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For a function I do this (Integrate over circular range):

NIntegrate[x*y Boole[Sqrt[x^2 + y^2] < 3], {x, 1, 5}, {y, 1, 5}]

Now, I want to do the Sum of the circular set of elements in a square array.


Thanks for your 1st answers and sorry to not have been clear enough.

Example: from a two-dimensional {n, m} array mat, I would like to sum all the elements included in a circle of radius r and located at {x0, y0}.

Something like this:

Sum[mat[[n]][[m]] Boole[Sqrt[(n-x0)^2 + (m-y0)^2] < r], {n, x0-r, x0+r}, {m, y0-r, y0+r}]

Would you do it differently?

And next step, I would like to increase r up to the Sum = mySum.

I hope I'm clearer.

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  • 2
    $\begingroup$ Related: mathematica.stackexchange.com/q/34827/1871 .You just need to Total@Flatten@#& after you extracting the desired elements. $\endgroup$ – xzczd Dec 9 '15 at 11:13
  • $\begingroup$ What is mySum? Is it some value you choose before-hand, and you want to find r such that the sum is equal (or at least close) to mySum? $\endgroup$ – march Dec 9 '15 at 17:11
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I'm not sure I understand the question fully, but in case you meant that:

You have a square $n$ by $n$ matrix mat, and that you want to compute the sum of elements which are within a circle inscribed in the matrix, then you can do:

Total[ DiskMatrix[(n-1)/2] * mat, 2]

This assumes that n is odd so the circle can be properly centred.

References: DiskMatrix, Total

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Not sure if this is what you meant. Do comment if this is not what you meant.

I would use Sum within Sum. Something like the following:

In[24]:= Sum[
  Sum[x*y Boole[Sqrt[x^2 + y^2] < 3], {x, 1, 5, 0.1}], {y, 1, 5, 
   0.1}]/100

Out[24]= 6.4424

With smaller step size...

In[25]:= Sum[
  Sum[x*y Boole[Sqrt[x^2 + y^2] < 3], {x, 1, 5, 0.01}], {y, 1, 5, 
   0.01}]/10000

Out[25]= 6.1591

As you decrease the step size, you should see the result approaching that found by NIntegrate, namely 6.125

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