text: # Parametric region cannnot be automatically discretized/not a valid region to plot
  
  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 ?
  
  • 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.
  – mjw
  Commented Mar 15, 2019 at 17:01
  • @mjw Thanks. Actually, it is an x not a z.
  – Pon
  Commented Mar 15, 2019 at 17:23
  • Please describe in mathematical terms the region you want to plot. Is it a 2D or 3D region?
  – mjw
  Commented Mar 15, 2019 at 17:40
  • 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]
  – Pon
  Commented Mar 15, 2019 at 23:21
  • If it is a 2D region, what are the abscissa and ordinate variables? It could only be two of these three, right?: $\{t,x,z\}$
  – mjw
  Commented Mar 17, 2019 at 2:30
  
  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}]
  
  
  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}]
  
  
  The union of all of the regions is the region with $$z=0$$:
  
   ParametricPlot[{x, -2 Sin[t]}, {x, 0, 1}, {t, 0, 1}]
  
  
  This is a rectangle of width 1 and height $$2 \sin 1 \approx 1.68294$$.
  
  • 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 ?
  – Pon
  Commented Apr 18, 2019 at 15:14
  • What do you think the resulting plot will look like, assuming we can use RegionPlot[ParametricPlot[]]?
  – mjw
  Commented Apr 18, 2019 at 16:00
  • 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.
  – Pon
  Commented Apr 19, 2019 at 15:41
  • 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).
  – mjw
  Commented Apr 19, 2019 at 17:06