I am trying to integrate a figure composed of a piecewise parametric curve with both ends bound by a straight, vertical line segment. I have displayed the curves on top of each other using the Show function because I could not display the straight line using the ParametricPlot, code shown:


enter image description here

However, now that I am trying to find the area within, I cannot figure out how to accomplish it. I have tried creating a Region, but the issue of one curve being Cartesian and the other Parametric seems to prevent mixing them successfully, despite Region accomodating multiple elements. For the same reason, I cannot use a ParametricRegion. What methods would facilitate creating a closed area that would allow me to then take the area, whether using the Area function on a type of Region or by taking an integral? Do I instead need to apply RegionUnion or rewrite my straight line in a very ugly parametric form, and figure out how to include it?

Example of incorrect attempt:Region[{InsideLoopA}, {Line[{{ 1.01 , 0.26},{1, 0.19 }}]}]


1 Answer 1


I think you're going about this in such a way as to make this harder than it is. Polygon represents a closed curve where the final point will connect to the first by default so we can just use that here. No need to add a line or anything.

The way this will work is as such, first get a tabular set of your noisy data. I'll do this in a slow way (with lots of RandomReal calls) but if possible vectorize your generating function to make it fast.

noisyFunction[t_] :=
 RandomReal[{.9, 1}]*{Cos[2*Pi*t], Sin[2*Pi*t]}

dats = BlockRandom[Table[noisyFunction[t], {t, 0, .5, .01}]];


enter image description here

We can see that this data was properly ordered from the get-go, but that's not a given, so you can use FindShortestTour to force an ordering if you need to. Then simply use Polygon to wrap this data and it can become a Region:

orderedDats = dats[[FindShortestTour[dats][[2]]]];

reg = Region[Polygon[orderedDats]]

enter image description here

At this point all the built-in stuff like Area will work:


  • $\begingroup$ I'm not sure I understand your solution. I am using an existing parametric piecewise equation; it is a set of stable functions resulting in a curve with an open side and not a set of data . I am trying to close it. While I think that Polygon looks like something that will fix the issue, I am not sure how to implement it. Did I misunderstand your answer? $\endgroup$
    – Mark
    Jan 13, 2020 at 4:02
  • $\begingroup$ @Mark I know you have a function so I showed you how to convert it to a set of points to be used with Polygon in the very first step. The InsideLoopA can be converted to a function or used directly as an argument to Table. Maybe it would have been clearer if I had written InsideLoopA:=RandomReal[{.9, 1}]*{Cos[2*Pi*t], Sin[2*Pi*t]} so that I could use dats = Table[InsideLoopA, {t, 0, .5, .01} $\endgroup$
    – b3m2a1
    Jan 13, 2020 at 4:03

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