Cylindrical Sector Surface Area F6Fb3A

1. **State the problem:** We need to find the exact surface area of a cylindrical sector with radius $6$ cm, central angle $240^\circ$, and height $10$ cm. 2. **Understand the shape:** The shape is a sector of a cylinder, which means it is a portion of a full cylinder cut by two radii forming a $240^\circ$ angle. 3. **Surface area components:** The total surface area consists of: - The curved lateral surface area of the sector - The two rectangular side faces formed by the height and the two radii edges - The top and bottom sector faces 4. **Formulas and calculations:** - Full cylinder lateral surface area: $2\pi r h$ - Sector lateral surface area: $\frac{\theta}{360^\circ} \times 2\pi r h$ - Area of full circle base: $\pi r^2$ - Area of sector base: $\frac{\theta}{360^\circ} \times \pi r^2$ - Area of each rectangular side face: radius $\times$ height Given: $$r = 6, \quad h = 10, \quad \theta = 240^\circ$$ 5. **Calculate lateral surface area of sector:** $$\text{Lateral area} = \frac{240}{360} \times 2\pi \times 6 \times 10 = \frac{2}{3} \times 120\pi = 80\pi$$ 6. **Calculate area of top and bottom sector faces:** $$\text{Area of one sector base} = \frac{240}{360} \times \pi \times 6^2 = \frac{2}{3} \times 36\pi = 24\pi$$ Since there are two bases (top and bottom): $$2 \times 24\pi = 48\pi$$ 7. **Calculate area of the two rectangular side faces:** Each side face is a rectangle with dimensions radius $6$ cm and height $10$ cm. $$\text{Area of one side face} = 6 \times 10 = 60$$ There are two such faces: $$2 \times 60 = 120$$ 8. **Sum all areas for total surface area:** $$\text{Total surface area} = 80\pi + 48\pi + 120 = 128\pi + 120$$ **Final answer:** $$\boxed{128\pi + 120 \text{ cm}^2}$$