1. **State the problem:** We have two concentric semicircles with inner radius $r=6$ cm and outer radius $R=15$ cm, and a total horizontal length of 216 cm. We need to find the volume of the solid formed by revolving the ring-shaped segment between these semicircles around the horizontal axis. 2. **Formula used:** The volume of a solid of revolution formed by revolving a semicircle of radius $r$ around its diameter is half the volume of a sphere of radius $r$, which is $$V = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.$$ For the ring-shaped segment, the volume is the difference between the volumes of the larger and smaller semicircle solids. 3. **Calculate volumes:** - Volume of larger semicircle solid: $$V_{large} = \frac{2}{3} \pi R^3 = \frac{2}{3} \pi (15)^3 = \frac{2}{3} \pi 3375 = 2250 \pi.$$ - Volume of smaller semicircle solid: $$V_{small} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6)^3 = \frac{2}{3} \pi 216 = 144 \pi.$$ 4. **Find volume of ring-shaped solid:** $$V = V_{large} - V_{small} = 2250 \pi - 144 \pi = (2250 - 144) \pi = 2106 \pi.$$ 5. **Interpretation:** The total horizontal length 216 cm is consistent with the diameter of the largest semicircle (2 × 15 = 30 cm) scaled or given for context, but the volume depends only on the radii. **Final answer:** $$\boxed{2106 \pi \text{ cubic centimeters}}.$$