1. **Problem Statement:** A hemispherical bowl of radius 7 cm is full of water. We need to find the area of the wet surface (the inner surface of the hemisphere in contact with water). Then, the water is emptied into a hollow cylinder with diameter 8 cm and height 7 cm. We must determine if the water will overflow the cylinder. 2. **Formulas and Important Rules:** - Surface area of a hemisphere (curved surface only) is given by $$2\pi r^2$$. - Volume of a hemisphere is $$\frac{2}{3}\pi r^3$$. - Volume of a cylinder is $$\pi r^2 h$$. - Diameter $$d$$ relates to radius $$r$$ by $$r=\frac{d}{2}$$. 3. **Step 1: Calculate the wet surface area of the hemisphere** Given radius $$r=7$$ cm, $$\text{Wet surface area} = 2\pi r^2 = 2\pi (7)^2 = 2\pi \times 49 = 98\pi \text{ cm}^2$$. 4. **Step 2: Calculate the volume of water in the hemisphere** $$V_{hemisphere} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (7)^3 = \frac{2}{3}\pi \times 343 = \frac{686}{3}\pi \text{ cm}^3$$. 5. **Step 3: Calculate the volume of the cylinder** Diameter $$d=8$$ cm, so radius $$r=\frac{8}{2}=4$$ cm, height $$h=7$$ cm. $$V_{cylinder} = \pi r^2 h = \pi (4)^2 \times 7 = \pi \times 16 \times 7 = 112\pi \text{ cm}^3$$. 6. **Step 4: Compare volumes to check for overflow** Volume of water $$= \frac{686}{3}\pi \approx 228.67\pi$$ cm³. Volume of cylinder $$= 112\pi$$ cm³. Since $$228.67\pi > 112\pi$$, the water volume is greater than the cylinder volume. 7. **Conclusion:** The water will overflow when poured into the cylinder because the volume of water from the hemisphere exceeds the cylinder's capacity. **Final answers:** - Wet surface area of hemisphere: $$98\pi \approx 307.88$$ cm². - Water will overflow the cylinder.