1. **Problem:** Name the geometric solid formed when a right triangle is rotated around the x-axis. The solid formed is a **cone** because rotating a right triangle around one of its legs (the x-axis) creates a cone. 2. **Problem:** Find the volume of the container formed by rotating the right triangle around the x-axis. Given formula for cone volume: $$V = \frac{1}{3} \pi r^2 h$$ From the graph: - Height $$h = 4$$ inches (along the x-axis from 2 to 6) - Radius $$r = 2$$ inches (maximum y-value) Calculate volume: $$V = \frac{1}{3} \pi (2)^2 (4)$$ $$V = \frac{1}{3} \pi \times 4 \times 4$$ $$V = \frac{1}{3} \pi \times 16$$ $$V = \frac{16}{3} \pi$$ cubic inches 3. **Problem:** Draw a cross-section of the cone parallel to the y-axis. A cross-section parallel to the y-axis is a triangle because slicing the cone vertically along the y-axis shows the original right triangle shape. 4. **Problem:** Draw a cross-section of the cone when slicing along the x-axis. A cross-section along the x-axis is a circle because slicing the cone horizontally produces circular cross-sections. 5. **Problem:** Find the volume of the solid formed when the right triangle is rotated around the y-axis. Rotating around the y-axis forms a different cone with: - Height $$h = 2$$ (distance along y-axis from 0 to 2) - Radius $$r = 4$$ (distance along x-axis from 2 to 6) Volume formula: $$V = \frac{1}{3} \pi r^2 h$$ Calculate volume: $$V = \frac{1}{3} \pi (4)^2 (2)$$ $$V = \frac{1}{3} \pi \times 16 \times 2$$ $$V = \frac{32}{3} \pi$$ cubic inches 6. **Problem:** Draw a cross-section of the solid from question 5 parallel to the x-axis. A cross-section parallel to the x-axis is a circle because slicing the cone horizontally produces circular cross-sections.