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Return to Part I of the course APMA0330

Glossary

Preface

This section addresses a buitiful application of Mathematica to plot figures with fillings. Therefore, this section presents numerous examples.

Plotting with filling

      We repeat the previously considered example for a piecewise linear function with filling:
Plot[2 - 2*x, {x, 0, 1}, FillingStyle -> Green, Filling -> Bottom]
        Piecewise linear function            Mathematica code

      Now we change the color of filling:
Plot[2 - 2*x, {x, 0, 1}, FillingStyle -> Green, Filling -> Bottom]
        Example of green color filling.            Mathematica code

      Now we change the color of filling:
Plot[{2, (x/3)}, {x, 0, 6}, Filling -> {1 -> {2}}, FillingStyle -> LightCyan]
        Example of light cyan color filling.            Mathematica code

      This specifies a specific filling to be used only for the first curve.
Plot[{Sin[2*x], Cos[3*x]}, {x,0,1}, Filling -> {1 -> 0.5}]
        Two curves with distinct fillings.            Mathematica code

     
Plot[{Sin[2*x], Cos[3*x]}, {x, 0, 1}, Filling -> Bottom]
        Region between sine and cosine functions.            Mathematica code

      Another color of filling:
Plot[{Sin[2*x], Cos[3*x]}, {x, 0, 1}, Filling -> {1 -> {2}}, FillingStyle -> LightGreen]
        Region between sine and cosine functions.            Mathematica code

      This code specifies a specific filling to be used only for the first curve.
Plot[{Sin[2*x], Cos[3*x]}, {x,0,1}, Filling -> {1 -> 0.5}]
        Only one part of the region is specified.            Mathematica code

     
Plot[{x, x^2}, {x, 0, 1}, Filling -> {1 -> {2}}, FillingStyle -> Pink]
        Region between two curves.            Mathematica code

      Now we change the color of filling:
Show[PolarPlot[{1, 2}, {t, 0, Pi/4}], RegionPlot[ x^2 + y^2 >= 1 && x^2 + y^2 <= 4 && y/x <= 1, {x, 0, 2}, {y, 0, 2}, ColorFunction -> "DarkRainbow"]]
        Example of PolarPlot.            Mathematica code

     
Plot[{-1, 2}, {x, 1, 3}, Filling -> {1 -> {2}}, FillingStyle -> Purple, AspectRatio -> Automatic, AxesOrigin -> {0, 0}]
        Rectangle.            Mathematica code

     
Plot[{3, (x/2) - 1}, {x, 0, 2}, Filling -> {1 -> {2}}, FillingStyle -> Yellow]
        Figure with filling.            Mathematica code

     
RegionPlot[{x^2 + y^2 <= 9}, {x, -3, 3}, {y, -3, 3}, PlotStyle -> LightBrown]
        Figure with filling.            Mathematica code

     
Show[PolarPlot[{1, 2}, {t, Pi/2, Pi}], RegionPlot[x^2 + y^2 >= 1 && x^2 + y^2 <= 4, {x, -2, 0}, {y, 0, 2}, ColorFunction -> "Rainbow"]]
        Part of a circle with filling.            Mathematica code

      To lift a traingle over the horizontal axis, type:
tr = Plot[{1., Max[1, Min[2 x + 1, 4 - x]]}, {x, 0, 3}, AspectRatio -> 1/2, Ticks -> {{0, 1, 2, 3}, {0, 1, 2, 3}},
Filling -> Axis, FillingStyle -> Blue, Axes -> True, AxesStyle -> Directive[Thick]]
a = Graphics[{{Opacity[0], White, tr[[1]]}, GeometricTransformation[{Opacity[1], Blue, tr[[1]]}, TranslationTransform[{0, 1}]]}]
Show[a, Axes -> True, AxesStyle -> Black, AspectRatio -> 0.75]
        Triangle is lifted over the axis.            Mathematica code

 

RegionPlot[Sin[x y] > 0, {x, -1, 1}, {y, -1, 1},
FrameTicksStyle -> Directive[FontOpacity -> 0, FontSize -> 0]]
When plotting, you still see frameticks data:
rp = RegionPlot[x^2 + y^3/4 < 2 && x + y < 1, {x, -2, 2}, {y, -2, 2}, FrameTicks -> Automatic]
First extract the frameticks information and change the labels to blank:
newticks = Last@First[AbsoluteOptions[rp, FrameTicks]];
While Mathematica complains about that Ticks: {Automatic,Automatic} is not a valid tick specification, it still does its job. next we type:
Show[rp, FrameTicks -> newticks]
You may also try
newticks = Last@First[AbsoluteOptions[rp, FrameTicks]];
newticks[[All, All, 2]] = "";
but Mathematica will complain again and out will be the same.

