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Homework Help: Yet another conservative field question

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to calculate:
    [tex]\oint_{\Gamma} \vec{F}\cdot d \vec{r}[/tex]

    \vec{F} = \frac{-y \vec{i} + x \vec{j}}{x^2+y^2}
    where [itex]\Gamma[/itex] is the positive direction circle:
    a. x2 + y2 = 1
    b. (x-2)2 + y2 = 1

    2. Relevant equations

    [tex]\int_{C} \nabla f \cdot d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a))[/tex]
    and/or (??)
    [tex]\int_{C} \nabla f \cdot d \vec{r} = \int_{a}^{b} \nabla f (\vec{r}(t)) \cdot \vec{r}'(t) dt [/tex]

    3. The attempt at a solution
    I'm totally lost on this one.
    I found out that [itex]f = -\arctan{\frac{x}{y}}[/itex], and that in polar coords:
    [tex]\vec{F} = \left( \frac{-\sin{\theta}}{r} , \frac{\cos{\theta}}{r} \right)[/tex]
    But what to do now?
    I tried doing the stuff required by the relevant equations but nothing seems to work, for both circles. What am I missing here?

    Thanks in advance.
  2. jcsd
  3. Sep 17, 2009 #2


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    Your second relevant equation can always, at least in principle, be used to calculate the line integral. But for an integral of the type

    [tex]\oint_C \vec F \cdot\, d\vec R[/tex]

    you always want to check if the theorems about conservative fields are valid. For example, in 2D with a closed path, are the hypotheses of Green's Theorem satisfied? If so, then what?

    And with respect to parameterizing one of those integrals and working it directly you might want to let t be the polar central angle. For example in the second problem you might try:

    [tex]x = 2+\cos(t),\ y = sin(t)[/tex]
  4. Sep 17, 2009 #3
    Well, in this case, everything should be valid.
    The potential is [itex]f = -\arctan{\frac{x}{y}}[/itex], and the path is closed. Which puts me right in the beginning of this question.
    Unless I got something wrong here.
  5. Sep 17, 2009 #4


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    There are hypotheses to the Green's and independence of path theorems. Read them carefully. If the hypotheses apply you can use the theorems and if they don't you have to calculate directly.
  6. Sep 17, 2009 #5
    The curve has to be positively oriented - yes.
    Curve has to be smooth - yes.
    Curve has to be simple (not cross itself) - yes.
    Curve has to be closed - yes, it's a circle.
    the partial derivatives have to be continuous D, which is the inside of the circle. Here is a problem - they're not, since (0,0) is undefined but it's inside the circle. This means that I can't use green's theorem, and I must calculate directly.
    Question is - how. What is my [itex]\vec{r}(t)[/itex]? (cost,sint)?
    In that case, my integral is:
    [tex]\int_{0}^{2 \pi} \frac{\sin^2{t} + \cos^2{t}}{r} dt = \int_{0}^{2 \pi} \frac{dt}{r}[/tex]
    Which is wrong (and what do I do about the r? assume it to be 1?).
  7. Sep 17, 2009 #6


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    You ask: What is my [tex]\vec{r}(t)[/tex] ? (cost,sint)?

    Does x = cos(t) y = sin(t) parameterize the circle? If r was some number a instead of 1 wouldn't you use x = a cos(t) y = a sin(t)?

    And can you use Green's theorem on the other circle or do you have to parameterize it too?
  8. Sep 17, 2009 #7
    ok so let's start over.
    my function is:
    \vec{F} = \frac{-y \vec{i} + x \vec{j}}{x^2+y^2}
    parameterizing the circle:
    x= a cost
    y= a sint
    a2 = x2 + y2
    [itex]\vec{r}(t) = (a \cos{t}, a\sin{t})[/itex]
    plugging it into the function:
    \vec{F}(\vec{r}(t)) = \left( \frac{-a\sin{t}}{a^2} , \frac{a\cos{t}}{a^2} \right) = \left( \frac{-\sin{t}}{a} , \frac{\cos{t}}{a} \right)
    now I need to differentiate the vector:
    [itex]\vec{r}'(t) = (-a \sin{t}, a\cos{t})[/itex]
    \oint_{\Gamma} \vec{F}\cdot d \vec{r} = \oint_{\Gamma} \left( \frac{-a\sin{t}}{a^2} , \frac{a\cos{t}}{a^2} \right) = \left( \frac{-\sin{t}}{a} , \frac{\cos{t}}{a} \right) \cdot (-a \sin{t}, a\cos{t}) = \int_{0}^{2\pi} dt = 2\pi - 0 = 2\pi[/tex]
    Which is the correct answer. Is this the way to do it?
    Seems so simple now.

    btw - it also means that the radius of the circle (a) doesn't matter anything, since it cancels out?

    I'll try the other circle now.
  9. Sep 17, 2009 #8
    Ok, for the second circle, trying to solve the integral directly with [itex]x = 2+\cos(t),\ y = sin(t)[/itex] seems to complicated.
    Since the function is not defined on (0,0), and the circle never reaches there (centered at 2,0 with radius 1) I can use green's theorem.
    \oint_{C} Pdx + Qdy = \iint_{D} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA
    in this case:
    [tex]\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}[/tex]
    [tex]\iint_{D} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA = 0[/tex]
    The answer should be 0, but I got the feeling that my method is not correct.

    Another way would be using the potential:
    [itex]f = -\arctan{\frac{x}{y}}[/itex]
    since the start and end points are the same, f(b)-f(a) = 0.
    again - i'm not sure that what i'm doing here is correct, since y=0 and that's a problem (division by zero).
  10. Sep 17, 2009 #9


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    2 pi is correct for the first one. But I have been trying to get you to see for yourself that a = 1. The circle is of radius 1. There should be no a in your work and it is just happenstance that the a's cancelled out.

    Your work for the second is correct. The reason your potential function seems to have a problem at y = 0 is that your potential function isn't correct. Check it.
  11. Sep 17, 2009 #10
    Ok, thanks a lot.
    Seems like I'm getting it right, even though I'm not always sure I am. :)
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