Draw a Filled-in Circle Mathematica

Preface

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

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Plotting with filling

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

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

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

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

		Plot[{Sin[2*10], Cos[iii*x]}, {x, 0, 1}, Filling -> Bottom]
Region between sine and cosine functions.		Mathematica lawmaking

		Some other color of filling: Plot[{Sin[2*x], Cos[3*10]}, {x, 0, 1}, Filling -> {ane -> {2}}, FillingStyle -> LightGreen]
Region betwixt sine and cosine functions.		Mathematica code

		This code specifies a specific filling to be used just for the first bend. Plot[{Sin[2*ten], Cos[3*ten]}, {x,0,1}, Filling -> {1 -> 0.5}]
Simply i part of the region is specified.		Mathematica lawmaking

		Plot[{10, x^ii}, {10, 0, ane}, Filling -> {one -> {two}}, FillingStyle -> Pinkish]
Region betwixt 2 curves.		Mathematica lawmaking

		At present we modify the colour of filling: Bear witness[PolarPlot[{i, 2}, {t, 0, Pi/four}], RegionPlot[ 10^2 + y^2 >= 1 && ten^2 + y^2 <= 4 && y/ten <= 1, {10, 0, 2}, {y, 0, 2}, ColorFunction -> "DarkRainbow"]]
Example of PolarPlot.		Mathematica code

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

		Plot[{three, (ten/ii) - one}, {x, 0, 2}, Filling -> {one -> {2}}, FillingStyle -> Yellow]
Figure with filling.		Mathematica code

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

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

		To lift a traingle over the horizontal axis, type: tr = Plot[{1., Max[one, Min[two ten + 1, 4 - x]]}, {ten, 0, 3}, AspectRatio -> i/2, Ticks -> {{0, 1, 2, 3}, {0, 1, ii, 3}}, Filling -> Axis, FillingStyle -> Blue, Axes -> True, AxesStyle -> Directive[Thick]] a = Graphics[{{Opacity[0], White, tr[[1]]}, GeometricTransformation[{Opacity[i], Bluish, tr[[i]]}, TranslationTransform[{0, ane}]]}] Show[a, Axes -> True, AxesStyle -> Black, AspectRatio -> 0.75]
Triangle is lifted over the centrality.		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^ii + y^three/four < 2 && x + y < i, {x, -2, 2}, {y, -2, 2}, FrameTicks -> Automated]

First extract the frameticks information and change the labels to blank:

newticks = Final@Commencement[AbsoluteOptions[rp, FrameTicks]];

While Mathematica complains nigh that Ticks: {Automatic,Automatic} is not a valid tick specification, information technology notwithstanding does its job. next nosotros type:

Evidence[rp, FrameTicks -> newticks]

You may as well try

newticks = Last@Start[AbsoluteOptions[rp, FrameTicks]];
newticks[[All, All, 2]] = "";

but Mathematica will complain again and out will be the same.

RegionPlot[1 < Abs[ten + I y] < 2, {x, -2, 2}, {y, -2, 2}, ImagePadding -> 1]

		Another example: Graphics3D[{Texture[img], EdgeForm[], Cylinder[{{0, 0, 0}, {0, 0, ii Pi}}, one]}, Boxed -> False]
Two pieces of a circumvolve.		Mathematica lawmaking

We plot half of the polygon:

poly = Polygon[Table[North[{Cos[due north *Pi/half-dozen], Sin[n*Pi/half dozen]}], {northward, 0, 6}]]
Graphics[{RGBColor[0.3, 0.5, ane], EdgeForm[Thickness[0.01]], poly}]

Evidence[%, Frame -> True]

Parametric Plots with fillings

---

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

		g1 = PolarPlot[Cos[two theta], {theta, Pi/four, 2 Pi - Pi/iv}]; g2 = ParametricPlot[ Cos[two theta] {Cos[theta], Sin[theta]} r, {theta, -Pi/4, Pi/4}, {r, 0, one}, Mesh -> False]; Show[g1, g2, PlotRange -> All]
Function of the curve r = cos(2 θ).		Mathematica code

		txt[t_, {x_, y_}] := Fashion[Text[t, {10, y}], FontSize -> 30, FontWeight -> Assuming] {xmin, xmax} = {-ane.425, i.425}; {ymin, ymax} = {-1.25, ane.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^ii)^(3/ii) <= ten^2 - y^2, {10, -0.02, one}, {y, -i, i}, PlotStyle -> ColorData["HTML", "Golden"], BoundaryStyle -> Directive[Thickness[0.025], ColorData["Legacy", "CadmiumOrange"]], Frame -> False, AspectRatio -> Automated, 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, Pointer[{{xmin, 0}, {xmax, 0}}], Arrow[{{0, ymin}, {0, ymax}}], txt[TraditionalForm[HoldForm[r == cos 2 t]], {-0.vi, one.0}], txt[TraditionalForm[HoldForm[t == Pi/iv]], {1.125, 0.925}], txt[TraditionalForm[HoldForm[t == -Pi/4]], {ane.125, -0.99}]}, ImageSize -> seven*72]
One cusp from the curve r = cos(two θ).		Mathematica code

