Calculate Integral of F*n on Square Curve C

[yolanda]

Homework Statement

First off, sorry this isn't symbolized correctly. I've wrestled with this for a half hour, so here it is in crude type:

Calculate \int_c F*n ds for F(x,y)=xi+yj across the square curve C with vertices (1,1), (-1,1), (-1,-1), and (1,-1).

Homework Equations

above

The Attempt at a Solution

I've found that this can be rewritten as \int_a^b F(r(t))*r'(t)) dt

I know my limits are from -1 to 1. I am having trouble parameterizing x and y. I have come up with x=t, but I don't know what y could be... y=^{+}_{-}t?

If someone wouldn't mind explaining this problem, and how they'd attack it, I sure would appreciate the help. Thanks in advance. A step by step solution would help most, but I'll take what I can get :)

 

[HallsofIvy]

Because the path is not "smooth", break it into smooth parts. That is, do the lines from
(1) (1,1) to (-1,1)
(2) (-1,1) to (-1,-1)
(3) (-1,-1) to (1,-1)
(4) (1,-1) to (1,10)
separately.

For example, the line from (1,1) to (-1,1) is simply "y= 1" with x going from 1 to -1.
You could just use x itself as parameter with x going from 1 to -1. Since y= 1, dy= 0.
If you don't like integrating from 1 down to -1, you could let x= -t so that dx= -dt and you are integrating with respect to t from -1 to 1.
Or you could take x= -2t+ 1 so that dx= -2dt and integrate with respect to t from 0 to 1. The crucial point is that at every point on this line y= 1 so dy= 0.

Do the same kind of thing for the other three lines.

 

[yolanda]

Thanks for your response.

Okay, that makes more sense. So, to get my final answer I should be adding the results of the 4 line integrals, right?

So far, for the line segment from (1,1) to (-1,1) I'm getting:

x=t y=1

r(t)=ti+j
r'(t)=i+0
F(r(t))=ti+j

So, \int^{-1}_{1}<ti,j>dot<i,0> = \int^{-1}_{1}t dt = t^{2}/2 |^{-1}_{1} = 0

How am I doing here? Everything look okay?