Volume of a ball in hyperbolic space I'm trying to derive the volume of the $n$-dimensional R-ball in hyperbolic space, $\mathcal{B}_\mathbb{H}(y; R) = \{x \in \mathbb{H}^n : \lVert x \rVert_y \le R\}$, using the Riemannian volume formula and integrating in the exponential chart at $y \in \mathbb{H}^n$: 
$$\mathcal{Vol}(\mathcal{B}_\mathbb{H}) = \int_\mathcal{X} d\mathcal{M} = \int_{B_y} \sqrt{\det g(\exp_y(u))} du,$$
where $B_y = \log_y \mathcal{B}_\mathbb{H}$, $g(x)$ is the metric tensor expressed in some orthonormal frame, and $\exp_y(\cdot)$ denotes the exponential map. I'm using [1, Sec 3.1] as a reference for this approach.
However, I'm getting a different result than what I expect, which is $\propto \int_0^R \sinh^{n-1}(r) dr$ (wiki), and I wonder what I'm doing wrong. I assume it's something that has to do with how I'm handling the metric tensor, but I couldn't pinpoint it precisely. Do I misinterpret the volume formula from above?

My approach is to use the Poincaré ball model because that's the model I'm familiar with where the metric tensor has a nice expression: $g(x) = \Big(\frac{2}{1 - \lVert x \rVert^2}\Big)^2 I_n$.  Because it only depends on $x$ via its extrinsic norm, the integral above separates per angular coordinates and radial distance. I want to compute the latter which amounts to computing $\sqrt{\det g(x)}$ for $x$ with $d(0, x) = r$ and then integrating over $r$ from $0$ to $R$. We have,
$$d(0, x) = \cosh^{-1}\Big(1 + 2 \frac{\lVert x \rVert^2}{1 - \lVert x \rVert^2}\Big) = r \implies \lVert x \rVert^2 = \frac{\cosh(r) - 1}{\cosh(r) + 1},$$
and
$$\sqrt{\det g(x)} = \Big(\frac{2}{1 - \lVert x \rVert^2}\Big)^n.$$
Finally, substituting the former into the latter gives $\sqrt{\det g(x)} = \big(\cosh(r) + 1\big)^n$ which is different than the expected $\sinh^{n-1}(r)$.

[1]: Pennec, Xavier. "Intrinsic statistics on Riemannian manifolds: Basic tools for geometric measurements." Journal of Mathematical Imaging and Vision 25.1 (2006): 127.
 A: This is a consequence of not converting into polar coordinates correctly. While in the ball model we do have
$$ ds^2 = \left( \frac{2}{1-\sum_i x_i^2} \right)^2 \sum_i dx_i^2 , $$
to calculate the volume of a ball, you first want to switch into polars; the calculation works in the same way as for ordinary spherical coordinates (put $x_1 = \rho \sin{\theta_1} \dotsb \sin{\theta_{n-1}} , \dotsc , x_n = \rho \cos{\theta_1} $, then $dx_i = () d\rho + \dotsb $), and you have
$$ ds^2 = \left( \frac{2}{1-\rho^2} \right)^2 (d\rho^2 + \rho^2 d\Omega^2), $$
where $d\Omega^2$ is the spherical volume element $d\theta_1^2 + \sin^2{\theta_1} \, d\theta_2^2 + \dotsb$, independent of $\rho$.
We can now go one of two ways: either find the appropriate integral, and change variables afterwards, or find an the geodesic distance and obtain the geodesic polar coordinates. I'll do the latter, since that's the way you've tried. As you note, the arclength in the polar direction is given by
$$ r = \int_0^{\rho} \frac{2}{1-t^2} dt = \log{\left(\frac{1+\rho}{1-\rho}\right)} = 2\arg\tanh{\rho} , $$
which gives
$$ \rho = \tanh{(r/2)} $$
as the appropriate substitution to make to obtain geodesic polars. In particular,
$$ d\rho = \frac{1}{2} \operatorname{sech}^2{(r/2)} \, dr , $$
so 
$$ \left( \frac{2}{1-\rho^2} \right)^2 d\rho^2 = \left( \frac{2}{1-\tanh^2{(r/2)}} \right)^2 \left(\frac{1}{2} \operatorname{sech}^2{(r/2)} \right)^2 dr^2 = dr^2 . $$
Meanwhile, the rest of the metric is
$$ \left( \frac{2\rho}{1-\rho^2} \right)^2 d\Omega^2 = \sinh^2{r} \, d\Omega^2 . $$
Hence we end up with the metric of the hyperboloid model, so the determinant is obviously the same.
On the other hand, integrating directly in polar coordinates gives
$$ \iint_{B_R} \sqrt{\det{g}} d\rho d\Omega = (\dots) \int_0^{P} \left( \frac{2}{1-\rho^2} \right)^n \rho^{n-1} \, d\rho ,  $$
where $P$ is the value of $\rho$ that gives the appropriate geodesic distance $R$, namely $P = \arg\tanh{(R/2)}$. Making the same substitution as above, it is straightforward to show that again we end up with $\sinh^{n-1}{r} \, dr$ in the integrand.
