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Verification using Green theorem

Md Noor
Posted 1 year ago

Compare {xy-x^2} dx+x^2ydy over the triangle bounded lines y=0,x=1,y=x and verify by Green's theorem.
How to solve this? Can't find any solution.
POSTED BY: Md Noor
3 Replies
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I am confused. I tried the new SurfaceIntegrate, and found that these inputs

SurfaceIntegrate[Curl[vf, {x, y}], Element[{x, y}, reg]]
Integrate[Curl[vf, {x, y}], Element[{x, y}, reg]]

give different outputs. So what does SurfaceIntegrate do in dimension 2?
POSTED BY: Gianluca Gorni

Trace shows the Integrate it computes:

Trace[
 SurfaceIntegrate[Curl[vf, {x, y}], {x, y} \[Element] reg],
 _Integrate
 ]
(* Integrate[-x + 2 x y, 
    {x, y} \[Element] Line[{{0, 0}, {1, 0}, {1, 1}, {0, 0}}], 
    GenerateConditions -> Automatic] *)

The docs say (in dimension $n$):

For solid (of dimension n) and bounded RegionQ objects R, take the surface to be the region boundary (RegionBoundary[R]) and the normal orientation to be pointed outward.

So it's integrating the scalar field Curl[vf, {x, y}] around the boundary of the "solid" (dimension 2 region in ${\Bbb R}^2$)/

Embed the problem in 3D and it works fine:

SurfaceIntegrate[
 Curl[vf, {x, y}] {0, 0, 1},
 {x, y, z} \[Element] Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}}]]
(* -1/12 *)
POSTED BY: Michael Rogers
Denis Ivanov
Denis Ivanov
Posted 1 year ago

Our vector field is

Clear[vf, x, y, reg, bnd];
vf = {x y - x^2, x^2 y};

Our region and its boundary are:

reg = Polygon[{{0, 0}, {1, 0}, {1, 1}}];
bnd = RegionBoundary[r]
(*Line[{{0, 0}, {1, 0}, {1, 1}, {0, 0}}]*)

Left hand side of Green theorem is line integral over boundary. We need to break boundary in separate Lines, find integrals and Total:

Total[Table[
  Integrate[
   vf . (#2 - #1) & @@seg
   , {x, y} \[Element] Line@seg]
  , {seg, Partition[First@bnd, 2, 1]}]]
(*-(1/12)*)

Right hand side of Green theorem is 2D integral of Curl over region:

Integrate[Curl[vf, {x, y}], {x, y} \[Element] reg]
  (*-(1/12)*)
POSTED BY: Denis Ivanov
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