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Math Help - [SOLVED] Area of a hyperbolic circle

  1. #1
    Senior Member Pinkk's Avatar
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    [SOLVED] Area of a hyperbolic circle

    Prove that a hyperbolic circle with hyperbolic radius R in the hyperbolic upper half plane is 2\pi (\cosh R - 1)

    The problem suggests to place a coordinate system at the Euclidean of the circle and work out the double integral in polar coordinates. However, and I believe I must be messing up somewhere, if we let the Euclidean center of the circle be (h,k) and have Euclidean radius \rho<k, then we have in polar coordinates centered at the Euclidean center x= \rho\cos\theta + h and y=\rho\sin\theta + k, and so we have the hyperbolic area of the circle being:

    \int_{0}^{2\pi}\int_{0}^{\rho}\, \frac{rdr d\theta}{(r\sin\theta + k)^{2}}

    However, this integral seems to be undefined because along the line somewhere, I get a \cot\theta, which is undefined at 0 and 2\pi.

    Are my parameterizations wrong? Or are they correct and this integral does exist? Should I not deal with the Euclidean radius and approach it some other way? Thanks.
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  2. #2
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    Quote Originally Posted by Pinkk View Post
    However, this integral seems to be undefined because along the line somewhere, I get a \cot\theta, which is undefined at 0 and 2\pi.
    It is perfectly well defined: you integrate \frac{1}{y^2} on a bounded domain where y>k-\rho>0 so this is a bounded function on a bounded subset... I can even say that the integral is less than \frac{1}{(k-\rho)^2}\pi \rho^2.
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  3. #3
    Senior Member Pinkk's Avatar
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    So my limits of integration are incorrect? I use this identity:

    \int \frac{x}{(x+a)^2}dx = \frac{a}{a+x}+\ln |a+x|+C

    for the first integral (the dr part) upon which I get:
    \int_{0}^{2\pi} \frac{\frac{k}{\rho\sin\theta + k} + \ln\frac{\rho\sin\theta + k}{k} - 1}{\sin^{2}\theta}d\theta

    So this is defined? If so, what does this integral equal? When I use integral calculators, a \cot\theta appears, and therefore the whole expression becomes undefined. So either my limits for integration are incorrect or these integral calculators are all wrong. Thanks.
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  4. #4
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    Quote Originally Posted by Pinkk View Post
    So my limits of integration are incorrect? I use this identity:

    \int \frac{x}{(x+a)^2}dx = \frac{a}{a+x}+\ln |a+x|+C

    for the first integral (the dr part) upon which I get:
    \int_{0}^{2\pi} \frac{\frac{k}{\rho\sin\theta + k} + \ln\frac{\rho\sin\theta + k}{k} - 1}{\sin^{2}\theta}d\theta

    So this is defined?
    Yes, it is; you can see that by considering an asymptotic expansion to order 2 when \theta\to 0.

    But this integral is indeed complicated; maybe you shouldn't use polar coordinates. Using the initial Cartesian coordinates, the area is (letting y=\rho+z)

    2\int_{-\rho}^\rho \frac{\sqrt{\rho^2-z^2}}{(k+z)^2}dz.

    This may be easier to compute (at least it looks nicer).
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  5. #5
    Senior Member Pinkk's Avatar
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    Quote Originally Posted by Laurent View Post
    Yes, it is; you can see that by considering an asymptotic expansion to order 2 when \theta\to 0.

    But this integral is indeed complicated; maybe you shouldn't use polar coordinates. Using the initial Cartesian coordinates, the area is (letting y=\rho+z)

    2\int_{-\rho}^\rho \frac{\sqrt{\rho^2-z^2}}{(k+z)^2}dz.

    This may be easier to compute (at least it looks nicer).
    This does not really help much; integral calculator shows that imaginary parts come up in the integral.

    Is there an easier way to prove that the hyperbolic area is 2\pi [\cosh(R) - 1]?

    EDIT: Nevermind, found a proof that made sense.
    Last edited by Pinkk; May 4th 2010 at 06:11 PM.
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