Parameter vectors outside integration function

I am trying to calculate a double integral with three varibales in the form of a vector AB and another scalar B also as a paremeter. If I write the routine so that AB and B are inside the integration fucntion:

Inty3=integral2(@Func3,0,1,0,@(y) y)
function fy3 = Func3(z,y)
              AB=[1 2 3];
              BB=1;
              fy0=sqrt((AB(1)*z+AB(2)).^2+(BB*z+AB(3)).^2); 
              fyy=((1+fy0).^2)./(1-fy0); % F(y) in Eq.17  
              fAy=(AB(1)*z+AB(2))./(1-fy0);   
              fy3=fyy.*fAy;                  
end

the integral is clauclated without a problem, but if AB and B are moved outside of Fun3, i.e.,

AB=[1 2 3];
  BB=1;
    Inty3=integral2(@Func3,0,1,0,@(y) y)
      function fy3 = Func3(z,y)
                    fy0=sqrt((AB(1)*z+AB(2)).^2+(BB*z+AB(3)).^2); 
                    fyy=((1+fy0).^2)./(1-fy0); % F(y) in Eq.17  
                    fAy=(AB(1)*z+AB(2))./(1-fy0);   
                    fy3=fyy.*fAy;                  
      end

then I receive the following error message:

Not enough input arguments.

Error in DoubleInt>Func3 (line 15)
              fy0=sqrt((AB(1)*z+AB(2)).^2+(BB*z+AB(3)).^2);

Error in integral2Calc>integral2t/tensor (line 228)
        Z = FUN(X,Y);  NFE = NFE + 1;

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 DoubleInt (line 10)
Inty3=integral2(@Func3,0,1,0,@(y) y)

Answer 1

[Star Strider]

You need to change your call to ‘Func3’ in the second example:

Inty3=integral2(@(z,y) Func3(z,y,BB,AB),0,1,0,@(y) y)

[Saeid]

I tried it and it didn't work :-( Did you try that exact same format yourself with my program?

[Star Strider]

Try this and it will :-)

AB=[1 2 3];
BB=1;
Inty3=integral2(@(z,y) Func3(z,y,AB,BB),0,1,0,@(y)y)
    function fy3 = Func3(z,y,AB,BB)
        fy0=sqrt((AB(1)*z+AB(2)).^2+(BB*z+AB(3)).^2);
        fyy=((1+fy0).^2)./(1-fy0); % F(y) in Eq.17
        fAy=(AB(1)*z+AB(2))./(1-fy0);
        fy3=fyy.*fAy;
    end

Inty3 =
     3.2745

[Saeid]

Perfect, thanks!

[Star Strider]

My pleasure!