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Area and line integrals

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm trying to set up a couple integrals.

    Suppose you have a closed loop and you want to find its area. You don't know what the shape of the loop is. All you know is that the length of the loop is L.


    3. The attempt at a solution

    These are my integrals. I just need to know if they are set up correctly:

    [tex]A=\int_{0}^{2\pi}\int_{0}^{r}f(r,\theta)r dr d\theta[/tex]
    [tex]L=\int_{0}^{2\pi}f(r,\theta)r d\theta[/tex]

    Where A is the area enclosed in the loop, and L is the length of the loop. And [tex]f(r,\theta)[/tex] is the equation of the line.

    Is that correct? (It's been such a log time since I've done this.)

    Thanks.
     
  2. jcsd
  3. Sep 2, 2008 #2

    Dick

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    I think if it's been a long time you might want to look this stuff up before you post. The first one looks more like a volume than an area, the second looks more like an area of a cross section at constant r than a length. The first one is not even grammatical. You never want to have the same variable in the limits of the integral as the integration variable. The integration variable is a dummy.
     
  4. Sep 2, 2008 #3
    I did. My calculus book didn't have exactly what I was looking for, so I had to "figure it out."

    How so? It's in two dimensions, like area. And if [tex]f(r,\theta)=1[/tex], then the integration works out to [tex]\pi r^2[/tex] - the area of a circle, which is what I expect.


    Then how would you do it?
     
  5. Sep 2, 2008 #4
    And the second one, if [tex]f(r,\theta)=1[/tex], works out to [tex]2\pi r[/tex], which is the circumference, as I expected.
     
  6. Sep 3, 2008 #5

    HallsofIvy

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    What exactly is [itex]f(r, \theta)[/itex]? You say it is the "equation of the line" but that can't be true. It is not an equation!
     
    Last edited: Sep 3, 2008
  7. Sep 3, 2008 #6
    That's what I'm trying to figure out and why I'm here asking for help.
     
  8. Sep 3, 2008 #7

    Dick

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    Maybe you could give us the exact question you are trying to solve?
     
  9. Sep 3, 2008 #8
    Find the maximum area A enclosed by a line of Length L using Lagrange multipliers.
     
  10. Sep 3, 2008 #9

    HallsofIvy

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    Sigh. A "line of length l" does not enclose an area. I might think you were talking about finding the closed path of given length that encloses maximum area but that requires the calculus of variations, not Lagrange multipliers. What does the problem really say?
     
  11. Sep 3, 2008 #10
    Here you go - quoted directly:

    I guess I better go back and tell the professor that his problem is flawed, since A "line of length l" does not enclose an area.
     
  12. Sep 3, 2008 #11

    Dick

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    Hmmm. I didn't know that you could prove the isoperimetric inequality that way, but it looks like you might. Take the curve to be of the parametric form f(t)=(x(t),y(t)). The length is the integral of sqrt(x'(t)^2+y'(t)^2) and the area is the integral of (1/2)*(x(t)*y'(t)-x'(t)*y(t)) for a closed curve (up to a sign).
     
    Last edited: Sep 3, 2008
  13. Sep 3, 2008 #12
    There's no time dependency in this problem.
     
  14. Sep 3, 2008 #13
    Looking a little closer, it appears as f=A, because:

    So that means my integral is going to have to be something like [tex]f=A=\int_{area} (something) da[/tex]
     
  15. Sep 3, 2008 #14

    Dick

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    t isn't time. It's a parameter that describes the curve. As in the unit circle is described by x(t)=cos(t) and y(t)=sin(t) where t is in [0,2pi]. I think you are intended to use Euler-Lagrange, not Lagrange multipliers (though the two are related). Review the Euler-Lagrange equations. You do need calculus of variations.
     
  16. Sep 3, 2008 #15
    Unfortunately, I have to use Lagrange multipliers. That's the point of the exercise.

    And the professor said it would be easier to use polar coordinates. We also can't assume that the line is a circle at the beginning. The line encloses an arbitrary shaped area. And the solution is supposed to show that it's a circle. If I start off with the equation of a circle, then I haven't shown anything (professor's words).
     
  17. Sep 3, 2008 #16

    Dick

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    The circle was just an example. Write a polar curve as r=f(theta) (so the t in the problem is theta). Then the area is the integral of (1/2)*f(theta)^2*dtheta and the length is the integral of sqrt(f(theta)^2+f'(theta)^2)*dtheta. Sure, you want to put a Lagrange multiplier on the arc length, since it's your constraint. I still would review Euler-Lagrange. I don't see how the polar coordinates are making it easier, but maybe I'm missing something.
     
  18. Sep 3, 2008 #17
    Looks like I was looking in the wrong place in my calculus book. I found the following equations for area and length.

    If the equation of my line is [tex]r=g(\theta)[/tex], then my area integral is [tex]A=\int \frac{1}{2} g^2(\theta) d\theta[/tex], and all the way around would, of course, be [tex]A=\oint \frac{1}{2} g^2(\theta) d\theta[/tex].

    The length of the line (all the way around) would be [tex]L=\oint \sqrt{g^2(\theta)+g'^2(\theta)} d\theta[/tex]
     
  19. Sep 3, 2008 #18
    So now setting the the Lagrange Method:

    [tex]\Lambda=\frac{1}{2}\oint g^2(\theta)d\theta + \lambda(\oint \sqrt{g^2(\theta)+g'^2(\theta)} d\theta)-L)[/tex]
     
  20. Sep 3, 2008 #19
    Now what? Is it:

    [tex]\frac{\partial\Lambda}{\partial g}=0[/tex]
    [tex]\frac{\partial\Lambda}{\partial \lambda}=0[/tex]

    OR

    [tex]\frac{\partial\Lambda}{\partial \theta}=0[/tex]
    [tex]\frac{\partial\Lambda}{\partial \lambda}=0[/tex]

    ?
     
  21. Sep 3, 2008 #20

    Dick

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    Now it's Euler-Lagrange. And it's the first set of equations if you interpret the partial derivative as a variation. You should be able to produce a differential equation for g that it's fairly easy to show g=constant satisfies for a choice of lambda. But there are other solutions. They are circles whose center is not at (0,0), but it's not clear to me how to prove that easily in polar coordinates. Can you?
     
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