# Thread: [SOLVED] Area of a hyperbolic circle

1. ## [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.

2. 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 $\frac{1}{(k-\rho)^2}\pi \rho^2$.

3. 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.

4. 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).

5. 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.