A question using trapz integration

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Lilach Saltoun 2015-8-4

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编辑： Lukas Bystricky 2015-8-4

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Hello all

I have a problem using the trapz integration. I’m trying to integrate this integral (in the attached file) using trapz twice when I’m trying to approximate the analytic solution to z = 0.

(For arctan(R/Z) far from pi/2 it is working ok). I’m taking z to be very small but final and it seems that the code I wrote doesn't handles it very well; I guess it is because tan diverges in pi/2.

(the analytic solution is pi*R^2)

Maybe there is another way to solve this problem?

Thanks

(In the next step I need to perform this integral where in the integrand I have a function which is dependent on theta and phi, And that is why I can’t just take the area of the circle)

The code I wrote:

theta_val_new  = (round(linspace(0,pi./2,100).*100)/100)';
phi_val_new = linspace(0,2.*pi,120);
theta_val_mat = repmat(theta_val_new,1,120);
    function_to_integrate = abs(tan(theta_val_mat)./(cos(theta_val_mat)).^2) .*(30./tan(theta_val_new(end,1))).^2;
 energy1 = (trapz(phi_val_new,trapz(theta_val_new,function_to_integrate))

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Lilach Saltoun 2015-8-4

Hi Torsten

I can't that because I need to integrate over the gamma function, and it is numerically calculated (from anther code I've created) for theta and phi. I can't express gamma in Polaric coordinate system, only in the azimuthal part of the sperical coordinates system.

the real integral i need to calculate is:

Lukas Bystricky 2015-8-4

编辑：Lukas Bystricky 2015-8-4

I see you added a second integral that can't be integrated exactly. Do you know if gamma is 0 at theta = pi/2 (for all phi)? If it isn't, then that integral is unbounded since Rz/cos^2(arctan(R/z)) goes to infinity as z goes to 0.

If you do wish to do this integral numerically for other values of z I agree with Torsten that integral2 is a much better route.

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