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(Line) Integral

  1. Feb 27, 2006 #1
    I have been working on the following line integral:



    [tex] \int_{T}^- {(-x^2y)dx + (y^2x)dy}[/tex]

    where T is the closed curve consisting of the semi-circle x^2 + y^2 = a^2 (y>0) and the segment (-a,a)

    I will tackle this in two steps:

    1)

    solve x^2 + y^2 = a^2 (y>0) for y and substitute into Integral

    this gives [tex] \int_{-a}^a {- x^2 * root (a^2-x^2) }dx[/tex]

    2) integrating along the x-axis - gives zero.

    My problem: How can one solve the above integral in part 1? Help is much appreciated!
     
    Last edited: Feb 27, 2006
  2. jcsd
  3. Feb 27, 2006 #2

    quasar987

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    Mmmh. It seems you forgot about the integral wrt to y.

    What does [itex] \int_{C}y^2xdy[/itex] transforms into after your paramatrization by x?

    Add it in your calculations, and you'll see that you don't have any integral to evaluate.
     
  4. Feb 27, 2006 #3
    if I substitute in I receive:


    (a^2-x^2)x*dy

    but dy= 0 (y^2 + x^2 = a^2 ........ 2y * dy/dx + 0 = O .......dy/dx =0)

    so therefore I left it out

    but I have the impression that I do this wrong. thanks for your help!
     
  5. Feb 27, 2006 #4

    quasar987

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    This line is complete giberrish to me. What I say is, just use what you know about parametrization: if x(t), y(t) [itex]t_1 \leq t \leq t_2[/itex] is your parametrization, then the line integral [itex]\int_C P(x,y)dx + Q(x,y)dy[/itex] is

    [tex]\int_{t_1}^{t_2}P(x(t),y(t))\frac{dx}{dt}dt + \int_{t_1}^{t_2}Q(x(t),y(t))\frac{dy}{dt}dt [/tex]

    This is what you have here, except that the parametrization you chose has x for a param. variable, instead of the "outsider" t. Your parametrization is simply x(x) = x, y(x) = root{a²-x²}, [itex]-a \leq x \leq a[/itex]. That's what you did so far, except you forgot about the integral wrt y. Now can you do it?
     
    Last edited: Feb 27, 2006
  6. Feb 27, 2006 #5
    sorry about being so slow...but I still don't understand it.

    this is the new integral:

    [tex]\int_{-a}^{a}(-x^2) root( a^2 - x^2) dx + \int_{a}^{a}-x * root (a^2-x^2)dx = \int_{-a}^{a} ((-x^2-x)* root(a^2-x^2)dx [/tex]

    where [tex] \frac {dy} {dx} = \frac {-2x} {2* root (a^2-x^2)} [/tex]

    but how can one go on from here?

    Also, since the second integral is an odd function couldn't one just ignore it, since it's value is zero anyways (I just realised that)?
     
    Last edited: Feb 27, 2006
  7. Feb 27, 2006 #6
    Why don't you just use the plane polar coordinates? It might reduce the algebra.

    Put [itex]x = r\cos t [/itex] and [itex]y = r\sin t[/itex]
    So [itex]dx = -r\sin t dt[/itex] & [itex]dy = r\cos t dt[/itex] for t in the range [itex]0[/itex] to [itex]\pi[/itex].
    You have the radius r = a.

    Hope this helps a little...
     
    Last edited: Feb 27, 2006
  8. Feb 27, 2006 #7

    benorin

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    use instead the parameterization [tex]x(t)=a\cos{t},y(t)=a\sin{t},0\leq t\leq \pi[/tex] so that [tex]dx=-a\sin{t}dt,dy=a\cos{t}dt[/tex] the line integral in part (1) becomes

    [tex] \int_{T} {(-x^2y)dx + (y^2x)dy} = \int_{t=0}^{\pi} (a^4\cos ^2{t}\sin ^2{t} + a^4\cos ^2{t}\sin ^2{t} )dt = 2a^4\int_{t=0}^{\pi} \cos ^2{t}\sin ^2{t}dt [/tex]

    you can get it from there
     
    Last edited: Feb 27, 2006
  9. Feb 27, 2006 #8

    benorin

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    By the way, use \sqrt{a^-x^2} instead of root(a^2-x^2)
     
  10. Feb 27, 2006 #9
    Wow, you have practically solved the problem :biggrin:.
     
  11. Feb 27, 2006 #10

    siddharth

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    which is against PF guidelines ...

    Benorin, while your contribution is greatly appreciated, can you try not to give away complete solutions to the question?
     
  12. Feb 27, 2006 #11

    benorin

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    Yes, but I likely should not have.
     
  13. Feb 27, 2006 #12

    benorin

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    I fixed it. Sorry: my bad.
     
  14. Feb 27, 2006 #13

    Tom Mattson

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    Thank you very much, siddharth. :smile: :approve:
     
  15. Feb 27, 2006 #14

    quasar987

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    Hey mmm37. Sorry to have lead you to believe the integrals wouls cancel. Instead, they add, which makes a mess. You should go with what Reshma and benorin suggested.
     
  16. Feb 27, 2006 #15
    Thanks very much all of you! I don't know what I would do without your help! It is very much appreciated!

    I am working on solving the integral (in polar coordinates) and I will post the answer once I got it.

    Mel
     
    Last edited: Feb 27, 2006
  17. Feb 28, 2006 #16
    On a slightly related note!

    There is a theorem that applies that is quite handy, called Green's Theorem (I wholeheartedly recommend looking it up)

    Basically, if dQ/dx does not equal dP/dy (meaning the integral isn't exact), then the integral is equal to the double integral over the bounded region of dQ/dx-dPdy. In this case, that leaves you with the double integral of x^2+y^2, which when converted to polar is a trivial integral. I really suggest looking up Green's (and of course, stoke's) theorem.
     
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