# Power Series for Volume of Balls in Riemannian Manifold

 P: 14 I'm trying to work out the following problem: Find the first two terms of the power series expansion for the volume of a ball of radius r centered at p in a Riemannian Manifold, M with dimension n. We are given that $$Vol(B_r(p)) = \int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v$$ (I'm ignoring the cut locus distance because we are only interested in small r anyway.) I assume that $S = S^{n-1} = \{v \in T_pM : |v| = 1 \}$. Clearly the constant term is 0, because a ball of radius 0 has no area in any manifold, so I think that they are really asking for the first two non-zero terms. To find the next term we need to calculate the derivative of $Vol(B_r(p))$ wrt r. $$\frac{\mathrm{d}}{\mathrm{d}r} Vol(B_r(p)) = \frac{\mathrm{d}}{\mathrm{d}r}\int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v = \int_S \frac{\mathrm{d}}{\mathrm{d}r} \left(\int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t\right) \mathrm{d}v = \int_S \det(d(exp_p)_{rv})r^{n-1} \mathrm{d}v$$ I can pass the derivative under the integral sign because S is compact and the inner integral is a continuous function of v. (Right?) So the second term is 0 also, unless n=1. Here's where I run into trouble. I'm not sure how to differentiate $\det(\mathrm{d}(exp_p)_{rv})$. I do know that $\mathrm{d}(exp_p)_{tv} \cdot tw = J(t)$ where $J(t)$ is the Jacobi Field along $\gamma_v$ defined by $J(0)=0$ and $J'(0)=w$. My guess is that the solution uses this along with some linear algebra trick to find the derivative. Thanks in advance for your help!
 P: 14 Thanks. That gets me the first term, and at least gives me an educated guess for the second term. I'm guessing that $\frac{d}{dr} \left( d(\exp_p)_{rv} \right) = ric_p(v)$. When you integrate this over the whole ball you end up getting the scalar curvature which is what was supposed to happen.