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When the integral of total derivative[tex] \oint d U = 0 [/tex]?,

  1. Sep 23, 2011 #1
    when the integral of total derivative[tex] \oint d U = 0 [/tex]????,

    when the integral of total derivative
    [tex] \oint d U = 0 [/tex]????, and ¿why is zero?



    [tex] \oint d U \neq 0 [/tex] ??
  2. jcsd
  3. Sep 23, 2011 #2
    Re: question

    Alex: are you integrating the identity? And, what is the region over which you are
  4. Sep 24, 2011 #3

    rude man

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    Re: question

    The closed line integral dU = 0 if dU is an exact differential. U is then a state function and by definition if you end up where you started you get zero.

    Think of a closed line integral of gravitational or electrostatic force.
  5. Sep 25, 2011 #4
    Re: question

    An example , a classical one, I think, of a form that is not exact is that of w=dz/z

    defined on the circle |z|=1, embedded in R-{(0,0)}, or C-{(0,0)} .
  6. Sep 25, 2011 #5
    Re: question

    my problem is

    [tex] \oint d( \frac{dA}{dy} exp(A) ) [/tex]

    where [tex]A=|y|[/tex]
    but, y is a angular coordinate...betwen [tex] )-\infty, \infty([/tex]
    but my problem is that
    [tex] \frac{d|y|}{dy} = -1 , y \in )-\pi,0(, )\pi,2\pi(, etc[/tex]
    [tex] \frac{d|y|}{dy} = 1 , y \in )0,\pi(, )2\pi,3\pi(, etc[/tex]
    [tex] \frac{d|y|}{dy} = undefinided , y =-\pi,0,\pi.....k \cdot \pi[/tex]
    but, if i integrate

    [tex] \oint exp(|y|) [/tex] ¿why is it Non zero, if i too started and end in the same point (0 and 2pi por example)
    Last edited: Sep 25, 2011
  7. Sep 27, 2011 #6
    Re: question

    By a theorem; I think Stokes or one of its corollaries, your last integral equals zero iff , f (exp|y| , in our case) as a differential form, is exact, or, equivalently, if there is an F
    in the region of definition, with dF= exp|y|.

    I'm sorry, I still don't fully understand why y is positive in some regions, and negative in others. Do you have an explicit formula for it?

    Still, if you know that your y is defined as you said, the only reason I can see for why
    it is not zero, is that , it does not have a global antiderivative.
  8. Oct 1, 2011 #7
    Re: question

    question :

    can i to say [tex] d \theta = \frac{1}{x^2+y^2} (-y dx +x dy ) [/tex]
    since i am to integrate on a circle, and , since

    [tex] x = 1 \cdot \cos (\theta) ; y = 1 \cdot \sin (\theta) [/tex]

    and then

    [tex] \oint e^{|\theta|} d \theta = \oint e^{| \sin^{-1} y|} \frac{1}{x^2+y^2} (-y dx +x dy ) \neq 0 [/tex] , since is not globally defined in [tex] x, y = 0,0 [/tex] ???????????????????????????
    Last edited: Oct 1, 2011
  9. Oct 1, 2011 #8
    Re: question

    The standard way (the one I know : ) ) of doing line integrals/ contour integration is
    by parametrizing the contour:

    x=cosθ , so that dx=cosθdθ

    y=sinθ , so that dy=sinθdθ

    And then integrate from 0 to 2∏ ; since the singularity happens at (0,0), which is
    not in the contour, you don't need to worry about this. Still, another way of testing

    dθ:= (xdy -ydy)/(x2+y2)

    is exact, is by integrating around a close contour. If the form is exact, it would then
    integrate to 0, by an extension of the fundamental theorem of calculus.
  10. Oct 2, 2011 #9
    Re: question

    firts , sorry by many questions,

    in my case, i need to integrate on te circle

    but in a text says:

    [tex](xdy -ydy)/(x^ 2+y^ 2)[/tex] it is not exact, since its integral along the unit circle is not 0...... [tex] \int ^\pi_{-\pi} d\theta = 2 \pi[/tex]....
    this is a argument for says that [tex] \oint e^{|\theta|} d\theta [/tex] is not zero????????????????
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