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The problem asks how much work is required to pump all the water from a circular swimming pool with a diameter of 20 feet, sides 5 feet high, and water filled to a depth of 4 feet. We are given that the weight of water is 62.5 lb/ft³.

### Solution Outline

1. **Dimensions and Parameters**:
   - Diameter = 20 ft → Radius \( r = \frac{20}{2} = 10 \) ft
   - Water depth \( h_w = 4 \) ft
   - Pool height \( h_p = 5 \) ft
   - Weight density of water \( \rho = 62.5 \) lb/ft³

2. **Work Formula**:
   The work needed to pump water is calculated using:
   \[
   W = \int_{0}^{h_w} \rho \cdot A(y) \cdot (h_p - y) \, dy
   \]
   where:
   - \( A(y) = \pi r^2 \) is the cross-sectional area of the water at height \( y \) (which is constant in this case since the pool is cylindrical),
   - \( h_p - y \) is the distance each slice of water must be lifted to get over the edge of the pool,
   - \( \rho \) is the weight density of water,
   - Limits of integration are from 0 to the water depth \( h_w = 4 \).

### Step-by-Step Calculation

1. **Cross-sectional area**:  
   The cross-sectional area of the pool is the area of a circle with radius 10 feet:
   \[
   A = \pi \times (10)^2 = 100\pi \text{ ft}^2
   \]

2. **Work integral setup**:
   The work to pump a thin slice of water at height \( y \) is:
   \[
   dW = \rho \cdot A \cdot (h_p - y) \, dy
   \]
   Substituting the values:
   \[
   dW = 62.5 \cdot 100\pi \cdot (5 - y) \, dy
   \]

3. **Evaluate the integral**:
   Now, integrating from 0 to 4:
   \[
   W = 6250\pi \int_0^4 (5 - y) \, dy
   \]
   This integral can be evaluated step by step.

Let's calculate it.The total work required to pump all the water out of the pool is approximately **235,619 ft-lb** (rounded to the nearest integer).

Would you like further details or have any questions? Here are some related questions you might find helpful:

1. How is work related to the force and distance in physics?
2. How would the work change if the water depth increased to 5 feet?
3. What would happen if the radius of the pool changed?
4. Can the formula for work be applied to non-cylindrical shapes?
5. How does the weight of the water affect the total work required?

**Tip**: When dealing with physics-based integrals, ensure the limits of integration match the physical boundaries of the problem.

A circular swimming pool has a diameter of 20 ft, the sides are 5 ft high, and the depth of the water 4 ft. How much work (in ft-lb) is required to pump all of the water out over the side? (Use the fact that water weighs 62.5 lb/ft³. Round your answer to the nearest integer.)

Calculate the work required to pump water from a circular swimming pool with a diameter of 20 ft and a water depth of 4 ft. This problem involves concepts of calculus, integration, and the work-energy theorem. Suitable for students in Grades 11-12.

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