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Homework Help: Calculate line integral across square curve with vertices (1,1) (-1,1) (-1,-1) (1,-1)

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    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 [tex]\int[/tex]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).

    2. Relevant equations


    3. The attempt at a solution

    I've found that this can be rewritten as [tex]\int[/tex]ab 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=[tex]^{+}_{-}[/tex]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 :)
  2. jcsd
  3. Dec 8, 2009 #2


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    Re: Calculate line integral across square curve with vertices (1,1) (-1,1) (-1,-1) (1

    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)

    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.
  4. Dec 8, 2009 #3
    Re: Calculate line integral across square curve with vertices (1,1) (-1,1) (-1,-1) (1

    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


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

    How am I doing here? Everything look okay?
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