How do I compute a line integral of a function over a helix?
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
So I need to find line the integral of
F=<(e^z)*(y^2), 2(e^z)xy, (e^z)x(y^2)> over a helix parametized as
x=2cost
y=2sint
z=t/5
for 0<= t <= 5pi.
I've no clue how to do this. I don't think I quite conceptually understand the requirement either.
Here's an attempt, however:
t=0:0.1:5*pi;
x=2.*cos(t);
y=2.*sin(t);
z=t./5;
myfunction =@(x,y,z) [(exp(z))*(y.^2); 2.*(exp(z)).*x.*y; (exp(z)).*x.*(y.^2)];
integral3(myfunction,-2,2,-2,2,0,pi);
Unfortunately, it did not work
0 commentaires
Réponses (1)
CARLOS RIASCOS
le 3 Avr 2018
Hello brother, here is a code I did with symbolic mathematics using the mathematical definition of line integral with a vector field F. Postscript: The integral gives 0 LOL.
syms t
%Parametrization of the Curve:
x=2*cos(t); y=2*sin(t); z=t/5;
%Vector field:
F = [exp(z)*(y^2), 2*exp(z)*x*y, exp(z)*x*(y^2)];
%Dot product F*dr:
D = F*[diff(x,t); diff(y,t); diff(z,t)];
%Integral of line respect of t (dt):
I = int(D,t,0,5*pi);
i=double(I);
%disp:
disp('Value:')
disp(i)
2 commentaires
CARLOS RIASCOS
le 3 Avr 2018
Yes, It must be because the vector field F, is conservative, therefore its line integral on a closed curve in this case an ellipse is zero. But the code is fine, I tested it with exercises and the results of the code matched the results of the exercises. I hope my code will help you.
Voir également
Catégories
En savoir plus sur Calculus dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!