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Matlab Triple Integration Error. Thank You

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sun on 28 Dec 2014
Commented: sun on 28 Dec 2014
Dear guys, I am trying to do a triple integration. But I am not very understand the meaning of error message. could you take a look for it? Thank you so much
The code is
clear all;
%%%== just some parameters here ========
la1=1/(pi*500^2); la2= la1*5;
p1=25; p2=p1/25;
syms r u1 u2 u3
powe= -2
seta= 10^powe;
%%%=== parameters end ================
yi = @(u3,u2,u1) exp(-u3.*(1+2.*...
(pi./4 - atan(10.*(u3.^2./u1.^2 + u3.^2./u2.^2 + 1).^(1./2))./2 )./...
((( (u3./u1).^(a./2) + (u3./u2).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u3.^(a./2)./...
((( (u3./u1).^(a./2) + (u3./u2).^(a./2) + 1 ).^(2./a)).^(a./2)) );
maxF2 =@(u2) u2;
maxF3 =@(u3) u3;
out2 = integral3(yi, 0, Inf , 0, maxF3 , 0, maxF2)
As you see, u3 is [0, Inf], u2 is [0, u3], u1 is [0, u2]. Error is showing me as
Error using @(u2)u2
Too many input arguments.
Error in integral3>@(y)ZMAXXY(x(1)*ones(size(y)),y) (line 142)
@(y)ZMAXXY(x(1)*ones(size(y)),y), ...
Error in integral2Calc>integral2t/tensor (line 191)
top = YMAX(x);
Error in integral2Calc>integral2t (line 56)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 10)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral3/innerintegral (line 138)
Q1 = integral2Calc( ...
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 84)
[q,errbnd] = vadapt(@AToInfInvTransform,interval);
Error in integral3 (line 122)
Q = integralCalc(@innerintegral,xmin,xmax,integralOptions);
Error in ref7_equ11n3 (line 33)
out2 = integral3(yi, 0, Inf , 0, maxF3 , 0, maxF2)
What's meaning of 'Error using @(u2)u2 Too many input arguments.' ?? If I change u2,u3's range to real number (e.g. out2 = integral3(yi, 0, Inf , 0, 1000 , 0, 1000). ), then it's working. out2 will be a real number. Thanks for your time!

Accepted Answer

Shoaibur Rahman
Shoaibur Rahman on 28 Dec 2014
Replace out2 by
out2 = integral3(yi, 0, Inf , 0, @(u3) u3 , 0, @(u3,u2) u2)
If you want to use function handles as the integration limit, then
ymax should be function x, in your case u3
zmax should be a function x and y, in your case u3 and u2, although u3 has no effect
sun on 28 Dec 2014
Dear ShoaiburRahman, I see this information as below,
You also can specify zmax to be a function handle (a function of x,y) when integrating over a nonrectangular region.
Does this mean zmax MUST be a function of x,y? hmmm. at first time, I think this is not necessary.

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