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Line Integral

  1. Jul 9, 2012 #1
    Consider that the curve C (IR3) is the intersection between x2+2z2=2 and y=1.

    Calculate the line integral:

    MmJ6C.png


    Note: The curve C is traversed one time in the counter-clockwise direction (seen from the origin of IR3).

    My attempt:

    http://i.imgur.com/5EsJO.png

    Can someone check? Thanks! :)
     
    Last edited by a moderator: Jul 9, 2012
  2. jcsd
  3. Jul 10, 2012 #2

    haruspex

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    If I have the question right, a lot of simplification is possible.
    Given y = 1, the integral wrt y is zero, and y can be set to 1 elsewhere.
    I also notice there's integral x.dz - integral z.dx. Hmmm... what do you think that might simplify to? Think of the geometric meaning, or of integration by parts.
     
  4. Jul 10, 2012 #3
    But it is still correct?
     
  5. Jul 10, 2012 #4

    haruspex

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    The answer looks right. I haven't verified all the working (because it can all be done in couple of lines).
     
  6. Jul 10, 2012 #5
    Thanks! Btw, I understand your simplifications, but I have to do this way... ;)
     
  7. Jul 10, 2012 #6

    HallsofIvy

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    You have curve C defined in an xyz- coordinate system but you have u, y, and z in your integral. Did you intend
    [tex]\int_C yzdx+ x^2dy- (x+ z)dz[/tex]?

    Assuming that, since y= 1, a constant, dy= 0 so we can ignore the middle term.

    I would be inclined to parameterize the curve by [itex]x= \sqrt{2}cos(\theta)[/itex], [itex]y= 1[/itex], [itex]z= sin(\theta)[/itex].

    Then [itex]dx= -\sqrt{2}sin(\theta)d\theta[/itex] and [itex]dz= cos(\theta)d\theta[/itex].

    The integral becomes
    [tex]\int_0^{2\pi} sin(\theta)(-\sqrt{2}sin(\theta)d\theta)- (\sqrt{2}cos(\theta)+ sin(\theta))cos(\theta)d\theta[/tex]
    [tex]= \int_0^{2\pi}[-\sqrt{2}(sin^2(\theta)+ cos^2(\theta))- \sqrt{2}sin(\theta)cos(\theta)]d\theta[/tex]
    [tex]= -\sqrt{2}\int_0^{2\pi}[1+ sin(\theta)cos(\theta)]d\theta[/tex]

    The answer you give, t/2- (1/4)sin(2t), can't possibly be correct because it is a function of t and there was not even a "t" in the problem! Did you intend to evaluate that at t= 0 and [itex]2\pi[/itex]?
     
    Last edited: Jul 10, 2012
  8. Jul 10, 2012 #7

    haruspex

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    The way I read it, the equations involving t are explanatory notes. The answer it gives earlier is "= 0".
     
  9. Jul 10, 2012 #8

    HallsofIvy

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    Thank you. I had to keep scrolling from side to side!
     
  10. Jul 10, 2012 #9
    Yeah, my answer is 0
     
  11. Jul 10, 2012 #10

    HallsofIvy

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    I've gone back and edited my first response, replacing the incorrect "2"s with [itex]\sqrt{2}[/itex]. However, I still do not get 0 as the integral.
     
    Last edited: Jul 10, 2012
  12. Jul 10, 2012 #11
    I found my mistake. I wrote:
    It's not 0, you're right. Thanks a lot! ;)
     
  13. Jul 10, 2012 #12

    haruspex

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    And I've found mine :blushing:.
    As HofI, I reasoned that we can get rid of the y terms, leaving:
    [itex]\int_C zdx- (x+ z)dz = \int_C (zdx-xdz) + zdz[/itex]
    Around a closed loop, zdz goes to zero (at each z, a dz cancels a -dz).
    [itex]\int_C zdx[/itex] and [itex]\int_C xdz [/itex] are each the area inside the loop. So I thought these cancelled too. But I overlooked that one will effectively run the opposite direction around the loop, so will add, not cancel. So the whole reduces to twice the area of the ellipse.
     
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