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

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

[darova]

[sawdiia_too]

I missed a 0 here but not in my original code.

[sawdiia_too]

fixed it

[darova]

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)) ;

[sawdiia_too]

I tried it and it says XMAX must be a floationg point scalar.

Answer 1

[darova]

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)) ;

[sawdiia_too]

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))

[sawdiia_too]

It now works when I put @(y,x). Thank you!

Answer 2

[Torsten]

Phi_ij1 =@(i,j) integral2(@(y,x) B_zi(i,x,y),0,y_min1(j),0,@(y)x_max1(y,j));

[sawdiia_too]

I already did it but I made a mistake with the indices above, it now works. Thank you!!

[Torsten]

Note that @(y,x) B_zi(...) instead of @(x,y) B_zi(...) because you changed the order of x- and y-integration.

This is wrong in Darova's answer.

[sawdiia_too]

Yes I reversed it.

[Torsten]

Note also that you only need to call integral2 once because you can calculate the difference of the two integrals as one integral with new y-limits as shown in my answer.

[sawdiia_too]

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?

[Torsten]

integral 0 -> ymax1 - integral ymin1 -> ymax1 = integral 0 -> ymin1 (outer integral)

[sawdiia_too]

Ok got it!! Thanks again