# Double Integral using Integral2 Error

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jack carter on 17 Apr 2017
Answered: Shahid Hasnain on 15 Nov 2018
I am trying to write code for the attached equation in matlab but I am getting errors.
My code is as under;
phi=0.0:0.01:90;
P=sqrt((pi^2*Ef*df^3*rounot*S*(1+n))/(4)+(pi^2*Ef*df^3*Gd)/(2));
p_phi=sin(phi);
p_z=2/Lf;
m1=P*exp(f*phi); %multiplier,inside the integral part of eq#8.
m2=(4*Vf)/(pi*df^2); %multiplier, outside the integral part of eq#8.
myfun_eq8=@(z,phi) m1.*p_phi.*p_z;
I= integral2(myfun_eq8,0,1,0,pi/2);
sigma_b=m2*I;
I am getting the following error, can someone help me correct it.
Error using integral2Calc>integral2t/tensor (line 241) Integrand output size does not match the input size.
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 integral2 (line 106) Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in myeqsetlin (line 160) I= integral2(myfun_eq8,0,1,0,pi/2);
Andrew Newell on 18 Apr 2017
I notice that p(z) is a constant, so integrating p(z)dz gives simply cos(phi). Thus, you really have a one-dimensional integral to solve numerically.
jack carter on 18 Apr 2017
Thanks Andrew for your reply. I was writing a series of equations in a function file to run a simulation after completion. That is why there seem to had variation in yours and mine results. I wrote delta as S in my script.
I re-ran the above mentioned script separately and I have received the same error as you mentioned in your first reply for m1. If I define as a range of numeric values (S=0:0.01:30), the function still does not work. But if I select S=1; then I get the same set of integral errors.
Your simplification of the equation i.e. p(z) as a constant is helpful

Andrew Newell on 18 Apr 2017
Based on the above discussion, the critical part of code you need is as follows:
myfun_eq8=@(phi) exp(f*phi).*cos(phi).*sin(phi);
I= integral(myfun_eq8,0,pi/2);
All the other terms can stay outside of the integral.
jack carter on 19 Apr 2017
Thank you Andrew for your help

David Goodmanson on 17 Apr 2017
Edited: David Goodmanson on 18 Apr 2017
Hi jack, I believe this is occurring because your definition of myfun_eqn8 uses the two variables z,phi, but the integrand m1.*p_phi.*p_z is not explicitly a function of either of those variables. Matlab does not know how to create the integrand for arbitrary z,phi. You need to define m1, p_phi,p_z explicitly in myfun_eqn8.
Once this gets working the first and third lines of code will probably become superfluous anyway, but right now you are taking the sine of angles in degrees. You can convert to radians or use the sind function there.
David Goodmanson on 18 Apr 2017
Edited: David Goodmanson on 18 Apr 2017
ok then it does appear to reduce to a one-dimensional integral in the variable phi, and the answer for that only depends on the constant f.
jack carter on 19 Apr 2017
Thank you David for your help

Shahid Hasnain on 15 Nov 2018
I have following expression with x=[1 2 3 4 5], y=[1 2 3 4 5],
L = 100; % Length of the box
p=1;
q=1;
Ui = @(x,y) sin(pi*x).*sin(pi*y); % An initial distribution for u, i.e. the u(x,y,0)
Vi = @(x,y) sin(pi*x).*sin(pi*y); % An initial distribution for v, i.e. the v(x,y,0)
Wi = Ui(x,y) + Vi(x,y); % Initial condition for w, i.e. w(x,y,0)
% Computation of the b-coefficients
if p < Ntrunc
if q < Ntrunc
my_integrand = @(x,y) 4*Wi.*sin(p.*x).*cos(q.*y)/L^2;
bd(p,q) = integral2(my_integrand(x,y),0,L,0,L);
q = q+1;
end
p = p+1;
end
I got following error,
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 integral2 (line 106)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Correction will be highly appreciated.