Geodesic Polar Coordinates and Area of a Circle in Hyperbolic Space

Question

I am trying to calculate the area of a circle of radius $R$ in the hyperbolic plane. I know that the answer is supposed to be $4\pi(\sinh(\frac{R}{2}))^2$. However, I am not getting this. To calculate the area, I am trying to use hyperbolic polar coordinates. I know that $d\mu(z)=(2\sinh r)drd\phi$, so if $B_R$ is the ball of radius $R$, then we have that the area is given by: $$ \iint_{B_R}d\mu(z)=\int_0^\pi\int_0^R(2\sinh r)drd\phi=2\pi(\cosh(R)-1) $$ However, this does not seem correct to agree with what several books and the internet have told me that the area of a circle in hyperbolic space is. Is it the case that $2\pi(\cosh(R)-1)=4\pi(\sinh(\frac{R}{2}))^2$ or where did I go wrong.

As a related aside, I want to similarly calculate the perimeter of a hyperbolic circle, and I similarly wish to use geodesic polar coordinates, and I know that $ds^2=dr^2+(2\sinh r)^2d\phi^2$, but the squaring of the differentials always confuses me, so I think I should do something along the lines of the perimeter is given by $$ \int_{\partial B_R}ds=\int_{\partial B_R}\sqrt{dr^2+(2\sinh r)^2d\phi^2} $$ but I'm not entirely sure what I'm supposed to be integrating in this case, and I fear that perhaps I might not get $2\pi \sinh(R)$ which is what the world has seemed to agreed that the perimeter of a hyperbolic circle is.

Any and all help is greatly appreciated.

Note: In regards to the perimeter calculation, I figured it out. We have that $$ \int_{\partial B_R}ds=\int_0^\pi\sqrt{(2\sinh r)^2+(\frac{dr}{d\phi})^2}d\phi $$

Now in geodesic polar coordinates, we have that our curve is $r=R$, so we will get that $\frac{dr}{d\phi}=0$, and this is just $$ \int_0^\pi 2\sinh(R)=2\pi\sinh(R) $$

Answer 1

In fact

$2\pi(\cosh R-1)=4\pi(\sinh^2(R/2))$

is true. It comes from applying the hyperbolic double-argument identity

$\cosh(2u)=\cosh^2u+\sinh^2u=2\cosh^2u-1=1+2\sinh^2u.$

See for instance Wolfram MathWorld.

We can also get the form $4\pi\sinh^2(R/2)$ directly from the integral. To this end apply another double-argument formula to render

$\sinh(r)=2\sinh(r/2)\cosh(r/2)$

and plug in

$u = 2\sinh(r/2), du = \cosh(r/2)dr.$

This converts the integral to

$2\int_0^{2\pi}\int_0^{2\sinh(R/2)}u(du)d\theta$

which will integrate directly to give $4\pi\sinh^2(R/2).$