Solving Complex Line Integrals
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Josée Mallah
el 21 de Mzo. de 2021
Comentada: Bjorn Gustavsson
el 22 de Mzo. de 2021
Hello everyone!
How to define a circle (e.g. | z-1 |=3) as an integral path using Waypoints?
For example, C here is a square contour:
C = [1+i -1+i -1-i 1-i];
q2 = integral(fun,1,1,'Waypoints',C)
How to define C as a circle?
Or else, how could I define the same circle ( | z-1 |=3) instead of the unit circle in the code below?
g = @(theta) cos(theta) + 1i*sin(theta);
gprime = @(theta) -sin(theta) + 1i*cos(theta);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)
Thank you
1 comentario
darova
el 21 de Mzo. de 2021
THe question is unclear. Do you have any picture or something? What are you trying to integrate?
Respuesta aceptada
Bjorn Gustavsson
el 22 de Mzo. de 2021
Wouldn't it be simplest to just multiply the radius of the polar representation of your unit-circle with the desired radius?
g = @(theta) 3*cos(theta) + 1i*3*sin(theta);
gprime = @(theta) -3*sin(theta) + 1i*3*cos(theta);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi);
If you have some more complex curve sometimes those too can be more straightforwardly represented in polar coordinates. Then
You might also be lucky enough to have an analytical function - which makes integration around a closed loop in the complex plane that much easier.
HTH
2 comentarios
Bjorn Gustavsson
el 22 de Mzo. de 2021
Great.
It's just to remember that any single-variable parameterization of your perimeter is good enough as long as you have some way to calculate dl too.
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