# Double Integral using Integral2 Error

조회 수: 7(최근 30일)
jack carter 2017년 4월 17일
답변: Shahid Hasnain 2018년 11월 15일
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);
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jack carter 2017년 4월 18일
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

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### 채택된 답변

Andrew Newell 2017년 4월 18일
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.
##### 댓글 수: 1표시숨기기 없음
jack carter 2017년 4월 19일
Thank you Andrew for your help

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### 추가 답변(2개)

David Goodmanson 2017년 4월 17일
편집: David Goodmanson 2017년 4월 18일
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.
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jack carter 2017년 4월 19일
Thank you David for your help

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Shahid Hasnain 2018년 11월 15일
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.

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