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So I want to numerically compute the integral of a long complicated expression over a specified domain (in this case an ellipse). I know how to use a Boole function to sample within the ellipse, but I want to pass a list of points within this ellipse into the function and have the integral be evaluated at each of these specific points. An example would be:

$\frac{1}{(x \cos(\theta)+y \sin(\theta))^\frac{5}{2}} $

So I want to integrate over θ from 0 to 2 π and I want to use different points (x, y) within the specified domain when integrating. Can anyone help me with this?

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Humm... This integral is zero. So the answer to your question is zero for any x, y

$$\frac{1}{(x \cos(\theta)+y \sin(\theta))^\frac{5}{2}}$$

Clear[theta, x, y];
f = 1/(x Cos[theta] + y Sin[theta])^(5/2);
int = Integrate[f, theta];

int /. theta -> 2 Pi

Mathematica graphics

int /. theta -> 0

Mathematica graphics

(int /. theta -> 2 Pi) - (int /. theta -> 0) 
(* 0  *)

Update

From comment below that the function used was just an example, here is one way to make a function to use NIntegrate for different x,y values. Changed the original function a little bit

ClearAll[theta, x, y, f];
f[{x_?NumericQ, y_?NumericQ}, theta_?NumericQ]:= 1/(x Cos[theta]^2+y Sin[theta])^(5/2);

pts = RandomReal[{0, 1}, {5, 2}];(*10 points*)

int = NIntegrate[Evaluate@f[#, theta], {theta, 0, 2 Pi}] & /@ pts

Mathematica graphics

This is a made up function just for illustration. Replace your actual function by f above. If this still not what is being asked, please feel free to follow up.

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  • $\begingroup$ Or divergent... $\endgroup$
    – Michael E2
    Commented Feb 11, 2014 at 4:33
  • $\begingroup$ I'm sorry if I was unclear, but that isn't the actual function. The actual function is honestly too long to type, how did you get mathematica output directly into your post? Once I know that I can just post the result directly $\endgroup$
    – Mr. Frobenius
    Commented Feb 11, 2014 at 4:55
  • $\begingroup$ @Nasser Cool, that is helpful. Thank you. So does random real just choose a random set of ten points in your example? What if I want to create a grid of a specified resolution in the plane, and then evaluate the integral at each of these grid points? $\endgroup$
    – Mr. Frobenius
    Commented Feb 11, 2014 at 15:19
  • $\begingroup$ Let me comment that the approach depends strongly on the concrete form of the expression to be integrated. The one above, for example, admits passing to a single parameter: (x Sin[t]+y Cos[t])^5/2==x^5/2*(Sin[t]+z*Cos[t])^5/2where z=y/x. If this (or at least, something like this) is the case, the problem becomes especially simple. $\endgroup$
    – Alexei Boulbitch
    Commented Aug 29, 2014 at 20:02

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