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Thread: integral

  1. #1
    Junior Member
    May 2009


    Help me calculate this integral
    $\displaystyle \int_0^{2\pi} \sqrt{a^2\sin^2 x+ b^2 \cos^2 x}dx$
    Thanks so much.
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    Circumference of Ellipse

    There is no known method for solving this integral explicitly. Your best bet is using a numerical method. Consider...

    The length of an arc defined by the vector equation $\displaystyle \vec{x}(t)$ from t=A to t=B is $\displaystyle L=\int_A^B\sqrt{\left(\frac{dx}{dt}\right)^2+\left (\frac{dy}{dt}\right)^2}dt$

    So the integral you have posed is equivalent to asking the circumference of the ellipse defined by $\displaystyle \vec{x}(t)=(-a\cos t,b\sin t)$. Take the derivatives to see what I mean. As you probably know, the circumference of an ellipse cannot be given in closed form.

    You can either use an infinite series or approximation, which can be found here: Ellipse - Wikipedia, the free encyclopedia
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