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Line integral of the second kind

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the polar curve [itex]r = 3\sqrt{\cos{2\varphi}},\ -\frac{\pi}{4}\leq\varphi\leq\frac{\pi}{4}[/itex]. Find the area of the surface enclosed by the curve using line integral of the second kind.

    2. Relevant equations


    3. The attempt at a solution
    According to Green's theorem: if [itex]F(x,y)[/itex] and [itex]G(x,y)[/itex] are continous on a smooth closed graph ([itex]r(\varphi)[/itex] in this case), then [itex]\oint_L F(x,y)\mathrm{d}x + G(x,y)\mathrm{d}y = \iint_D G_x(x,y) - F_y(x,y)\mathrm{d}x\mathrm{d}y[/itex].

    The textbook, however, is very ineloquent as to what exactly [itex]F[/itex] and [itex]G[/itex] are and there are no examples given with polar curves. Do I have to parameterize the curve? I don't understand. :wideeyed:
     
    Last edited: May 11, 2015
  2. jcsd
  3. May 11, 2015 #2

    Svein

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    This looks like a slightly offbeat version of Stokes' theorem.
    upload_2015-5-11_18-31-13.png This is the curve. Observe:
    1. x = r*cos(φ), y = r*sin(φ)
    2. sin(-φ) = -sin(φ), so the part above the x-axis is equal to the part below
    3. You do not need any fancy theorems to do this integral.
     
  4. May 11, 2015 #3

    Ray Vickson

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    The area is ##\iint_D 1 \, dx dy##, so you need to devise ##F(x,y), G(x,y)## that satisfy ##G_x - F_y = 1 ## everywhere (or, at least, inside the curve). Can you think of some convenient functions ##F,G## that have that property?
     
  5. May 11, 2015 #4

    Zondrina

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    The area of the region is what you mean.

    The area of a polar region ##r = f(\theta)## is given by:

    $$\frac{1}{2} \int_a^b f^2(\theta) \space d\theta$$

    This will allow you to solve the question quite easily as you already have the limits.

    On a side note, I think you are referring to the reverse implication of Green's theorem, which allows you to compute areas. Since the area of a region ##D## is given by:

    $$\text{Area}(D) = \iint_D \space dA$$

    We wish to choose the functions ##P## and ##Q## satisfying Green's theorem such that:

    $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$$

    There are several possibilities for choosing these functions, which all lead to the same result for the area of a region ##D##. Lets choose ##P = -\frac{1}{2}y## and ##Q = \frac{1}{2}x##. The following result can then be deduced directly from Green's theorem:

    $$\text{Area}(D) = \frac{1}{2} \oint_C xdy - y dx = \iint_D \space dA$$

    If you were to "go backwards" from ##r = f(\theta)## by using polar co-ordinates, you could also use the above result to get the exact same answer, i.e:

    $$\text{Area}(D) = \frac{1}{2} \oint_C xdy - y dx = \frac{1}{2} \int_a^b f^2(\theta) \space d\theta = \iint_D \space dA$$
     
  6. May 13, 2015 #5
    Would the area then be
    [tex]A = \int_{ -\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{r(\varphi)} r\mathrm{d}r\mathrm{d}\varphi[/tex]
    Edit: Yes, it is, thanks, I get it now.
     
    Last edited: May 13, 2015
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