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Homework Help: Green's theorem question

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    For a > 0, let [itex]C_a[/itex] be the circle [itex]x^2 + y^2 = a^2[/itex] (counter-clockwise orientation). Let [tex]\textbf{F} : R^2[/tex] \ {0} [tex]\rightarrow R^2[/tex] be the following vectorfield:
    [tex]\textbf{F}\left(x,y\right) = F_1\left(x,y\right)\textbf{i} + F_2\left(x,y\right)\textbf{j}[/tex]

    Also given:
    [tex]\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = \frac{1}{\sqrt{x^2+y^2}}[/tex]
    [tex]\oint_{C_1} \textbf{F} \cdot d \textbf{r} = 1[/tex]

    [tex]\oint_{C_a} \textbf{F} \cdot d \textbf{r}[/tex]
    for arbitrary a > 0.

    2. Relevant equations
    Green's theorem:
    [tex]\oint_{C} \textbf{F} \cdot d \textbf{r} = \iint_R \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \, dA[/tex]

    3. The attempt at a solution
    It is obvious that we should use Green's theorem, even if it's not explicitly mentioned in the question, but I fear that I'm using it where it is not valid...

    Using Green's theorem directly I calculate:
    (R is the interior (surface) of the circle C_a)
    [tex]\oint_{C_a} \textbf{F} \cdot d \textbf{r} = \iint_R \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \, dA = \iint_R \frac{dA}{\sqrt{x^2+y^2}}[/tex]
    [tex]= \iint_R \frac{dA}{a} = \frac{1}{a} \times \text{surface of R} = \pi a[/tex]

    This answer is wrong, and my question is actually why?
    I don't need the actual answer to the question (I have it right here in fact) but I need to know why I cannot use green's theorem like this.

    I can see two possible reasons:
    1. F needs to be smooth (0 is not included in the domain of F)
    2. The [tex]\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}[/tex] part needs to be smooth (it's now undefined at 0)

    Which is the right reason? Or are they equivalent? I can't remember my teacher telling us F needs to be smooth but I expect he simply forgot...
  2. jcsd
  3. Nov 11, 2008 #2


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    Homework Helper
    Gold Member

    Hmmmm... does [tex]\frac{1}{\sqrt{x^2+y^2}}[/tex] Really equal [itex]\frac{1}{a}[/itex] everywhere in your region....or just on the boundary of the region?:wink:
  4. Nov 11, 2008 #3
    Oh wow, that was probably the worst mistake I ever made LOL!

    Thanks for spotting that... :p
  5. Nov 12, 2008 #4
    from the definition of the problem and since the line integral would not be defined at r=0 my idea is that the line integral is not 0 but [tex] 2\pi [/tex]

    it is a similar problem to 'Cauchy integral formula' on the complex plane but know we miss the ' i'
  6. Nov 12, 2008 #5
    Actually the answer is 1 + 2 pi (a - 1)

    Let D be the region enclosed by the curves [tex]C_1[/tex] and [tex]C_a[/tex].
    For a < 1 we have:
    [tex]\oint_{C_1} \textbf{F} \cdot \textbf{dr} - \oint_{C_a} \textbf{F} \cdot \textbf{dr} = \iint_D \frac{1}{\sqrt{x^2+y^2}}\,dx\,dy = \int_0^{2\pi} \int_a^1 dr\,d\theta = 2\pi \left(1 - a\right)[/tex]
    And since the first integral on the left hand side is 1 (see problem statement) we have:
    [tex]\oint_{C_a} \textbf{F} \cdot \textbf{dr} = 1 - 2\pi(1-a) = 1 + 2\pi (a - 1)[/tex]

    And a similar argument for a > 1 yields the same value.
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