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How to explain a difference in solutions of integrals between MATLAB and MAPLE?

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Dear community,
after solving an integral via MAPLE 2022 (see the solution in blue),
I tried to reproduce the calculation in MATLAB R2021b as a numerical and symbolic expression:
%% integral: numerical expression
%empirical coefficents
vrel = 0.000018;
alpha = 0.08;
beta = -6.75;
h =1;
d =1;
%function
fun = @(theta) (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);
%integral
F = integral(fun,0,pi/2)
F = -0.0322
%% integral: symbolic expression
%system parameters
syms theta alpha beta d vrel
%function
fun = (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);
%substiution of empirical coefficents
fsub = subs(fun,{alpha, beta, d, vrel}, {0.0814,-6.75,1,0.000018});
%integral
f = int(fsub,theta,0,pi/2);
%evaluation for complex number
c = subs(f, 1+i);
F = double(c)
F = -0.0328
However, the solution between MATLAB and MAPLE differs by 8 magnitudes. My colleauge and me double-checked for typos but couldn't figure out any. I verified the result from MAPLE via comparison to a simplfied solution for the integral.
I am curious to learn about my mistake(s) when adapting the integral into MATLAB!
  3 Comments
Tim Hammer
Tim Hammer on 10 Nov 2022
Hi Jiri,
thanks for your suggestion! I checked it, and it seems that both softwares evaluate the function correctly:
MAPLE:
MATLAB:
%empirical coefficents
vrel = 0.000018;
alpha = 0.08;
beta = -6.75;
h =1;
d =1;
%function
fun = @(theta) (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);
close all
figure
fplot(@(theta) fun(theta),[0 1.6])
Bjorn Gustavsson
Bjorn Gustavsson on 10 Nov 2022
After following @Jiri Hajek's suggestion one gets a smoothly varying curve starting at approximately -0.054 at theta=0 that aproaches zero at pi/2. This curve is everywhere in this interval 0<theta<=1 under the line y = -0.05 + 0.05*x. Therefore the integral has to have a smaller value than the area of the triangle with corners at [0,0], [0,-0.05] and [1,0] which should be -0.05*1/2 or -0.025.

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Accepted Answer

Jiri Hajek
Jiri Hajek on 10 Nov 2022
Nicely done graphs! So, just by looking at the graphs, it is clear that:
  • The integral of your function is a negative value, with lower limit on the value being the area of the graph, i.e. about 1,6x(-0,05)= - 0,08
  • Comparing your original results from MATLAB with the above, it seems that both provide good estimates, although it remains to see which one is more precise
  • The MAPLE result is a complete nonsense, 8 orders of magnitude off the target.You have to search for error in your MAPLE implementation, not in MATLAB.

More Answers (2)

Torsten
Torsten on 10 Nov 2022

Alan Stevens
Alan Stevens on 10 Nov 2022
This is what I get in Maple:

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