Shading between polar graphs

Question

Since PolarPlot doesn't support Filling, what is the best way to shade or fill the a region between two polar curves?

For instance, how would I generate a version of the following graph with the region inside the first curve but outside the second curve filled?

PolarPlot[{{1, -1} Sqrt[2 Cos[t]], 2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}]

Answer 1

You have a (or more) curves. If you don't use PolarPlot you could use ParametricPlot instead but you would have to make the transformation from polar coordinates by yourself.

Knowing this, you could think about what your functions mean. For instance 2 (1 - Cos[phi]) is just the radius of your curve for a given phi. If you want to draw the region outside your curve, the only thing you have to do is (attention, I'm mixing polar and Cartesian coord.):

Check a every point $\{x,y\}$ whether the radius $\sqrt{x^2+y^2}$ is larger than $2(1-\cos(\varphi))$ where $\varphi=\arctan(y/x)$.

Using this, your filling can be achieved with RegionPlot and your graphics

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
 ]

If you encounter dark mesh lines in the filling and want to get rid of them, please read the question of david here. You then have to include

Method -> {"TransparentPolygonMesh" -> True}

as option.

Answer 2

Even if Filling were an option in PolarPlot, you won't be able to create such a plot because for 2D graphics, Filling just blindly fills along the y-axis, whereas you need to check for an inequality here.

That said, here's another approach that's in the same spirit as halirutan's, but you don't have to convert to Cartesian, etc.

eqns[t_] := { Sqrt[2 Cos[t]], 2 (1 - Cos[t])};
region = PolarPlot[Evaluate@eqns[t], {t, -π, π}, 
    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, -π, π}, PlotStyle -> colors], 
    ListLinePlot[pts, PlotStyle -> colors, Filling -> Axis, FillingStyle -> LightGreen], 
    PlotRange -> All
]

Answer 3

Just another way:

<< VectorAnalysis`;
{rho, t, z} = CoordinatesFromCartesian[{x, y, z}, Cylindrical]
Quiet@Show[
  PolarPlot[{ Sqrt[2 Cos@t], 2 (1 - Cos@t)}, {t, -Pi, Pi}],
  RegionPlot[ Sqrt[2 Cos@t] > rho > 2 (1 - Cos@t), {x, 0, 2}, {y, -1, 1}]]

Answer 4

Yet another way. It's similar to belisarius's solutions but doesn't require an inverse coordinate transformation

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]]

Answer 5

You can parameterize your polar functions on to discs, and then shade appropriately.

ρ[t_] := Sqrt[2 Cos[t]];
σ[t_] := 2 (1 - Cos[t]);
ParametricPlot[{{r Cos[t] ρ[t], r Sin[t] ρ[t]}, {r Cos[t] σ[t], r Sin[t] σ[t]}}, 
    {t, -π, π}, {r, 0, 1}, PlotStyle -> {{Opacity[.5], Red}, {Opacity[1], White}}, 
    Mesh -> None, PlotRange -> All]

Answer 6

How about

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 -> {LightGray, Polygon[pts]},
 PlotStyle -> Thick]

Answer 7

Here's a meager attempt that will soon be humiliated by Heike's answer. ;-)

g = PolarPlot[Evaluate@{{1, -1} Sqrt[2 Cos[t]], 
    2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}];

Graphics[{
   {Pink, Rectangle[Scaled[{0, 0}], Scaled[{1, 1}]]},
   Thread@{{White, Green, Blue}, 
     Reverse@Cases[g, Line[x__] :> Polygon[x], ∞]}
   }];

Show[g, %, g]

---

Seeing as I misunderstood the question, here is an admittedly fragile way to shade the correct region:

g = PolarPlot[Evaluate@{{1, -1} Sqrt[2 Cos[t]], 2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}];

f = Graphics[Thread@{{Yellow, White, White}, Cases[g, Line[x__] :> Polygon[x], ∞]}];

Show[g, f, g]

Answer 8

Combining ListLinePlot, Filling and Overlay:

   pnts = Cases[
      PolarPlot[{{1, -1} Sqrt[2 Cos[t]], 2 (1 - Cos[t])}, {t, -\[Pi], \[Pi]}],
     _Line, {0, Infinity}];
   Table[ListLinePlot[{Join[pnts[[1, 1]], pnts[[2, 1]]], pnts[[3, 1]]}, 
     Filling -> fllng, PlotStyle -> {Red, Green},  AspectRatio -> 1], 
   {fllng, {Automatic, {1 -> {Axis, White}}, {2 -> {Axis, White}}, {1 -> {Axis, Red}},
    {2 -> {Axis, Green}}}}]

gives

With Overlay and

  GraphicsRow[{Overlay[%[[{4, 3}]]], Overlay[%[[{5, 2}]]]}]

we get

Answer 9

I've used an general and light method, using polygons:

regionpoly[cv1_,cv2_,a_,b_,n_: 30,op_:0.5,col_: Green] := Module[{polys, ds},
       ds = N[(b - a)/n];
       polys = Table[Polygon[{
           cv1 /. {t -> dt}, 
           cv1 /. {t -> dt + ds}, 
           cv2 /. {t -> dt + ds},
           cv2 /. {t -> dt}, 
           cv1 /. {t -> dt} }],
           {dt, a, b - ds, ds}];
       (*Return*)
       {EdgeForm[], FaceForm[col], Opacity[op], polys}
       ];

    r1[t_] = Sqrt[2 Cos[t]]*{Cos[t], Sin[t]};
    r2[t_] = 2 (1 - Cos[t])*{Cos[t], Sin[t]};

    Show[{
      Graphics[  regionpoly[r1[t], r2[t], -Pi/3, Pi/3] ],
      PolarPlot[{Sqrt[2 Cos[t]], 2 (1 - Cos[t])}, {t, 0, 2 Pi}]                     
      }, Axes -> True, AxesStyle -> Arrowheads[0.04]]

Answer 10

Show[PolarPlot[{-Sqrt[2 Cos[t]],Sqrt[2 Cos[t]],2(1-Cos[t])},{t,-\[Pi],\[Pi]}],ParametricPlot[r {Cos[t],Sin[t]},{t,-Pi/3,Pi/3},{r,Min[a={Sqrt[2 Cos[t]],2(1-Cos[t])}],Max[a]}],PlotRange->All]