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I want to integrate a function f[x,y] over a region like this.

enter image description here

Is there any way to define the region using the vertices of the purple polygon?

Then I can subsequently perform NIntegrate[f[x,y],{x, y} ∈ region]

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Here is an example in 12.2.

poly = Polygon[{{0, 0}, {1/2, Sqrt[3]/2}, {1, 1/Sqrt[3]}, {1, 0}}];
NIntegrate[Log[x + y + 1], {x, y} \[Element] poly]

0.366623

Let us verify it by

Integrate[Log[x + y + 1], {x, y} \[Element] poly]

-((36 - 12 Sqrt[3] + 12 Log[2] + 228 Sqrt[3] Log[2] + 138 Log[3] + 54 Sqrt[3] Log[3] + 9 Log[4] - 3 Sqrt[3] Log[4] + 48 Sqrt[3] Log[6] - 2 Log[8] - 2 Sqrt[3] Log[8] + 2 Sqrt[3] Log[9] - 48 Sqrt[3] Log[2 - 2/Sqrt[3]] + 90 Log[2 - Sqrt[3]] + 54 Sqrt[3] Log[2 - Sqrt[3]] - 90 Log[3 - Sqrt[3]] - 54 Sqrt[3] Log[3 - Sqrt[3]] - 180 Log[-1 + Sqrt[3]] - 108 Sqrt[3] Log[-1 + Sqrt[3]] + 72 Log[1 + Sqrt[3]] - 48 Sqrt[3] Log[1 + Sqrt[3]] - 36 Log[2 + Sqrt[3]] + 24 Sqrt[3] Log[2 + Sqrt[3]] - 18 Log[3 + Sqrt[3]] - 90 Sqrt[3] Log[3 + Sqrt[3]] - 48 Log[6 + Sqrt[3]] - 52 Sqrt[3] Log[6 + Sqrt[3]] - 72 Log[3 + 2 Sqrt[3]] + 48 Sqrt[3] Log[3 + 2 Sqrt[3]] + 36 Log[9 + 5 Sqrt[3]] - 24 Sqrt[3] Log[9 + 5 Sqrt[3]])/(8 Sqrt[ 3] (19 + 11 Sqrt[3]) (-45 + 26 Sqrt[3])))

N[%]

0.366623

Addition. NIntegrate produces [a different] the same result if the vertices are taken couunter-clockwise as

poly1 = Polygon[{{1, 1/Sqrt[3]}, {1/2, Sqrt[3]/2}, {0, 0}, {1, 0}}];
NIntegrate[Log[x + y + 1], {x, y} \[Element] poly1]

[0.17812]0.366623

shows.

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    $\begingroup$ Does it matter whether I define the points in the function Polygon[] in clockwise or counter-clockwise manner? $\endgroup$ Apr 1, 2021 at 5:18
  • $\begingroup$ Now I am confused about the different results for the clockwise and counterclockwise. Why is it different? Which one is the actual value of the integral over the region? $\endgroup$ Apr 1, 2021 at 7:06
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    $\begingroup$ I get the same result (0.366623) when I integrate over poly1 and poly $\endgroup$ Apr 1, 2021 at 7:16
  • $\begingroup$ @ArishmanPanigrahi: Yes, you are right. Executing that on a fresh kernel, I obtain the same result. $\endgroup$
    – user64494
    Apr 1, 2021 at 8:54

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