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Subject: Re: Double Numerical Integration
Date: Fri, 23 Jan 2009 16:37:01 +0000 (UTC)
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Roger,
Yes you are right, i am integrating over a square for each iteration along the xy axes. In the original volume, the value of z is constanlty evolving. In the case of the cuboid, the value of z remains constant. So it is practical to integrate only the 'L' portion for each subsequent iteration. 
The equation of z = (1-theta)^-0.5 * exp(-0.5*(1-theta^2)^-1 * (x^2 + y^2)) * exp(0.5*(x^2 + y^2)), theta = constant. This function to imagine it simply looks like a nice peak from Ben Nevis ;)
My problem is further compounded by the fact that i am forced to calculate the volume each time over the rectangular region but i need to assign any z value obtained  equal to zero when the value of x has crossed a certain threshold value, which again changes for each limit decrease/increase. 
I have also played around with tolerances, and so far the tol of 1e-6 is proving to be the minimum necessary for good results. 
I have tried to overcome the above problem by normalizing the value of z, so that the largest value of z does not exceed 1. But again, i gain a penalty by increasing the tolerance target to 1e-10. Slight improvement in speed only ;((
Any bright ideas...
Regards,
V