Orientation of Inti.parametric_curve()

[fmonteghetti]

I have a question about how Inti.parametric_curve is oriented.

Let us consider the droplet shape:

   line_droplet = let
       Inti.parametric_curve(0, 2*pi) do (s,)
           SVector(
               (2/sqrt(3)) * sin(s/2),
               - sin(s)
           )
       end
   end
   # Create the mesh
   Γ = Inti.Domain(line_droplet)
   msh = Inti.meshgen(Γ; meshsize=meshsize)
   qrule = Inti.GaussLegendre(; order=qorder)
   Q = Inti.Quadrature(msh, qrule)

Q gives the expected outward normal.

However, if I consider the droplet shape with a re-entrant corner:

   line_droplet = let
       Inti.parametric_curve(0, 2*pi) do (s,)
           SVector(
               (3/2) * sin(3*s/2),
               - sin(s)
           )
       end
   end
   # Create the mesh
   Γ = Inti.Domain(line_droplet)
   msh = Inti.meshgen(Γ; meshsize=meshsize)
   qrule = Inti.GaussLegendre(; order=qorder)
   Q = Inti.Quadrature(msh, qrule)

The normal vector in Q is the inward normal.

My question is:

• Is there a way to tell Inti that "this curve is the outer boundary of my domain and I want the outward normal" ?

At first I thought I could fix the normal by changing the orientation of the curve (i.e. using Inti.parametric_curve(2*pi,0)), but this has no effect on the normal.

[fmonteghetti]

#93 does fix the orientation issue.