# Creating a domain for NDSolveValue via ParametricRegion

A circular arc $$R_2$$ can be defined parametrically as $$R_2 = \langle x(s),y(s) \rangle : s\in[-s_0,s_0]$$ (see code below for specific $$x,y$$ definitions) where $$s_0$$ is given (I must make the arc this way, for reasons that are beyond this question).

The total domain I want to make is a box with a circular arc as the top lid instead of a flat lid. The base of the box is $$R_1 = \langle x(s),-2 \rangle : s\in[-s_0,s_0]$$. I plot them both so you can see the shape, but does anyone know how to fill in the total domain from $$R_1$$ to $$R_2$$?

Ultimately, I will create a mesh on this box-ish region to solve a PDE, if that's helpful.

\[Alpha] = \[Pi]/3;
s0[\[Alpha]_] = ArcSin[Cos[\[Alpha]]]/Cos[\[Alpha]];
x[s_, \[Alpha]_] = Sin[Cos[\[Alpha]] s]/Cos[\[Alpha]];
y[s_, \[Alpha]_] = (1 - Cos[Cos[\[Alpha]] s])/Cos[\[Alpha]];
\[ScriptCapitalR]1 =
ParametricRegion[{x[s, \[Alpha]], -2}, {{s, -s0[\[Alpha]],
s0[\[Alpha]]}}];
\[ScriptCapitalR]2 =
ParametricRegion[{x[s, \[Alpha]],
y[s, \[Alpha]]}, {{s, -s0[\[Alpha]], s0[\[Alpha]]}}];
RegionPlot[{\[ScriptCapitalR]1, \[ScriptCapitalR]2}]


Edit: so I used the approach below by Erlich Neumann, but there is still a major issue: Mathematica is not registering the roof of the parametric region as a true circular arc. It is evident the parametric form $$\langle x(s),y(s) \rangle$$ is a piece of the circle $$x^2+(y-r)^2=r^2$$. However, this is not the case, as the code below shows the circle and also the domain created through ParametricRegion, which are not coincident.

\[Alpha] = \[Pi]/3;
r = 1/Cos[\[Alpha]];
s0 = ArcSin[Cos[\[Alpha]]]/Cos[\[Alpha]];
x[s_] := Sin[Cos[\[Alpha]] s]/Cos[\[Alpha]];
y[s_] := (1 - Cos[Cos[\[Alpha]] s])/Cos[\[Alpha]];
Clear[h]
h = Solve[Integrate[(y[s] - h) x'[s], {s, 0, s0}] == 1, h][[All, 1,
2]][];
\[CapitalOmega] =
ParametricRegion[{x[s], z}, {{s, -s0, s0}, {z, h, y[s]}}];
mesh = DiscretizeRegion[\[CapitalOmega], {{-1, 1}, {h, 1}}];
Show[mesh,
ContourPlot[x^2 + (y - r)^2 == r^2, {x, -1, 1}, {y, -0, 1}],
Axes -> True]


Sorry I didn't understand your code in detail. But I can present you my version which gives the region you describe:

(*s0=\[Alpha]/Sin[\[Alpha]]*)
(*x=Sin[s Sin[\[Alpha]]]/Sin[\[Alpha]]*)
(*y=(1 - Cos[s Sin[\[Alpha]]])/Sin[\[Alpha]])*)

\[Alpha] = Pi/3
R = ParametricRegion[ { Sin[s Sin[\[Alpha]]]/Sin[\[Alpha]],y},
{{s, -(\[Alpha]/Sin[\[Alpha]]), \[Alpha]/Sin[\[Alpha]]},
{y, -2, (1 - Cos[s Sin[\[Alpha]]])/Sin[\[Alpha]]}}]

Show[DiscretizeRegion[R, {{-1, 1}, {-2, 1}}], Axes ->True] • Which version Mathematica are you using? I'm in 11.3. Are you sure you gave the same code? When I clear my Kernel and copy your code my output is a non-uniform mesh (very highly concentrated on the sides) and no points at the outer ends of the arc, so the arc ends about $x = 0.9$. I'd post a picture but don't know how. – Josh McCraney Jan 16 at 22:15
• @ Josh McCraney Sorry, my fault: The definitions s0,x,y` are only informative. I edited my answer. – Ulrich Neumann Jan 17 at 7:33
• Below is the picture I get when using Ulrich Neumann's code. ![enter image description here](i.stack.imgur.com/GkgqH.png) – Josh McCraney Jan 17 at 14:50
• The definitions were fine. But see the picture I posted; it's nothing like yours. – Josh McCraney Jan 17 at 14:51
• Can confirm: 11.3 is buggy, and will produce picture I've shown. I contacted Wolfram about the issue. – Josh McCraney Jan 17 at 19:54