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For a mapping $\gamma(x,y)$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ we can plot the image of some two dimensional region with ParametricPLot as, for instance,

\[Gamma][x_, y_] := {x^2, x*y^3}
 ParametricPlot[\[Gamma][x, y], Element[{x, y}, Disk[]], PlotRange -> All]

being the domain, say, a disk. But if I want to plot just the image of the boundary as

ParametricPlot[\[Gamma][x, y], Element[{x, y}, Circle[]], PlotRange -> All]

then ParametricPlot has troubles understanding the dimensionality and the number of variables. Then, how can one make a parametric plot, being the domain some curve, as in the previous example ?

Note: I do not want to parametrize the circle as $(\cos\theta,\sin\theta)$ and then replace back to get $\gamma(\theta)$, because this may not be possible for more complicated curves without closed, analytical formula.

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  • $\begingroup$ ParametricPlot[\[Gamma][x, y], Element[{x, y}, Disk[]], PlotRange -> All, BoundaryStyle -> Red, PlotStyle -> None] $\endgroup$
    – cvgmt
    Commented Sep 20, 2022 at 14:17
  • $\begingroup$ @cvgmt True. But my question comes from the fact that the function is some numerical, slowly running bunch of things and running for the whole interior, which I don't need, is unnecessary, so just want to compute the values of the boundary. $\endgroup$ Commented Sep 20, 2022 at 14:19
  • $\begingroup$ reg = ParametricRegion[{\[Gamma][x, y], {x, y} \[Element] Circle[]}, {x, y}]; Region[Style[reg, Red]] $\endgroup$
    – cvgmt
    Commented Sep 20, 2022 at 14:42
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    $\begingroup$ @cvgmt - what version are you using? With v13.1 on a Mac, I don't get a plot $\endgroup$
    – Bob Hanlon
    Commented Sep 20, 2022 at 15:17
  • $\begingroup$ @BobHanlon I am using v13.1 on Win 11 and v13.1 on Manjaro Linux. $\endgroup$
    – cvgmt
    Commented Sep 20, 2022 at 22:39

2 Answers 2

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  • draw a circle and mapping it by γ[x_, y_] = {x^2, x*y^3}
Clear[circle,γ];
circle= RegionPlot[DiscretizeRegion[Circle[], AccuracyGoal -> 3]];
γ[x_, y_] = {x^2, x*y^3};
circle /. {x_Real, y_Real} -> γ[x, y]

enter image description here

  • Use ParametricRegion as in comment.
reg = ParametricRegion[{γ[x, y], {x, y} ∈ 
     Circle[]}, {x, y}];
RegionPlot[DiscretizeRegion[reg, MaxCellMeasure -> .0001], 
 AspectRatio -> Automatic]

enter image description here

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You may first generate a parametrized region. From the result you then calculate the boundary using "RegionBoundary". Here is your example:

reg = ParametricRegion[{\[Gamma][x, y], {x, y} \[Element] Disk[]}, {x,
    y}]
Region[Style[reg, Red]] (*show the region*)
Region[RegionBoundary[reg]] (* show the border *)

enter image description here

If you want to increase the performance, you may only calculate a small strip. The drawback is, that the border is then 2 sided:

reg = ParametricRegion[{\[Gamma][x, y], {x, y} \[Element] 
     RegionDifference[Disk[{0, 0}, 1], Disk[{0, 0}, 0.998]] }, {x, y}];
Region[Style[reg, Red]] (*show the region*)
Region[RegionBoundary[reg]] (* show the border *)

enter image description here

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  • $\begingroup$ Thanks. But my question comes from the fact that the function is some numerical, slowly running bunch of things and running for the whole interior, which I don't need, is unnecessary, so I just want to compute the values of the boundary $\endgroup$ Commented Sep 20, 2022 at 18:34
  • $\begingroup$ Look at my answer again. I added way to decrease the amount of calculation to my answer. $\endgroup$ Commented Sep 20, 2022 at 19:01
  • $\begingroup$ Nice. That's an ingenious solution. $\endgroup$ Commented Sep 20, 2022 at 19:20

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