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I am trying to make a polar plot of a two variable function, which has a radial and an angular part. For instance Y[r,φ]=r^2 Cos[φ].

I can make this plot using ContourPlot, but is there a way to do it using PolarPLot? My ContourPlot is

Y[r_, f_] := r^2 Cos[f]
ContourPlot[Y[Sqrt[x^2 + y^2], ArcTan[x, y]], {x, 0, 1}(*Radius till 1*), {y, 0, 60}(*Angle from 0 to 60*)]

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  • $\begingroup$ The argument of Cos, ArcTan etc. is assumed to be in radians, multiply by Degree to convert from degrees. $\endgroup$
    – VLC
    Commented Feb 22, 2013 at 13:49
  • $\begingroup$ Well, the help on PolarPlot[] states "Use ContourPlot and RegionPlot for implicit curves and regions:" $\endgroup$
    – Dr. belisarius
    Commented Feb 22, 2013 at 14:13
  • $\begingroup$ @VLC:Thank's for that! I didn't know it!!! $\endgroup$
    – Thanos
    Commented Feb 22, 2013 at 14:16
  • $\begingroup$ @belisarius: I've seen, but I would like to plot this two variable function in the most efficient way...and I am not sure on how to do it without polar plot... $\endgroup$
    – Thanos
    Commented Feb 22, 2013 at 14:58
  • $\begingroup$ Well, I just posted an answer :) $\endgroup$
    – Dr. belisarius
    Commented Feb 22, 2013 at 14:59

1 Answer 1

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PolarPlot[] is a handicapped plotting function (it doesn't support Filling, for example). Much easier with other functions:

h[r_, f_] := r^2 Cos[f]
Quiet@Show[
        ContourPlot[h[Sqrt[x^2 + y^2], ArcTan[x, y]], {x, 0, 1}, {y, 0, 1},
                  RegionFunction -> Function[{x, y, f}, 0 < ArcTan[x, y] < Pi/3 && x^2 + y^2 < 1],
                  Contours -> 10,  AspectRatio -> 1], 
         Graphics@Circle[]]

Mathematica graphics

Another possibility:

Plot3D[h[Sqrt[x^2 + y^2], ArcTan[x, y]],
 {x, -1, 1},
 {y, -1, 1},
 AspectRatio -> 1,
 ColorFunction -> "SunsetColors",
 MeshFunctions -> {#3 &},
 Mesh -> 7,
 PlotStyle -> Directive[Specularity[White, 50], Opacity[0.8]]]

Mathematica graphics

You may even draw your Pi/3 angle on the surface:

Plot3D[h[Sqrt[x^2 + y^2], ArcTan[x, y]],
 {x, -1, 1},
 {y, -1, 1},
 AspectRatio -> 1,
 ColorFunction -> "SunsetColors",
 MeshFunctions -> {UnitBox@(ArcTan[#2, #1] - Pi/3) &},
 Mesh -> 7,
 PlotStyle -> Directive[Specularity[White, 50], Opacity[0.8]]]

Mathematica graphics

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  • $\begingroup$ I believe the first plot should be fine!!! Thank you! $\endgroup$
    – Thanos
    Commented Feb 22, 2013 at 15:33

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