		Another version: Show[{RegionPlot[(x^2 + y^2)^(1/ii) <= Cos[2 ArcTan[y/x]], {ten, 0, 1}, {y, -i/2, 1/ii}], PolarPlot[Cos[2 theta], {theta, 0, ii Pi}, PlotStyle -> Red]}, PlotRange -> All] or Show[{RegionPlot[(x^2 + y^ii)^(3/2) <= x^2 - y^2, {x, 0, 1}, {y, -1/ii, i/2}], PolarPlot[Cos[ii theta], {theta, 0, 2 Pi}, PlotStyle -> Red]}, PlotRange -> All] or PolarPlot[Cos[ii theta], {theta, 0, 2 Pi}, PlotStyle -> Red, Prolog -> RegionPlot[(ten^ii + y^ii)^(3/2) <= x^ii - y^2, {10, 0, ane}, {y, -1/2, 1/2}][[i]]] or s = PolarPlot[Cos[2 theta], {theta, 0, two Pi}]; s1 = PolarPlot[Cos[two theta], {theta, -Pi/4, Pi/four}] /. Line -> Polygon; Evidence[s, s1]
1 cusp from the curve r = cos(ii θ).		Mathematica code

Polar plots with fillings

---

I fashion 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 -> five]; NumberForm[#, {8, v}] &@Area[regiontoplot] NumberForm[Integrate[1, {10, y} \[Element] regiontoplot], {8, v}]; pt = RegionCentroid[regiontoplot]; Show[ParametricPlot[t {Cos[t], Sin[t]}, {t, 0, 6 Pi}], regiontoplot, Graphics[{PointSize[Large], Red, Indicate[pt]}]]
Archimede'due south spiral with a dot.		Mathematica lawmaking

		Show[PolarPlot[ Evaluate[{{1, -ane} Sqrt[ii Cos[t]], 2 (one - Cos[t])}], {t, -\[Pi], \[Pi]}], RegionPlot[ Sqrt[x^2 + y^ii] > 2 (i - Cos[ArcTan[x, y]]) && Sqrt[ten^two + y^ii] < Re@Sqrt[2 Cos[ArcTan[10, y]]], {x, -ii, 2}, {y, -3, iii}], 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[ four (ane - Cos[t])^2 < r^ii < 2 Cos[t], {r, 0, three}, {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 -> Role[{x, y, t, r}, {#ane > #ii} & @@ Re[eqns[t]] // Get-go]]; pts = Cases[region, Line[x___] :> x, Infinity]; colors = {Darker@Green, Bluish}; 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

		Yous can parameterize your polar functions on to discs, and then shade appropriately. \[Rho][t_] := Sqrt[2 Cos[t]]; \[Sigma][t_] := ii (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, i}, PlotStyle -> {{Opacity[.5], Reddish}, {Opacity[i], White}}, Mesh -> None, PlotRange -> All]
Shading between polar graphs.		Mathematica lawmaking

		Some other version: dt = Pi/99; pts = Join[ Tabular array[2 (ane - Cos[t]) {Cos[t], Sin[t]}, {t, 0, -Pi/iii + dt, -dt}], Table[Sqrt[2 Cos[t]] {Cos[t], Sin[t]}, {t, -Pi/3, Pi/3, dt}], Table[ii (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 betwixt polar graphs with contours.		Mathematica code

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

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

Graphics[{Mode[Disk[{0, 0}], Greenish], Opacity[0.5], Pink, Deejay[{1, 0}]}]

We tin plot Venn diagrams using the following subroutine

VennDiagram2[n_, ineqs_: {}] :=
Module[{i, r = .6, R = ane, five, grouprules, x, y, x1, x2, y1, y2, ve},
five = Table[Circle[r {Cos[#], Sin[#]} &[ii Pi (i - one)/due north], R], {i, north}];
{x1, x2} = {Min[#], Max[#]} &[ Flatten@Replace[five, Circumvolve[{xx_, yy_}, rr_] :> {20 - 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)^ii + (y - yy)^two < rr^2;
grouprules[x_, y_] = ineqs /. Tabular array[With[{is = i}, Subscript[_, is] :> ve[ten, y, is]], {i, n}];
Show[If[MatchQ[ineqs, {} | False], {}, RegionPlot[grouprules[x, y], {ten, x1, x2}, {y, y1, y2}, Axes -> False]],
Graphics[v], PlotLabel -> TraditionalForm[Supplant[ineqs, {} | Imitation -> \[EmptySet]]], Frame -> Faux]]

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}]]
Evidence[a12, a1, b1]

or

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

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Source: https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/ch1/venn.html

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