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Re: Shading in polar plot

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107879] Re: [mg107857] Shading in polar plot
  • From: "David Park" <djmpark at comcast.net>
  • Date: Mon, 1 Mar 2010 08:05:40 -0500 (EST)
  • References: <19509473.1267436979165.JavaMail.root@n11>

PolarPlot doesn't have a Filling option. But the plot is not too difficult
with the Presentations package. To obtain the fills, we first draw r versus
t between the limits where the lemniscates intersects the circle, which can
be solved for. We can use filling there. Then we transform the result from
the t-r plane to the x-y plane using DrawingTransform. We do this for each
side, combine with a PolarDraw and we have your plot.

Needs["Presentations`Master`"]

trfill = Draw[{Sqrt[6 Cos[2 t]], Sqrt[3]}, {t, -\[Pi]/6, \[Pi]/6},
   Filling -> {1 -> {2}},
   FillingStyle -> LightBlue];
Draw2D[
 {PolarDraw[{Sqrt[3], Sqrt[6 Cos[2 t]], -Sqrt[6 Cos[2 t]]}, {t, 0, 2 Pi}],
  trfill /. DrawingTransform[#2 Cos[#1] &, #2 Sin[#1] &],
  trfill /. DrawingTransform[-#2 Cos[#1] &, #2 Sin[#1] &]},
 AspectRatio -> Automatic,
 Frame -> True,
 ImageSize -> 300]  


David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/  


From: Jon Joseph [mailto:josco.jon at gmail.com] 

All: I would like to shade the area outside of the circle Sqrt[3] and 
inside the lemniscate r^2=6Cos[2 t]. Can this be done using PolarPlot 
and mesh functions or is something else needed?

PolarPlot[{Sqrt[3], Sqrt[6 Cos[2 t]], -Sqrt[6 Cos[2 t]]}, {t, 0, 2 Pi}]

Shows the curves without the shading. Jon




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