A sphere is melted and molded into a solid cylinder. If the radius of both solids is equal, then find the ratio of the total surface area of the sphere to the total surface area of the cylinder.

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**Solution:**

1. Let the radius of both the sphere and the cylinder be *r*.

2. Volume of the sphere:
   $$V_{sphere} = \frac{4}{3} \pi r^3$$

3. Volume of the cylinder:
   $$V_{cylinder} = \pi r^2 h$$

4. Since the sphere is melted and molded into the cylinder, volumes are equal:
   $$\frac{4}{3} \pi r^3 = \pi r^2 h$$

5. Simplify to find height *h* of the cylinder:
   $$h = \frac{4}{3} r$$

6. Total surface area (TSA) of the sphere:
   $$TSA_{sphere} = 4 \pi r^2$$

7. Total surface area of the cylinder (including top and bottom):
   $$TSA_{cylinder} = 2 \pi r (r + h)$$

8. Substitute *h*:
   $$TSA_{cylinder} = 2 \pi r \left(r + \frac{4}{3} right) = 2 \pi r 	imes \frac{7}{3} r = \frac{14}{3} \pi r^2$$

9. Ratio of TSA of sphere to TSA of cylinder:
   $$\frac{4 \pi r^2}{\frac{14}{3} \pi r^2} = \frac{4}{1} 	imes \frac{3}{14} = \frac{12}{14} = \frac{6}{7}$$

**Answer:** The ratio is 6 : 7.

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