A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. The volume of steel used in making the bowl is

### Concept & Volume of a Hollow Hemisphere The material volume of a hollow object is found by subtracting the internal empty volume from the total external volume. $$V = \frac{2}{3}\pi(R^3 - r^3)$$ Where $R$ is the outer radius and $r$ is the inner radius. ### Step-by-Step Solution 1. **Identify Radii:** Inner radius ($r$) = $4\text{ cm}$. Thickness = $0.5\text{ cm}$. Outer radius ($R$) = $r + \text{thickness} = 4 + 0.5 = 4.5\text{ cm}$. 2. **Apply the Formula:** $V = \frac{2}{3}\pi(4.5^3 - 4^3)$ 3. **Calculate Cubes:** $4.5^3 = \left(\frac{9}{2}\right)^3 = \frac{729}{8} = 91.125$ $4^3 = 64$ Difference: $91.125 - 64 = 27.125$ 4. **Calculate Final Volume:** $V = \frac{2}{3} \times \frac{22}{7} \times 27.125$ $V = \frac{44}{21} \times 27.125$ $V = \frac{1193.5}{21} \approx 56.833...\text{ cm}^3$ ### Exam Strategy & Shortcut To simplify multiplication, use fractions: $27.125 = \frac{217}{8}$. $V = \frac{2}{3} \times \frac{22}{7} \times \frac{217}{8} = \frac{44}{21} \times \frac{217}{8}$. Notice $217$ divided by $7$ is exactly $31$. $V = \frac{22 \times 31}{3 \times 4} = \frac{11 \times 31}{6} = \frac{341}{6} \approx 56.83$. ### Common Pitfall Simply multiplying the inner surface area by the thickness ($\approx 0.5 \times 2\pi r^2$) is an approximation and yields an incorrect lower value, as it ignores the expanding volume of the thicker outer layer. ### Final Answer Therefore, the correct answer is **56.83 cm^3**.