Someone asked me this question, and I was embarrassed to not know the answer: is the volume of Moduli space with respect to the Teichmuller metric finite? The answer is "yes" when we replace Teichmuller metric with WeilPetersson metric, but the geometry of the two spaces is quite different.

The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric. The reason is the theorem of Royden, that the Kobayashi metric on Teich(S) coincides with the Teichmuller metric, and the fact that the moduli space $M(S)$ associated to S has a nice compactification $\overline{M(S)}$, the DeligneMumford compactification. The argument goes as follows: For a stable curve Z in $\overline{M(S)}$ with k nodes, you can find a neighborhood U of Z such that U is locally $\Delta^n /G$, where $\Delta$ is the unit disc in $\mathbb{C}$ and $G$ is a finite group. Then $U\cap M(S)$ is locally isomorphic to $((\Delta^{*})^{k} \times \Delta^{nk} )/ G$. The volume of $(\Delta^{*})^{k} \times \Delta^{nk} $ near the origin is finite in the Kobayashi metric. Since inclusions contract the Kobayashi metric it follows that there is a small neighborhood V of $Z \in \overline{M(S)}$ such that volume of $V\cap M(S)$ is finite. The result now follows by compactness of $\overline{M(S)}$. You can look at Curt McMullen's paper : http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf for more details and references. (Proof of Theorem 8.1) 


The answer is again yes. See the proof of theorem 8.1 in Curtis McMullen's paper "The moduli space of Riemann surfaces is Kahler hyperbolic". 

