A glaciologist is modeling a glacier as a truncated cone with a top radius of $3r$ and a bottom radius of $r$, with a height of $h$. If the volume of the glacier is $V$, express $V$ in terms of $r$ and $h$ and find the volume when $r = 2$ and $h = 5$.

["A glaciologist modeling a glacier as a truncated cone (frustum of a cone) seeks an accurate formula for its volume, essential for estimating ice mass and melt projections. The glacier is modeled with a bottom radius of $ r $, top radius of $ 3r $, and vertical height $ h $. We aim to derive the volume $ V $ of this shape in terms of $ r $ and $ h $, then compute $ V $ for $ r = 2 $ and $ h = 5 $.", "A frustum of a cone has volume given by the classical formula: \[V = \frac{1}{3} \pi h (R_{\ ext{bottom}}^2 + R_{\ ext{bottom}} R_{\ ext{top}} + R_{\ ext{top}}^2)\] where $ R_{\ ext{bottom}} $ and $ R_{\ ext{top}} $ are the radii of the base and top surfaces, respectively.", "Substituting $ R_{\ ext{bottom}} = r $ and $ R_{\ ext{top}} = 3r $: \[V = \frac{1}{3} \pi h \left( r^2 + r(3r) + (3r)^2 \right)\] \[V = \frac{1}{3} \pi h \left( r^2 + 3r^2 + 9r^2 \right) = \frac{1}{3} \pi h (13r^2)\] \[V = \frac{13}{3} \pi r^2 h\]", "This is the general expression for the volume of the frustum-shaped glacier.", "Now, substitute $ r = 2 $ and $ h = 5 $: \[V = \frac{13}{3} \pi (2)^2 (5) = \frac{13}{3} \pi \cdot 4 \cdot 5 = \frac{13}{3} \pi \cdot 20 = \frac{260}{3} \pi\]", "Thus, the volume is $ \frac{260}{3} \pi $ cubic units when $ r = 2 $ and $ h = 5 $.", "This model provides glaciologists with a computationally efficient yet physically meaningful approximation for glacier volumes, especially useful in remote regions where direct measurement is challenging. By representing complex ice formations as frustums, researchers can better estimate ice loss and contribute to climate change predictions.", "Final answer: \[\boxed{\frac{260}{3}\pi}\]"]