MATLAB: Numerical evaluation of multiple integral not working

distancedistance between pointsintegralintegral2MATLABmultiple integralpoints

Hi guys,
I'm trying to numerically calculate the integral after Equation (1) from here: Distances In Bounded Regions
The final purpose is to calculate the average distance between all points in a region.
The result of the integral should be 0.5214.
I'm calculating it by calling integral2 function like this:
clear all;
f = @(theta,r)(1-r.*sin(theta)).*(1-r.*cos(theta)).*r.^2;
ymax = @(theta)1./cos(theta);
xmax = pi/8;
8*integral2(f, 0, xmax, 0, ymax)
However I'm getting the wrong answer. What's the problem with my code?

Best Answer

Really? Lets do it symbolically.
syms r theta
kernel = (1-r.*sin(theta)).*(1-r.*cos(theta)).*r.^2;
rint = int(kernel,r,[0,1/cos(theta)])
rint =
1/(12*cos(theta)^3) - sin(theta)/(20*cos(theta)^4)
sbar = 8*int(rint,theta,[0,pi/8])
sbar =
log(tan((5*pi)/16))/3 - 16/(15*(2^(1/2) + 2)^(3/2)) + (2*2^(1/2))/(3*(2^(1/2) + 2)^(3/2)) + 2/15
vpa(sbar)
ans =
0.24810023714331117778996075103314
So the symbolic toolbox questions the result in that paper.
pretty(sbar)
   /    / 5 pi \ \
log| tan| ---- | |
   \    \  16  / /            16                2 sqrt(2)        2
------------------ - ------------------- + ------------------ + --
         3                           3/2                  3/2   15
                     15 (sqrt(2) + 2)      3 (sqrt(2) + 2)
What did integral2 produce as a result?
f = @(theta,r)(1-r.*sin(theta)).*(1-r.*cos(theta)).*r.^2;
ymax = @(theta)1./cos(theta);
xmax = pi/8;
8*integral2(f, 0, xmax, 0, ymax)
ans =
       0.248100237144312
Which happens to be quite the same as that produced by the symbolic solution.
So I don't know where that article you found on the internet came from. But it seems to have at least one flawed claim in it. I won't spend the time to check their previous results, or the derivation that went into it. Anyone can post anything online, and with no validation of their results, you get what you pay for.
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