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Dynamic filling of a region of a polar plot



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to plot filling under a curve?Filling only part of a plotFilling between ParametricPlot and PlotGenerating hatched filling using Region functionalityFilling a Manipulate PlotHow can I increase the font size in PlotLabel?Filling under plotFilling the region under a lineFilling in parametric plotHow to shade a region using Filling function










6












$begingroup$


I would like to shade area of region as a function of angle using PolarPlot.
Here is my attempt.



With[
pts =
Cases[PolarPlot[1 + 2 Sin[θ], θ, 0, 2 π], _Line, 0, Infinity][[1, 1]],
Manipulate[
Show[
ListLinePlot[0, 0, pts[[n]], pts[[1 ;; n]],
Filling -> 2 -> Axis, LightBlue, 1 -> Axis, LightBlue,
PlotRange -> -2, 2, -0.5, 3.2, AspectRatio -> 1,
PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
ImageSize -> 500, AxesStyle -> Directive[Black, 18],
PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
PolarPlot[1 + 2 Sin[θ], θ, 0, 2.2 π,
AspectRatio -> 1, PlotStyle -> Black, AbsoluteThickness@3]],
n, 1, Length @ pts, 1]]


enter image description here



enter image description here



Two thing I would like to achieve:



  1. I don't want to see the yellow highlited region.

  2. When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.

Any suggestion..










share|improve this question











$endgroup$
















    6












    $begingroup$


    I would like to shade area of region as a function of angle using PolarPlot.
    Here is my attempt.



    With[
    pts =
    Cases[PolarPlot[1 + 2 Sin[θ], θ, 0, 2 π], _Line, 0, Infinity][[1, 1]],
    Manipulate[
    Show[
    ListLinePlot[0, 0, pts[[n]], pts[[1 ;; n]],
    Filling -> 2 -> Axis, LightBlue, 1 -> Axis, LightBlue,
    PlotRange -> -2, 2, -0.5, 3.2, AspectRatio -> 1,
    PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
    ImageSize -> 500, AxesStyle -> Directive[Black, 18],
    PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
    PolarPlot[1 + 2 Sin[θ], θ, 0, 2.2 π,
    AspectRatio -> 1, PlotStyle -> Black, AbsoluteThickness@3]],
    n, 1, Length @ pts, 1]]


    enter image description here



    enter image description here



    Two thing I would like to achieve:



    1. I don't want to see the yellow highlited region.

    2. When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.

    Any suggestion..










    share|improve this question











    $endgroup$














      6












      6








      6


      1



      $begingroup$


      I would like to shade area of region as a function of angle using PolarPlot.
      Here is my attempt.



      With[
      pts =
      Cases[PolarPlot[1 + 2 Sin[θ], θ, 0, 2 π], _Line, 0, Infinity][[1, 1]],
      Manipulate[
      Show[
      ListLinePlot[0, 0, pts[[n]], pts[[1 ;; n]],
      Filling -> 2 -> Axis, LightBlue, 1 -> Axis, LightBlue,
      PlotRange -> -2, 2, -0.5, 3.2, AspectRatio -> 1,
      PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
      ImageSize -> 500, AxesStyle -> Directive[Black, 18],
      PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
      PolarPlot[1 + 2 Sin[θ], θ, 0, 2.2 π,
      AspectRatio -> 1, PlotStyle -> Black, AbsoluteThickness@3]],
      n, 1, Length @ pts, 1]]


      enter image description here



      enter image description here



      Two thing I would like to achieve:



      1. I don't want to see the yellow highlited region.

      2. When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.

      Any suggestion..










      share|improve this question











      $endgroup$




      I would like to shade area of region as a function of angle using PolarPlot.
      Here is my attempt.



      With[
      pts =
      Cases[PolarPlot[1 + 2 Sin[θ], θ, 0, 2 π], _Line, 0, Infinity][[1, 1]],
      Manipulate[
      Show[
      ListLinePlot[0, 0, pts[[n]], pts[[1 ;; n]],
      Filling -> 2 -> Axis, LightBlue, 1 -> Axis, LightBlue,
      PlotRange -> -2, 2, -0.5, 3.2, AspectRatio -> 1,
      PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
      ImageSize -> 500, AxesStyle -> Directive[Black, 18],
      PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
      PolarPlot[1 + 2 Sin[θ], θ, 0, 2.2 π,
      AspectRatio -> 1, PlotStyle -> Black, AbsoluteThickness@3]],
      n, 1, Length @ pts, 1]]


      enter image description here



      enter image description here



      Two thing I would like to achieve:



      1. I don't want to see the yellow highlited region.

      2. When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.

      Any suggestion..







      plotting filling






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 20 at 1:56







      Okkes Dulgerci

















      asked Apr 19 at 23:46









      Okkes DulgerciOkkes Dulgerci

      5,5261919




      5,5261919




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          This is what you need:



          Manipulate[ParametricPlot[
          r (1 + 2 Sin[θ]) Cos[θ], Sin[θ],
          θ, 0, thmax,
          r, 0, 1,
          PlotRange -> -2.25, 2.25, -0.5, 3.5,
          PerformanceGoal -> "Quality"
          ], thmax, 0.01, 2 Pi]


          Mathematica graphics






          share|improve this answer











          $endgroup$













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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            This is what you need:



            Manipulate[ParametricPlot[
            r (1 + 2 Sin[θ]) Cos[θ], Sin[θ],
            θ, 0, thmax,
            r, 0, 1,
            PlotRange -> -2.25, 2.25, -0.5, 3.5,
            PerformanceGoal -> "Quality"
            ], thmax, 0.01, 2 Pi]


            Mathematica graphics






            share|improve this answer











            $endgroup$

















              5












              $begingroup$

              This is what you need:



              Manipulate[ParametricPlot[
              r (1 + 2 Sin[θ]) Cos[θ], Sin[θ],
              θ, 0, thmax,
              r, 0, 1,
              PlotRange -> -2.25, 2.25, -0.5, 3.5,
              PerformanceGoal -> "Quality"
              ], thmax, 0.01, 2 Pi]


              Mathematica graphics






              share|improve this answer











              $endgroup$















                5












                5








                5





                $begingroup$

                This is what you need:



                Manipulate[ParametricPlot[
                r (1 + 2 Sin[θ]) Cos[θ], Sin[θ],
                θ, 0, thmax,
                r, 0, 1,
                PlotRange -> -2.25, 2.25, -0.5, 3.5,
                PerformanceGoal -> "Quality"
                ], thmax, 0.01, 2 Pi]


                Mathematica graphics






                share|improve this answer











                $endgroup$



                This is what you need:



                Manipulate[ParametricPlot[
                r (1 + 2 Sin[θ]) Cos[θ], Sin[θ],
                θ, 0, thmax,
                r, 0, 1,
                PlotRange -> -2.25, 2.25, -0.5, 3.5,
                PerformanceGoal -> "Quality"
                ], thmax, 0.01, 2 Pi]


                Mathematica graphics







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago









                m_goldberg

                89k873200




                89k873200










                answered 2 days ago









                C. E.C. E.

                51.4k3101207




                51.4k3101207



























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