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Total length of ellipse

  1. Jun 12, 2005 #1
    I'm asked to show that the total length of the ellipse
    x = a sin x
    y = b cos x, a>b>0 is

    [tex] L = 4a \int_{1}^{pi/2} \sqrt{1-e^2sin^2x} dx
    where e is the eccentricity of the ellipse (e = c/a, where c = [tex]\sqrt{a^2-b^2}[/tex]

    I've tried a whole bunch of algebraic and trignometric manipulation but I'm getting the feeling I'm overlooking something. How would you approach this problem? I've tried working backwards and eventually got something of the form:

    [tex] L = \int_{1}^{pi/2} \sqrt{2b^2cos^2x+c^2cos^2x-b^2(cos 2x)} dx

    but i figured I should post here for ideas as well.
  2. jcsd
  3. Jun 12, 2005 #2


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    HINT:The total length is 4 times the arc in the first quadrant.Choose the wise parametrization and use the definition of the eccentricity (the modulus of the elliptic integral).

  4. Jun 12, 2005 #3


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    This can be a challenging problem. Checking out how http://home.att.net/~numericana/answer/ellipse.htm#elliptic [Broken] have solved this might give you some ideas on how to approach it.
    Last edited by a moderator: May 2, 2017
  5. Jun 12, 2005 #4


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    There's no big deal,Bob

    [tex] L_{\mbox{ellipse}}=4L_{\mbox{arc in the first quadrant}} [/tex] (1)

    Parametrization (polar elliptical)

    [tex] \left\{\begin{array}{c}x(\phi)=a\cos\phi \\ y(\phi)=b\sin\phi \end{array} \right [/tex] (2)

    [tex] \left\{\begin{array}{c} (dx)^{2}(\phi)=a^{2}\sin^{2}\phi \ (d\phi)^{2}\\ (dy)^{2}(\phi)=b^{2}\cos^{2}\phi \ (d\phi) ^{2}\end{array} \right [/tex] (3)

    [tex] L_{\mbox{arc in the first quadrant}} =\int \sqrt{(dx)^{2}+(dy)^{2}} =\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}\sin^{2}\phi+b^{2}\cos^{2}\phi} \ d\phi [/tex] (4)

    Now i'll leave it to the OP to finish it.


    P.S.His answer is wrong.
  6. Jun 13, 2005 #5
    wow. that was down and funky. I figured it out in four lines after an embarrassing number of scrap sheets before. Thanks people.

    I'll give several hints to those with the same question in the future:

    The integral is from 0 to 2pi originally, but you can cut it down to 0 to pi/2 and multiply by four. recall that c^2is really just 1 - b^2/a^2 and when at the end of a deserted road, question if you can multiply by 1 expressed in a twisty way.
  7. May 25, 2010 #6
    the length of any geometrical arc is given by [ intg of ( (dx .dx +dy.dy)^1/2 ) ] within proper limits... that is according to "pythogorean theorem"..

    on simplifying the above thing we get... intg [(1+(dy/dx)^2)^1/2) ].dx

    Now, comming to the problem i.e., length of an ellipse, calculate (dy/dx) from the standard equation in terms of "x" and put it in the above integral and solve thus obtained integral within proper limits....thas alll........... :)

    if u do tht u will get approx L= pi[ 3(a+b)- {(3a+b)(a+3b)}^1/2 ]
  8. May 29, 2010 #7
    By "within limits" i mean the limits of corresponding variable by which the integrand is multiplied i.e. the factor by which the integrand is multiplied which is a differential magnitude.... ( eg., limits of "x" if the integrand is multiplied by "dx" and limits of "theta" if the integrand is multiplied by "d(theta)" )
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