# Volume of a complete, simply connected Riemannian manifold of constant negative curvature

Given an $n$-dimensional complete simply connected Riemannian manifold of constant negative curvature $-1$, I need to show that $$\operatorname{vol} B_{r}(p) = \alpha_{n-1} \int_{0}^{r} \sinh^{n-1}(t)\,dt.$$ where the author uses $\alpha_{n-1}$ to denote the volume of $S^{n-1}$.

The hint is the use the following corollary:

For a Riemannian manifold $M$ and point $p \in M$, if $\exp_p |_{B_r (o_p)}$ is a diffeomorphism, then $$\operatorname{vol}B_r(p) = \int_{B_r(o_p)} \sqrt{\det g_{ij} \circ \exp_p} \,dx^1 \cdots dx^m = \int_{S^{m-1}} dS^{m-1} \int_0^r \theta(t,u)\,dt$$ where (if I'm seeing it correctly on the page before) $\{e_1, \ldots, e_{m-1}, u\}$ is an o.n.b for $T_pM$ and $\theta(t,u) = \sqrt{\det \langle J_i(t), J_j(t) \rangle}$ for Jacobi fields $J_i$ along the normal geodesic $\gamma_{u}$ satisfying $J_i (0), \nabla J_i(0)=e_i$.

So I guess my question is, in particular, about that $\theta(t,u)$ term. What do we know about the inner product of two Jacobi fields in relation to the curvature, and why is $\det(\langle J_i(t), J_j(t)\rangle) = \sinh^{2(m-1)}(t)$ (for $1\leq i,j\leq m-1$)?

From Jacobi field equation, $$J'' + R(c'(t),J)c'(t)=0$$
If $M$ has a constant curvature $-1$, if $c$ is a normal geodesic, and if $J=f_i(t)e_i(t)$ where $f(0)=0$ and $e_i(t)$ is a parallel orthonormal along $c(t)$ then $$f_i'' f_i-f_i^2=0$$ and $$f_i={\rm sinh}\ (t)$$
So$${\rm det}\ (g(J_i,J_j))= ({\rm sinh}\ (t))^{2(m-1)}$$