RegionPlot[1 < Abs[x + I y] < 2, {x, -2, 2}, {y, -2, 2}, ImagePadding -> 1]
      Another example:
Graphics3D[{Texture[img], EdgeForm[], Cylinder[{{0, 0, 0}, {0, 0, 2 Pi}}, 1]}, Boxed -> False]
        Two pieces of a circle.            Mathematica code

We plot half of the polygon:
poly = Polygon[Table[N[{Cos[n *Pi/6], Sin[n*Pi/6]}], {n, 0, 6}]]
Graphics[{RGBColor[0.3, 0.5, 1], EdgeForm[Thickness[0.01]], poly}]
Show[%, Frame -> True]

     
ill = Graphics[{LightRed, Polygon[Table[{0.99 Cos[theta], 0.99 Sin[theta]}, {theta, Pi + 0.3, 2 Pi - 0.25, 0.05}]]}];
circle = Graphics[{Black, Thick, Circle[{0, 0}, 1.0]}];
Show[circle, fill]
        Circle with inclusion.            Mathematica code

 

Parametric Plots with fillings

     
ParametricPlot[ Cos[2 theta] {Cos[theta], Sin[theta]} r, {theta, 0, 2 Pi}, {r, 0, 1}, Mesh -> False, PlotPoints -> {30, 2}]
        Four cusp curve r = cos(2 θ).            Mathematica code

     
g1 = PolarPlot[Cos[2 theta], {theta, Pi/4, 2 Pi - Pi/4}];
g2 = ParametricPlot[ Cos[2 theta] {Cos[theta], Sin[theta]} r, {theta, -Pi/4, Pi/4}, {r, 0, 1}, Mesh -> False];
Show[g1, g2, PlotRange -> All]
        Part of the curve r = cos(2 θ).            Mathematica 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[2 t], {t, 0, 2 Pi}, 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 2 t]], {-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]
        One cusp from the curve r = cos(2 θ).            Mathematica code

      Another version:
Show[{RegionPlot[(x^2 + y^2)^(1/2) <= Cos[2 ArcTan[y/x]], {x, 0, 1}, {y, -1/2, 1/2}], PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}, PlotStyle -> Red]}, PlotRange -> All]
or
Show[{RegionPlot[(x^2 + y^2)^(3/2) <= x^2 - y^2, {x, 0, 1}, {y, -1/2, 1/2}], PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}, PlotStyle -> Red]}, PlotRange -> All]
or
PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}, PlotStyle -> Red, Prolog -> RegionPlot[(x^2 + y^2)^(3/2) <= x^2 - y^2, {x, 0, 1}, {y, -1/2, 1/2}][[1]]]
or
s = PolarPlot[Cos[2 theta], {theta, 0, 2 Pi}];
s1 = PolarPlot[Cos[2 theta], {theta, -Pi/4, Pi/4}] /. Line -> Polygon;
Show[s, s1]
        One cusp from the curve r = cos(2 θ).            Mathematica code

 

Polar plots with fillings

One way to go around a problem to make plots with filling is to use ParametricPlot
     
parmplot = ParametricPlot[ r {Cos[t], Sin[t]}, {t, 0, Pi/2}, {r, 2 Pi + t, 4 Pi + t}, ColorFunction -> "RustTones"];
Show[ParametricPlot[t {Cos[t], Sin[t]}, {t, 0, 6 Pi}], parmplot]
        Archimede's spiral.            Mathematica code

     
cartreg = ImplicitRegion[ 2 \[Pi] < Sqrt[x^2 + y^2] - ArcTan[x, y] < 4 \[Pi] && 0 <= x <= 15 && 0 <= y <= 15, {x, y}];
regiontoplot = DiscretizeRegion[cartreg, AccuracyGoal -> 5];
NumberForm[#, {8, 5}] &@Area[regiontoplot] NumberForm[Integrate[1, {x, y} \[Element] regiontoplot], {8, 5}];
pt = RegionCentroid[regiontoplot];
Show[ParametricPlot[t {Cos[t], Sin[t]}, {t, 0, 6 Pi}], regiontoplot, Graphics[{PointSize[Large], Red, Point[pt]}]]
        Archimede's spiral with a dot.            Mathematica code

     
Show[PolarPlot[ Evaluate[{{1, -1} Sqrt[2 Cos[t]], 2 (1 - Cos[t])}], {t, -\[Pi], \[Pi]}], RegionPlot[ Sqrt[x^2 + y^2] > 2 (1 - Cos[ArcTan[x, y]]) && Sqrt[x^2 + y^2] < Re@Sqrt[2 Cos[ArcTan[x, y]]], {x, -2, 2}, {y, -3, 3}], PlotRange -> All]
        Shading between polar graphs.            Mathematica code

