Volume of a geodesic ball This may be embarassingly simple, but I can't see it.
Let $M$ be a Riemannian manifold of dimension $n$; fix $x \in M$, and let $B(x,r)$ denote the geodesic ball in $M$ of radius $r$ centered at $x$.  Let $V(r) = \operatorname{Vol}(B(x,r))$ be the Riemannian volume of $B(x,r)$.  It seems to be the case that for small $r$, $V(r) \sim r^n$, i.e. $V(r)/r^n \to c$ with $0 < c < \infty$.  How is this proved, and where can I find it?
Given a neighborhood $U \ni x$ and a chart $\phi : U \to \mathbb{R}^n$, certainly $\phi$ has nonvanishing Jacobian, hence (making $U$ smaller if necessary) bounded away from 0.  So $\operatorname{Vol}(\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))) \sim r^n$.  But I do not see how to relate the pullback $\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))$ of a Euclidean ball to a geodesic ball in $M$.
 A: It's simple, all right.  As I realized not long after posting (and as Hans also suggested), the key is the exponential map.  The tangent space $T_x M$ gets an inner product space structure from the Riemannian metric; we can isometrically identify it with $\mathbb{R}^n$.  Now $\exp_x : \mathbb{R}^n \to M$ is a diffeomorphism on some small ball $B_{\mathbb{R}^n}(0,\epsilon)$; on this ball, straight lines map to length-minimizing geodesics (see Do Carmo, Riemannian Geometry, Proposition 3.6), and thus Euclidean balls map to geodesic balls of the same radius.  Taking $\epsilon$ smaller if necessary, we can assume the Jacobian of $\exp_x$ is bounded away from $0$ and $\infty$ on $B_{\mathbb{R}^n}(0, \epsilon)$; thus for $r < \epsilon$ we have that $\operatorname{Vol}(B(x,r))$ is comparable to $\operatorname{Vol}(B_{\mathbb{R}^n}(0,r)) \sim r^n$.
A: A Google search for the title of the question finds this article where the first five coefficients in the series expansion of the volume (in powers of $r$) are computed. The first term is the same as the Euclidean volume (proportional to $r^n$, in other words); then come higher-order corrections depending on the curvature.
