Re: Re: how fill PolarPlot?

*To*: mathgroup at smc.vnet.net*Subject*: [mg84455] Re: [mg84431] Re: [mg84420] how fill PolarPlot?*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Tue, 25 Dec 2007 06:28:05 -0500 (EST)*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst*References*: <20071224091350.HNWYO.62626.root@eastrmwml30>*Reply-to*: murray at math.umass.edu

Your response suggests that perhaps it's not possible directly to use polar coordinates themselves as arguments to the ColorFunction function. If so, that's a shame. Bob Hanlon wrote: > PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}, > ColorFunction -> > Function[{x, y}, If[-Pi/4 <= ArcTan[x, y] <= Pi/4, Red, Blue]], > ColorFunctionScaling -> False] > > > Bob Hanlon > > ---- Murray Eisenberg <murray at math.umass.edu> wrote: >> I was finally able to do this with Epilog->{Inset[RegionPlot[...]]}. >> >> Below is the entire code for the embellished plot I wanted. I am still >> unhappy with at the amount of work I had to do in order to adjust the >> ImageSize of the filled leaf and the thickness of its boundary so as to >> cover up the underlying blue boundary of that leaf from the PolarPlot. >> >> Some of that adjustment could probably be avoided by using a >> ColorFunction for the overall POlarPlot. But how does one set up >> ColorFunction for PolarPlot so as to specify using, say, one color for >> part of the plot and another for another part, depending on the value of >> theta alone? >> >> I found no example of ColorFunction in the documentation. I tried, e.g., >> >> PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}, >> ColorFunction -> >> Function[{theta,r}, If[-Pi/4 <= theta <= Pi/4, Red, Black]]] >> >> but that doesn't work as expected. >> >> The finished figure's code: >> >> txt[t_,{x_,y_}]:=Style[Text[t,{x,y}],FontSize->30,FontWeight->Bold] >> {xmin,xmax}={-1.425,1.425}; {ymin,ymax}={-1.25,1.25}; >> >> PolarPlot[Cos[2t],{t,0,2Pi}, PlotRange->{{xmin,xmax},{ymin,ymax}}, >> PlotStyle->{ColorData["Legacy","SteelBlue"], Thickness[0.007]}, >> Ticks->None, >> >> Epilog->{ >> Inset[RegionPlot[(x^2+y^2)^(3/2)<=x^2-y^2,{x,-0.02,1},{y,-1,1}, >> PlotStyle->ColorData["HTML","Gold"], >> BoundaryStyle->Directive[Thickness[0.025], >> ColorData["Legacy","CadmiumOrange"]], >> Frame->False,AspectRatio->Automatic, >> ImageSize->2.6*72], >> {0.5,0}], >> Black,Thick,Dashing[{0.045,0.03}], >> Line[{{0,0},{0.85,0.85}}],Line[{{0,0},{0.85,-0.85}}], >> Dashing[{}],Thick, >> Arrow[{{xmin,0},{xmax,0}}],Arrow[{{0,ymin},{0,ymax}}], >> txt[TraditionalForm[HoldForm[r==cos 2t ]],{-0.6,1.0}], >> txt[TraditionalForm[HoldForm[t==Pi/4]],{1.125,0.925}], >> txt[TraditionalForm[HoldForm[t==-Pi/4]],{1.125,-0.99}] >> }, >> ImageSize->7*72] >> >> Murray Eisenberg wrote: >>> I Mathematica 6 I have a PolarPlot, e.g., a 4-leaved rose: >>> >>> PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}] >>> >>> How can I fill the inside -- or, what I really want, just the leaf in >>> the right half-plane -- with some color? >>> >>> I note that Filling does not seem to be an option for PolarPlot (or for >>> what would be almost as good, ParametricPlot). >>> >>> I tried including the following (obtained by converting from : >>> >>> Prolog->RegionPlot[(x^2 + y^2)^(3/2) <= x^2-y^2, {x,-0.02,1},{y,-1,1}, >>> Frame->False, AspectRatio->Automatic] >>> >>> However, that led to a mysterious error message: >>> >>> $Aborted is not a Graphics primitive or directive. >>> >>> (Perhaps because of an incompatibility of a Prolog with cartesian >>> coordinates inside a polar coordinate plot??) >>> >> -- >> Murray Eisenberg murray at math.umass.edu >> Mathematics & Statistics Dept. >> Lederle Graduate Research Tower phone 413 549-1020 (H) >> University of Massachusetts 413 545-2859 (W) >> 710 North Pleasant Street fax 413 545-1801 >> Amherst, MA 01003-9305 >> > -- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305

**Follow-Ups**:**Re: Re: Re: how fill PolarPlot?***From:*Brett Champion <brettc@wolfram.com>