A solid spherical ball with a radius of $21 ext{ cm}$ is melted and recast into $n$ solid hemispherical bowls. The radius of each hemispherical bowl is half the radius of the original spherical ball. Find the total surface area of all the hemispherical bowls combined.

Question

A solid spherical ball with a radius of $21 	ext{ cm}$ is melted and recast into $n$ solid hemispherical bowls. The radius of each hemispherical bowl is half the radius of the original spherical ball. Find the total surface area of all the hemispherical bowls combined.

Accepted Answer

**Solution:**

1. Given the radius of the original solid sphere is $r = 21 	ext{ cm}$.

2. The radius of each hemispherical bowl is half of the original sphere's radius:
   $$r_{bowl} = \frac{21}{2} = 10.5 	ext{ cm}$$

3. Calculate the volume of the original sphere:
   $$V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (21)^3 = \frac{4}{3} \pi 	imes 9261 = 12348 \pi 	ext{ cm}^3$$

4. Calculate the volume of one hemispherical bowl:
   $$V_{bowl} = \frac{1}{2} 	imes \frac{4}{3} \pi r_{bowl}^3 = \frac{2}{3} \pi (10.5)^3 = \frac{2}{3} \pi 	imes 1157.625 = 771.75 \pi 	ext{ cm}^3$$

5. Number of hemispherical bowls formed:
   $$n = \frac{V_{sphere}}{V_{bowl}} = \frac{12348 \pi}{771.75 \pi} = \frac{12348}{771.75} = 16$$

6. Total surface area of one hemispherical bowl includes the curved surface area plus the base area:
   - Curved surface area of hemisphere:
     $$2 \pi r_{bowl}^2 = 2 \pi (10.5)^2 = 2 \pi 	imes 110.25 = 220.5 \pi 	ext{ cm}^2$$
   - Base area (circle):
     $$\pi r_{bowl}^2 = \pi 	imes 110.25 = 110.25 \pi 	ext{ cm}^2$$
   - Total surface area of one bowl:
     $$220.5 \pi + 110.25 \pi = 330.75 \pi 	ext{ cm}^2$$

7. Total surface area of all $n = 16$ bowls:
   $$16 	imes 330.75 \pi = 5292 \pi 	ext{ cm}^2$$

8. Using $\pi \approx 3.1416$:
   $$5292 	imes 3.1416 = 16632 	ext{ cm}^2$$

**Answer:** $16632 	ext{ cm}^2$

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