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How to properly create function for integration?

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Kaleem Graham
Kaleem Graham on 19 Sep 2021
Commented: Star Strider on 21 Sep 2021
Hi i'm trying to create functions that I can then use to compute a definite integral, here are the functions:
\\
Integral I'm trying to compute:
The Error i'm getting:
My Attempt:
syms theta_func
z_func = @(theta_func) exp(1i*pi*cos(theta_func));
E_func = @(theta_func) sin(theta_func).*(z_func(theta_func)-1).*((z_func(theta_func).^3)-1);
E_func = @(theta_func) abs(E_func(theta_func))^2;
expr = @(theta_func) E_func(theta_func).*sin(theta_func);
q = 2*pi*int(expr,theta,0,pi);
  1 Comment
Jan
Jan on 20 Sep 2021
Prefer to post code as text, not as screenshot.
You show your apporach. Does it work? We cannot try this by our own, because it is a screenshot only. Most of the readers will not take the time to type this code again, most of all because it is not clear, how this code is related to your question.

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Answers (1)

Star Strider
Star Strider on 20 Sep 2021
There are several problems with the posted code image.
First, ‘z’ is not defined anywhere, and it is not an argument to ‘E_func’, and the second ‘E_func’ line should be:
E_func = @(theta_func) abs(E_func(theta_func)).^2;
It may also not be necessary to use the Symbolic Math Toolbox for this, anyway.
.
  3 Comments
Star Strider
Star Strider on 21 Sep 2021
The way the code is written means that the Symbolic Math Toolbox functions are not appropriate.
Try this —
z_func = @(theta_func) exp(1i*pi*cos(theta_func));
E_func = @(theta_func) sin(theta_func).*(z_func(theta_func)-1).*((z_func(theta_func).^3)-1);
E_func = @(theta_func) abs(E_func(theta_func)).^2;
expr = @(theta_func) E_func(theta_func).*sin(theta_func);
q = 2*pi*integral(expr,0,pi) % Use 'integral' For Numeric Integration
q = 20.6011
.

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