The outer radius of a spherical container is 5 cm and the thickness of the container is 2 cm . Find the volume of the metal content of the shell .

Answer

To find the volume of the metal content of the spherical shell, we need to calculate the volume of the outer sphere and subtract the volume of the inner sphere. Let's go through the steps to solve the problem.

### Step-by-Step Solution:

1. **Identify the outer radius and thickness:**
- The outer radius \( R \) of the spherical container is given as 5 cm.
- The thickness of the container is given as 2 cm.

2. **Calculate the inner radius:**
- The inner radius \( r \) can be calculated by subtracting the thickness from the outer radius.
\[ r = R - \text{thickness} = 5 \, \text{cm} - 2 \, \text{cm} = 3 \, \text{cm} \]

3. **Calculate the volume of the outer sphere:**
- The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- Substituting the outer radius:
\[ V_{\text{outer}} = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi (125) = \frac{500}{3} \pi \, \text{cm}^3 \]

4. **Calculate the volume of the inner sphere:**
- Using the same formula for the inner sphere:
\[ V_{\text{inner}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = \frac{108}{3} \pi = 36 \pi \, \text{cm}^3 \]

5. **Calculate the volume of the metal content of the shell:**
- The volume of the metal content of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere:
\[ V_{\text{metal}} = V_{\text{outer}} - V_{\text{inner}} = \left(\frac{500}{3} \pi - 36 \pi\right) \]
- To combine these, convert \( 36 \pi \) to a fraction with a denominator of 3:
\[ 36 \pi = \frac{108}{3} \pi \]
- Now, substitute this into the equation:
\[ V_{\text{metal}} = \left(\frac{500}{3} \pi - \frac{108}{3} \pi\right) = \frac{392}{3} \pi \, \text{cm}^3 \]

6. **Final Calculation:**
- To get a numerical value, we can use \( \pi \approx 3.14 \):
\[ V_{\text{metal}} \approx \frac{392}{3} \times 3.14 \approx 410.67 \, \text{cm}^3 \]

### Final Answer:
The volume of the metal content of the shell is approximately \( \frac{392}{3} \pi \, \text{cm}^3 \) or about \( 410.67 \, \text{cm}^3 \).