Solving an integration problem
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Good morning,
I am tring to solve an integration problem. But Matlab don't give me the value. Can you help me please.
Here is the code:
--------------------
clc; clear all;
syms R t r c d x y z s2 D N g;
x = cos(t)+d./(c*r);
y = sin(t);
s1 = x^2+y^2;
z = cos(t)+d./(c*R);
s2 = z^2+y^2;
D = simplify(c^4*R^2*r*s1*s2)
N = (r-R)^2;
%function to integrate
g = expand(N/D)
%-first integral-%
f = @(t) g;
I = int(f, t, 0, 2*pi)
---------------------------
Best regards
Respuestas (2)
Ameer Hamza
el 12 de Mayo de 2018
You need to pass symbolic expression directly into int as follow
I = int(g, t, 0, 2*pi);
int() tries to find an analytical solution for the integration. And since there are so many symbols, it is highly unlikely that you will get a useful solution. Try pre-defining the values of r, c, d and R, and only define t as sym because it is variable of integration. So modify your code as follow
syms t;
d = 1; % give appropriate values
c = 10;
r = 1;
R = 200;
x = cos(t)+d./(c*r);
y = sin(t);
s1 = x^2+y^2;
z = cos(t)+d./(c*R);
s2 = z^2+y^2;
D = c^4*R^2*r*s1*s2;
N = (r-R)^2;
%function to integrate
g = N/D;
%-first integral-%
I = int(g, t, 0, 2*pi)
Babacar Ndiaye
el 16 de Mayo de 2018
3 comentarios
Walter Roberson
el 16 de Mayo de 2018
Do not use
g = @(r) R*I;
as that is a function handle that takes a single argument that it ignores. You should just
J = int(R*I, r, R, 1/2*pi)
However... there are only narrow circumstances under which the result is not one of the infinities or undefined. Are there restrictions you can place on the variables, such as particular ones being non-negative or real valued?
Ameer Hamza
el 17 de Mayo de 2018
Editada: Ameer Hamza
el 17 de Mayo de 2018
@Babacar, the integrand in your case is quite complex. It is very unlikely that you will be able to get an analytical solution. The best chance of solving this problem is to use a numerical optimization method.
Walter Roberson
el 17 de Mayo de 2018
I am finding analytical solutions under some reasonable assumptions. The analytic solutions are inf or undefined except for narrow cases such as R=π/2, but the exact sign is rather messy to express. Like is it -inf or is it +inf+undefined*1i depends on details of the relationship between the variables.
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