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On the help page of the SurfaceIntegrate, an example of Stokes' theorem is listed:

Compute the Curl g of a vector field f:

Clear["Global`*"];

f = {-y + x  y^2, x, z};

g = Curl[f, {x, y, z}];

The surface integral of g over an open surface is:

reg = ParametricRegion[{Cos[t]  Cos[p], Cos[t]  Sin[p], 
    Sin[t]}, {{p, 0, 2  Pi}, {t, 0, Pi/2}}];
SurfaceIntegrate[g, {x, y, z} \[Element] reg]

(*2 Pi*)

This is the same as the line integral of f over the boundary of the surface:

bound = ParametricRegion[{Cos[t], Sin[t], 0}, {{t, 0, 2  Pi}}];
LineIntegrate[f, {x, y, z} \[Element] bound]

(*2 Pi*)

Here is the ParametricRegion form of the bound directly provided.

My question is, if the ParametricRegion form of the boundary is not directly given, how can one obtain the boundary ParametricRegion form from a two-dimensional ParametricRegion surface?

It would be best if the boundary is also in the form of ParametricRegion, so it can be directly used for LineIntegrate.

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  • $\begingroup$ Try RegionBoundary[ your region ] $\endgroup$ Commented Apr 19 at 9:40
  • $\begingroup$ @UlrichNeumann: Does not work. According to the documentation, "RegionBoundary of a lower-dimensional region is the region itself". $\endgroup$
    – user64494
    Commented Apr 19 at 10:38
  • $\begingroup$ @lotus2019 It could be better to show region your are asking about. $\endgroup$ Commented Apr 19 at 16:09

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