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Line Integral

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data
    (i) Evaluate

    [tex]\int_C \dfrac{-ydx + xdy}{9x^2 + 16y^2} [/tex]

    when C is the ellipse

    [tex]\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1[/tex]

    (ii) Use the ans to (i) to evaluate the integral along C' = ellipse:

    [tex] \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I have done (i) but have no clue about (ii). great thnx for any help.
     
  2. jcsd
  3. Feb 12, 2008 #2

    HallsofIvy

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    Okay, what did you get for (i)? I can think of a number of ways that might help you answer (ii) but since you have shown no work at all I don't know which way would be appropriate for you. Do you know Green's Theorem?
     
  4. Feb 12, 2008 #3
    got pi/6 for i. didn't use green's for that. the other way is easier. for ii u can't use green's since it would either be too complicated or impossible to integrate. yes i know green's thm. sorry for not showing any work. i just don't know what to do for ii.
     
  5. Feb 12, 2008 #4

    HallsofIvy

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    That's not at all what I get for (i). And the problem with (ii) is I don't know what theorems or methods you have available. I do notice that the integrand is defined everywhere except at (0,0) which is inside both ellipses.
     
  6. Feb 13, 2008 #5
    checked it and still got the same. is it allowed to say 9x^2 + 16y^2 = 144 on that integral? that's something i'm using.

    for ii, if u could mention some of the methods u have in mind, i might recognize it as something given in class. thnx
     
  7. Feb 14, 2008 #6

    HallsofIvy

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    My apologies. I just screwed up (1) by copying part of the problem incorrectly. Yes, [itex]\pi/6[/itex] is the correct answer.

    My point about the integrand not being defined at (0,0) was that you make a cut from the outer ellipse to the inner, integrate around one ellipse, then up that cut to the other and back. Then you are integrating around a curve that does NOT have (0,0) in its interior. The integral around that path will be 0, showing that the integral around the two ellipses is the same.
     
  8. Feb 14, 2008 #7
    and why is it 0? what does the integrand not being defined at (0,0) has to do with that? by a cut do u mean to make a line connecting the 2 ellipses? could u explain more plz?
     
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