# How to calculate this surface integral with ImplicitRegion and RegionBoundary?

How can I calculate this surface integral with ImplicitRegion and RegionBoundary?

Clear["Global*"];
reg1 = Region[
ImplicitRegion[
x > 0 && y > 0 && z > 0 && x + y + z < 1, {x, y, z}]];
reg2 = RegionBoundary[reg1];
f[x_, y_, z_] := x*y*z;
Integrate[f[x, y, z], Element[{x, y, z}, reg2]]

(* Returns unevaluated *)

• NIntegrate[f[x, y, z], Element[{x, y, z}, reg2]] evaluates! Commented Mar 7, 2022 at 11:35
• reg2 = BoundaryDiscretizeRegion[reg1] and Integrate[f[x, y, z], Element[{x, y, z}, reg2]]
– Syed
Commented Mar 7, 2022 at 11:38
• @Syed What's your Mathematica version? v12.2 doesn't evaluate your code. Commented Mar 7, 2022 at 11:44
• @UlrichNeumann I see this.
– Syed
Commented Mar 7, 2022 at 11:50
• @Syed:reg3 = DiscretizeRegion[reg2] ; Integrate[x*y*z, Element[{x, y, z}, reg3]] performs 0.0144338. Commented Mar 7, 2022 at 16:36

One way is as directed in the Ulrich Neumann's comment.

Clear["Global*"];reg1 = Region[ImplicitRegion[
x >= 0 && y >= 0 && z >= 0 && x + y + z <= 1, {x, y, z}]];reg2 = RegionBoundary[reg1];
NIntegrate[x*y*z, {x, y, z} \[Element] reg2]


0.01443375672974058

RootApproximant[%]


1/(40 Sqrt[3])

The command of Maple

VectorCalculus:-SurfaceInt(x*y*z, [x, y, z] = Surface(<s, t, 1 - s - t>, [s, t] = Triangle(<0, 0>, <1, 0>, <0, 1>)))


sqrt(3)/120

confirms it (The integrals over the triangles lying in the coordinate planes equal zero.).

• I'd like to add that Mathematica is able to calculate this surface integral, making use of Integrate[x*y*(1 - x - y)*Sqrt[1 + D[1 - x - y, x]^2 + D[1 - x - y, y]^2], Element[{x, y}, Triangle[{{0, 0}, {1, 0}, {0, 1}}]]] which results in 1/(40 Sqrt[3]). Commented Mar 7, 2022 at 13:56
• Thanks a lot! @user64494 Commented Mar 8, 2022 at 7:58

Since this is a piece-wise surface,we need to calculate all the individual surface integral and sum up.

pts = {{0, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
triangles = Triangle /@ Subsets[pts, {3}];
Integrate[x*y*z, {x, y, z} ∈ #] & /@ triangles // Total


1/(40 Sqrt[3])

• How do you derive pts from reg2? Commented Mar 7, 2022 at 13:57
• Thanks a lot! @cvgmt Commented Mar 8, 2022 at 7:59