1. **State the problem:** We have a solid hemisphere H with radius $x$ cm and volume $6174\pi$ cm³. A bowl is made by removing a smaller solid hemisphere from H, leaving a uniform thickness of 2 cm. We need to find the total surface area of the bowl in terms of $\pi$. 2. **Find the radius $x$ of the original hemisphere:** The volume $V$ of a hemisphere is given by the formula: $$V = \frac{2}{3} \pi r^3$$ Given $V = 6174\pi$, substitute and solve for $x$: $$6174\pi = \frac{2}{3} \pi x^3$$ Cancel $\pi$ from both sides: $$6174 = \frac{2}{3} x^3$$ Multiply both sides by $\frac{3}{2}$: $$x^3 = 6174 \times \frac{3}{2} = 6174 \times 1.5 = 9261$$ Take the cube root: $$x = \sqrt[3]{9261} = 21$$ 3. **Determine the radius of the inner hemisphere:** The bowl has uniform thickness 2 cm, so the inner radius is: $$r_{inner} = x - 2 = 21 - 2 = 19$$ 4. **Calculate the total surface area of the bowl:** The bowl consists of: - The outer curved surface area of the larger hemisphere - The inner curved surface area of the smaller hemisphere - The flat circular base (the ring-shaped area between the two radii) The curved surface area of a hemisphere is: $$A_{curved} = 2 \pi r^2$$ Calculate outer curved surface area: $$A_{outer} = 2 \pi (21)^2 = 2 \pi \times 441 = 882 \pi$$ Calculate inner curved surface area: $$A_{inner} = 2 \pi (19)^2 = 2 \pi \times 361 = 722 \pi$$ Calculate the flat circular base area (ring area): $$A_{base} = \pi (21^2 - 19^2) = \pi (441 - 361) = \pi \times 80 = 80 \pi$$ 5. **Add all areas to get total surface area:** $$A_{total} = A_{outer} + A_{inner} + A_{base} = 882 \pi + 722 \pi + 80 \pi = 1684 \pi$$ **Final answer:** $$\boxed{1684 \pi \text{ cm}^2}$$