# [SOLVED] Area of a hyperbolic circle

• Apr 29th 2010, 08:28 PM
Pinkk
[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, 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.
• Apr 30th 2010, 04:18 AM
Laurent
Quote:

Originally Posted by Pinkk
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 $\frac{1}{(k-\rho)^2}\pi \rho^2$.
• Apr 30th 2010, 06:48 AM
Pinkk
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.
• Apr 30th 2010, 11:37 AM
Laurent
Quote:

Originally Posted by Pinkk
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).
• May 4th 2010, 03:54 PM
Pinkk
Quote:

Originally Posted by Laurent
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.