A hyperbolic 3-manifold has finite volume if and only if it is either closed or has toroidal boundary and it is not homeomorphic to $T^2\times I$.


This statement is from [3-Manifold Groups, page 18 (the link is editted)](https://www.uni-regensburg.de/Fakultaeten/nat_Fak_I/friedl/papers/3-manifold-groups-final-version-031115) by Matthias Aschenbrenner, Stefan Friedl and Henry Wilton, it seems that the three references in the book toward this statement only give partial results (when the boundary components are already cusps).


**Edits**: The precise statement in my opinion should be:


Let $M$ be a compact three dimensional manifold with incompressible toroidal boundary (possibly none). If the interior of $M$ admits a hyperbolic structure, then $M$ either has finite volume, or is homeomorphic to $T^2\times I$.


Thanks for any solutions or hints.



Volume of hyperbolic 3-manifolds with toroidal boundary
I think you are trying to ask the following question.



> 
> Suppose that $M$ is a compact connected oriented three-manifold. Suppose that $M^\circ$, the interior of $M$, admits a hyperbolic metric. Then when must this hyperbolic metric have finite volume?
> 
> 
> 


As Ryan points out, if $M$ is closed, then $M^\circ = M$ is compact and thus has finite volume. Also, as you note, the interior of $M = T^2 \times I$ admits (many) hyperbolic metrics, but all have infinite volume. There is another such manifold: namely the solid torus $D^2 \times S^1$.


I think that the place where you are confused (please correct me if I am wrong) is the case where $M$ has a boundary of higher genus. Here $M^\circ$ again always has infinite volume.


