We are trying to calculate the integrals in the attached file below. And we are having trouble doing it in matlab. We have tried using this for reference: https://se.mathworks.com/matlabcentral/answers/77571-how-to-perform-4d-integral-in-matlab but sadly to no sucess. We would appriciate any help we could get :)

Double area intergral (4D integral)

Torsten 2020-3-31

编辑：Torsten 2020-3-31

What problems do you encounter ? What errors do you get ?

In my opinion, the integral does not exist due to the singular part (0,0,a,a) with 0 <= a <= 0.5.

Birdman 2020-3-31

What is the result of this integral? Is this a homework question?

Viktor Könemann 2020-3-31

在 MATLAB Online 中打开

@Torsten We get a bunch of errors ;)

And this is the code we try to use:

clear all
%syms x y z w
L = 0.5;
%R = sqrt(x^2+y^2+(z-w)^2);
%cos1 = y/R;
%cos2 = x/R;
func = @(x,y,z,w) (x*y/(pi*(x^2+y^2+(z-w)^2)^2));
F = integral(@(x)integral3(@(y,z,w)func,0,L,0,L,0,L),0,L,'ArrayValued',true)
%Here are all the errors:
Error using integral2Calc>integral2t/tensor (line 231)
Input function must return 'double' or 'single' values. Found 'function_handle'.
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
    [q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral3/innerintegral (line 137)
        Q1 = integral2Calc( ...
Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);
Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral3 (line 121)
    Q = integralCalc(@innerintegral,xmin,xmax,integralOptions);
Error in View_factor>@(x)integral3(@(y,z,w)func,0,L,0,L,0,L)
Error in integralCalc/iterateArrayValued (line 156)
                fxj = FUN(t(1)).*w(1);
Error in integralCalc/vadapt (line 130)
            [q,errbnd] = iterateArrayValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in View_factor (line 12)
F = integral(@(x)integral3(@(y,z,w)func,0,L,0,L,0,L),0,L,'ArrayValued',true)

Viktor Könemann 2020-3-31

@Birdman

We are solving this problem in our bachelor thesis.

We are calculating view factors between two planes.

Viktor Könemann 2020-3-31

For reference, this is where the problem originates from. Where we are trying to solve equation (3), with a=b=c=d=0.5

David Goodmanson 2020-4-1

HI Torsten,

the x and y integrals are instantly doable, leaving (for the contribution evaluated at the lower limit x = y = 0) , the integrand

log((w-z)^2) which should be all right.

Torsten 2020-4-1

Good suggestion - I'm convinced.

To Viktor Koenemann:

viewfactor is approximately 0.05( without warranty ).

Viktor Könemann 2020-4-2

Thanks Torsten, that's the right answer. How did you do it in matlab? :)

Torsten 2020-4-2

编辑：Torsten 2020-4-2

Integrate analytically with respect to x, evaluate the integral at its limits.

Function to integrate has the form a*x/(x^2+b)^2

Integrate analytically with respect to y, evaluate the integral at its limits.

Functions to integrate have the form a*y/(y^2+b)

These two tasks can either be done with pencil and paper or using the symbolic toolbox.

Now you have a double integral depending on z and w that can be evaluated numerically using Matlab's integral2.

It contains a combination of functions of the form log((z-w)^2+a).

Double area intergral (4D integral)

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