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I can't think of such an example, which is a complete, non compact Riemannian manifold and has finite volume.

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2 Answers 2

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Take the graph of the function $x \mapsto e^{-x^2}$ and rotate it around the $x$-axis in $\mathbb{R}^3$. You get a submanifold of $ \mathbb{R}^3$ which can be endowed with the Euclidean metric of $ \mathbb{R}^3$. It's an easy exercise to show that it is complete, non compact and that its area is finite

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Any cusped (finite-volume --- so the volume in the cusp falls off exponentially, as the volume of the upper-half-space model of $\mathbb{H}^n$ falls of as $z\to\infty$) hyperbolic manifold. For example, the complement of a hyperbolic knot, or a punctured Riemann surface --- http://en.wikipedia.org/wiki/Hyperbolic_knot, http://en.wikipedia.org/wiki/Riemann_surface#Hyperbolic_Riemann_surfaces.

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