      Another plot:
Show[PolarPlot[{Sqrt[2 Abs[Cos[t]]], 2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}], RegionPlot[ 4 (1 - Cos[t])^2 < r^2 < 2 Cos[t], {r, 0, 3}, {t, -Pi, Pi}, PlotPoints -> 30] /. GraphicsComplex[a_, b__] :> GraphicsComplex[#1 {Cos[#2], Sin[#2]} & @@@ a, b]]
        Shading between polar graphs.            Mathematica code

      Using Cartesian coordinates:
eqns[t_] := {Sqrt[2 Cos[t]], 2 (1 - Cos[t])};
region = PolarPlot[Evaluate@eqns[t], {t, -\[Pi], \[Pi]}, RegionFunction -> Function[{x, y, t, r}, {#1 > #2} & @@ Re[eqns[t]] // First]];
pts = Cases[region, Line[x___] :> x, Infinity];
colors = {Darker@Green, Blue};
Show[PolarPlot[Evaluate@eqns[t], {t, -\[Pi], \[Pi]}, PlotStyle -> colors], ListLinePlot[pts, PlotStyle -> colors, Filling -> Axis, FillingStyle -> Purple], PlotRange -> All]
        Shading between polar graphs.            Mathematica code

      You can parameterize your polar functions on to discs, and then shade appropriately.
\[Rho][t_] := Sqrt[2 Cos[t]];
\[Sigma][t_] := 2 (1 - Cos[t]);
ParametricPlot[{{r Cos[t] \[Rho][t], r Sin[t] \[Rho][t]}, {r Cos[t] \[Sigma][t], r Sin[t] \[Sigma][t]}}, {t, -\[Pi], \[Pi]}, {r, 0, 1}, PlotStyle -> {{Opacity[.5], Red}, {Opacity[1], White}}, Mesh -> None, PlotRange -> All]
        Shading between polar graphs.            Mathematica code

      Another version:
dt = Pi/99;
pts = Join[ Table[2 (1 - Cos[t]) {Cos[t], Sin[t]}, {t, 0, -Pi/3 + dt, -dt}], Table[Sqrt[2 Cos[t]] {Cos[t], Sin[t]}, {t, -Pi/3, Pi/3, dt}], Table[2 (1 - Cos[t]) {Cos[t], Sin[t]}, {t, Pi/3, dt, -dt}]];
PolarPlot[{Sqrt[2 Cos[t]], 2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}, Prolog -> {Gray, Polygon[pts]}, PlotStyle -> Thick]
        Shading between polar graphs with contours.            Mathematica code

 

Venn Diagrams

Filling circles can be plotted using Graphics cammand. Graphics are represented as symbolic expressions, using
either"directives" or "styles":

Graphics[{Blue, Disk[{0, 0}], Opacity[0.7], Pink, Disk[{1, 0}]}]

Graphics[{Style[Disk[{0, 0}], Green], Opacity[0.5], Pink, Disk[{1, 0}]}]

 

We can plot Venn diagrams using the following subroutine

 

VennDiagram2[n_, ineqs_: {}] :=
Module[{i, r = .6, R = 1, v, grouprules, x, y, x1, x2, y1, y2, ve},
v = Table[Circle[r {Cos[#], Sin[#]} &[2 Pi (i - 1)/n], R], {i, n}];
{x1, x2} = {Min[#], Max[#]} &[ Flatten@Replace[v, Circle[{xx_, yy_}, rr_] :> {xx - rr, xx + rr}, {1}]];
{y1, y2} = {Min[#], Max[#]} &[ Flatten@Replace[v, Circle[{xx_, yy_}, rr_] :> {yy - rr, yy + rr}, {1}]];
ve[x_, y_, i_] := v[[i]] /. Circle[{xx_, yy_}, rr_] :> (x - xx)^2 + (y - yy)^2 < rr^2;
grouprules[x_, y_] = ineqs /. Table[With[{is = i}, Subscript[_, is] :> ve[x, y, is]], {i, n}];
Show[If[MatchQ[ineqs, {} | False], {}, RegionPlot[grouprules[x, y], {x, x1, x2}, {y, y1, y2}, Axes -> False]],
Graphics[v], PlotLabel -> TraditionalForm[Replace[ineqs, {} | False -> \[EmptySet]]], Frame -> False]]

Then we plot two Venn diagrams:

a12 = VennDiagram2[2, Subscript[A, 1] && Subscript[A, 2]]
a1 = Graphics[Text[dogs, {-0.9, 0}]]
b1 = Graphics[Text[brown, {0.9, 0}]]
Show[a12, a1, b1]

or

a32 = VennDiagram2[2, Not[Subscript[A, 1]] && Subscript[A, 2]]
a33 = VennDiagram2[2, Not[Not[Subscript[A, 1]] && Subscript[A, 2]]]

   

 

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