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Line Integral

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Let C be the arc of y=x2 from (0,0) to (1,1). Evaluate [tex]\int[/tex]xdx


    2. Relevant equations
    C1:
    x=t
    y=t2
    -1 [tex]\leq[/tex] t [tex]\leq[/tex] 1 so -1 [tex]\leq[/tex] x [tex]\leq[/tex] 1

    3. The attempt at a solution

    dx is x' dt, right?

    For some reason, I just can't figure this out. Any help?
     
    Last edited: Dec 9, 2008
  2. jcsd
  3. Dec 9, 2008 #2

    gabbagabbahey

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    ummm [tex]\int x dx[/tex] is not even a line integral; it's just an ordinary integral....do you mean [tex]\int_{\mathcal{P}} \vec{x} \cdot \vec{ds}[/tex]?
     
  4. Dec 10, 2008 #3
    No, that is what he has down...Except the C is where the p is on yours. I did not know how to do that. I figured it out, though.

    since dx=xdt and x=t,

    [tex]\int[/tex] from 0 to 1 of t2dt is 1/3.
     
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