mrcleanhands said:
I'm working on a green's theorem integral and I'm given a line (0,0) to (1,2) which i parameterise as (t,2t) but the answer is actually (1-t, 2(1-t)) and if I use my parameterisation on the integral I get a different answer!
Using (t, 2t), when t= 0 you get (0, 0) and when t= 1 you get (1, 2). Using (1- t, 2(1- t)), when t= 0 you get (1, 2) and when t= 1 you get (0, 0). Those both give exactly the same line, just
oriented oppositely.
The distinction is the
orientation of the line. You cannot just integrate 'on' the line, you must either integrate
from (0, 0)
to (1, 2) or from (1, 2) to (0, 0). If, using your parameterization, you integrate \int_0^1 f(x(t), y(x))dt you are integrating from (0, 0) to (1, 2). If, using the other parameterization, you integrate \int_0^1 f(x(t),y(t))dt, you are integrating from (1, 2) to (0, 0) and will get the
negative[/b[] of the previous result. You must think about which direction you want to go on the line and choose your limits of integration accordingly. If you want to integrate from (0, 0) to (1, 2) using the x= 1- t, y= 2(1- t) parameterization, since t= 1 gives (0, 0) and t= 0 gives (1, 2), you must integrate from t= 1 to t= 0: \int_1^0 f(x(t), y(t))dt.
