# Having problems with another line integral

• Mar 6th 2007, 09:20 PM
pakman
Having problems with another line integral
int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?
• Mar 7th 2007, 08:03 AM
ThePerfectHacker
Quote:

Originally Posted by pakman
int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?

I am not sure what you did, but you tought you can use the fundamental theorem of line integral you cannot.
Because the cross partials test fails.

The parametrization of the curve is,
x=t and y=2-t for 1<=t<=3.

That means,
INT(1,3)(x+2y)*(dx/dt)dt+INT(1,3)(x-2y)(dy/dt)dt
Insert the parametrization.

INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt

INT(1,3) 4-t dt + INT(1,3) 3t-4 dt

What ever that is should be the answer.
• Mar 7th 2007, 08:49 AM
Plato
Quote:

Originally Posted by ThePerfectHacker
INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt
INT(1,3) 4-t dt + INT(1,3) 3t-4 dt

There is a minor sign error there: (dy/dt)=(-1).