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When I try to plot this simple parametric region:

ParametricRegion[{x, Sin[z + t] + Sin[t - z]}, {{x, 0, 1}, {z, 0, 1}, {t, 0, 1}}];
RegionPlot[%]

I get errors telling me the "parametric region cannot be automatically discretized" and "is not a valid region to plot".

How can I plot this parametric region ?

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  • $\begingroup$ Is that an $x$ or a $z$? If you change $x$ to $z$, you get output. Also, please add a ] at the end of the first line. $\endgroup$
    – mjw
    Commented Mar 15, 2019 at 17:01
  • $\begingroup$ @mjw Thanks. Actually, it is an x not a z. $\endgroup$
    – Pon
    Commented Mar 15, 2019 at 17:23
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    $\begingroup$ Please describe in mathematical terms the region you want to plot. Is it a 2D or 3D region? $\endgroup$
    – mjw
    Commented Mar 15, 2019 at 17:40
  • $\begingroup$ It is a 2D region defined as the set of points of coordinates ( x , sin(t+z)+sin(t-z) ) when x, t and z run over the interval [0,1] $\endgroup$
    – Pon
    Commented Mar 15, 2019 at 23:21
  • $\begingroup$ If it is a 2D region, what are the abscissa and ordinate variables? It could only be two of these three, right?: $\{t,x,z\}$ $\endgroup$
    – mjw
    Commented Mar 17, 2019 at 2:30

1 Answer 1

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Let's not worry about $x$ for now. For a few values of $z$, we plot the curve representing $y$: 

 p = 3; Plot[Evaluate@Table[Sin[z - t] - Sin[z + t], {z, Range[0,p]/p}], {t, 0, 1}]

enter image description here

For various values of $z$, we can now plot the regions in $xy$-space.

 p = 3; 
 ParametricPlot[
   Evaluate@
     Table[{x, Sin[z - t] - Sin[z + t]}, {z, Range[0,p]/p}], {x, 0,1}, {t, 0, 1}]

enter image description here

The union of all of the regions is the region with $z=0$:

 ParametricPlot[{x, -2 Sin[t]}, {x, 0, 1}, {t, 0, 1}]

enter image description here

This is a rectangle of width 1 and height $2 \sin 1 \approx 1.68294$.

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  • $\begingroup$ Thanks for your answer. I understand your steps but this is not exactly what RegionPlot[ParametricRegion[{x, Sin[z + t] + Sin[t - z]}, {{x, 0, 1}, {z, 0, 1}, {t, 0, 1}}]] is supposed to do in one shot ? $\endgroup$
    – Pon
    Commented Apr 18, 2019 at 15:14
  • $\begingroup$ What do you think the resulting plot will look like, assuming we can use RegionPlot[ParametricPlot[]]? $\endgroup$
    – mjw
    Commented Apr 18, 2019 at 16:00
  • $\begingroup$ Not with ParametricPlot[] but with ParametricRegion[] . As far as I know, since Mathematica 10 we can define a region using ParametricRegion[] and then plot it with RegionPlot[]. For me, RegionPlot[ParametricRegion[]] is supposed to do the job, and I should get the same plot as you. For instance, if I try RegionPlot[ParametricRegion[{x, Sin[z + t]}, {{x, 0, 1}, {z, 0, 1}, {t, 0, 1}}]], it works perfectly in one shot. What I do not understand is why when I add a second term for the ordinate, namely Sin[t-z], it does not work anymore. $\endgroup$
    – Pon
    Commented Apr 19, 2019 at 15:41
  • $\begingroup$ You are right. That breaks it. ParametricRegion[] can handle it, but then when it is processed by RegionPlot[] it cannot handle it. By the way: $\sin(z-t)-\sin(z+t)=-2\cos z \sin t.$ It seems that the $z$ parameterization is redundant. I don't know if this has anything to do with Mathematica's inability to plot the sum (or product). $\endgroup$
    – mjw
    Commented Apr 19, 2019 at 17:06

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