Simply, to transform data of the form {{r, t, f} ..} into data of the form {{x, y, f} ..} you doN[{#[[1]] Cos[ #[[2]] ], #[[1]] Sin[ #[[2]] ], #[[3]]}]& /@ data, when applied to data taken from BesselJ[1, r/2] Cos[3 t] you get

What about when you just have an array of data, like this guy? In that case, you have a 2D array where each point in the array has known location, and in order to plot it, you have to turn it into the second form. I'm partial to MapIndexed, but there are other ways of doing it. Let's say your data is stored in an array where the rows correspond to the radial coordinate and the columns are the angular coordinate. Then to transform it, I'd use

R = 0.01;    (*radial increment*)
T = 0.05 Pi; (*angular increment*)
xformed = MapIndexed[ 
   With[{r = #2[[1]]*R, t = #2[[1]]*t, f = #1},
   {r Cos[t], r Sin[t], f}]&, data, {2}]//Flatten[#,1]&

which gives the same result.

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If you have an analytic solution, then you need to transform it to Cartesian coordinates, like above, but you use replacement rules, instead. For instance,

ContourPlot[ Evaluate[
    BesselJ[1, r/2]*Cos[3 t ] /. {r -> Sqrt[x^2 + y^2], t -> ArcTan[x, y]}], 
   {x, -5, 5}, {y, -5, 5}, PlotPoints -> 50, 
   ColorFunction -> ColorData["DarkRainbow"], Contours -> 25]

gives

Two things to note: 1) Evaluate is needed to ensure that the replacement is performed correctly, and 2) ArcTan[x, y] takes into account the quadrant that the point {x,y} is found in.