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Dynamic filling of a region of a polar plot

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]]

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

edited Apr 20 at 1:56

Okkes Dulgerci

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

$endgroup$

add a comment | 

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]]

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

edited Apr 20 at 1:56

Okkes Dulgerci

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

$endgroup$

add a comment | 

$begingroup$

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]]

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

edited Apr 20 at 1:56

Okkes Dulgerci

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

$endgroup$

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]]

share|improve this question

edited Apr 20 at 1:56

Okkes Dulgerci

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

share|improve this question

edited Apr 20 at 1:56

Okkes Dulgerci

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

5,5261919

asked Apr 19 at 23:46

Okkes DulgerciOkkes Dulgerci

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]

share|improve this answer

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

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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]

share|improve this answer

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

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]

share|improve this answer

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

$begingroup$

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]

share|improve this answer

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

$endgroup$

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]

share|improve this answer

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

edited Apr 20 at 2:59

m_goldberg

89.1k873200

edited Apr 20 at 2:59

m_goldberg

89.1k873200

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207

answered Apr 20 at 2:38

C. E.C. E.

51.4k3101207