1. **State the problem:** We have a large cone of height 39 mm and base radius 6 mm. It is split horizontally into a smaller cone on top (height 13 mm, radius 2 mm) and a frustum below (height 26 mm, top radius 2 mm, bottom radius 6 mm). We need to find the volume of the frustum. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ where $r$ is the radius and $h$ is the height. 3. **Volume of the large cone:** $$V_{large} = \frac{1}{3} \pi (6)^2 (39) = \frac{1}{3} \pi \times 36 \times 39 = 468 \pi$$ 4. **Volume of the smaller cone:** $$V_{small} = \frac{1}{3} \pi (2)^2 (13) = \frac{1}{3} \pi \times 4 \times 13 = \frac{52}{3} \pi$$ 5. **Volume of the frustum:** The frustum volume is the volume of the large cone minus the volume of the smaller cone: $$V_{frustum} = V_{large} - V_{small} = 468 \pi - \frac{52}{3} \pi = \left(468 - \frac{52}{3}\right) \pi = \frac{1404 - 52}{3} \pi = \frac{1352}{3} \pi$$ 6. **Calculate the numerical value:** Using $\pi \approx 3.1416$, $$V_{frustum} \approx \frac{1352}{3} \times 3.1416 = 450.67 \times 3.1416 \approx 1415$$ **Final answer:** The volume of the frustum is approximately **1415** cubic millimeters.