MATLAB Answers

How can I put an integral inside a for loop when the bounds depend on the loop's variable?

1 view (last 30 days)
sawdiia_too
sawdiia_too on 1 Apr 2020
Commented: sawdiia_too on 2 Apr 2020
Hi, I would like to do this integration where the j of cj is the loop variable.
I tried this but when I calculate Phi_ij1(1,1) for example, Matlab says "unrecognized function or variable 'y' ". Can you help me?
for j=1:13
x_max1 = @(y,j) (-sqrt(3)/3).*y + c(j);
y_min1 = @(j) (sqrt(3)/2)*c(j);
y_max1 = @(j) c(j)*sqrt(3);
Phi_ij1 = @(i,j) integral2(B_zi(i,x,y), 0, x_max1(y,j),0,y_max1(j)) - integral2(B_zi(i,x,y),0, x_max1(y,j),y_min1(j),y_max1(j)) ;
end

  5 Comments

Show 2 older comments
darova
darova on 1 Apr 2020
Try to specify variables integral depends on
Phi_ij1 = @(i,j) integral2(@(x,y)B_zi(i,x,y), 0, @(y)x_max1(y,j),0,y_max1(j)) - ...
integral2(@(x,y)B_zi(i,x,y), 0, @(y)x_max1(y,j),y_min1(j),y_max1(j)) ;

Sign in to comment.

Accepted Answer

darova
darova on 1 Apr 2020
Here is an example for integral3
According to this integration is done from z to x
You have only y. Looks like x=x(y) in your case
So you have to integrate by x first
Try to replace limits of integration
Phi_ij1 = @(i,j) integral2(@(x,y)B_zi(i,x,y), 0, y_max1(j)), 0, @(y)x_max1(y,j) - ...
integral2(@(x,y)B_zi(i,x,y), y_min1(j),y_max1(j), 0, @(y)x_max1(y,j)) ;

  2 Comments

sawdiia_too
sawdiia_too on 2 Apr 2020
In the example, the integration of x is from a scalar to a scalar while on mine the integrations are from functions to functions (one depending on y like you said x(y) and the other depending on j y(j))

Sign in to comment.

More Answers (1)

Torsten
Torsten on 2 Apr 2020
Phi_ij1 =@(i,j) integral2(@(y,x) B_zi(i,x,y),0,y_min1(j),0,@(y)x_max1(y,j));

  7 Comments

Show 4 older comments
sawdiia_too
sawdiia_too on 2 Apr 2020
Oh that's clever. So I have to put all the limits at the same time in the same integral2? Because I don't see y_min1 in your answer. How does it work?

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by