A circular swimming pool has a diameter of 20 ft, the sides are 6 ft high, and the depth of the water is 5 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/ft^3

To calculate the work required to pump all the water out of the swimming pool, we'll use the following approach:

### 1. Setup
- The pool has a diameter of 20 ft, which gives it a radius \( r = \frac{20}{2} = 10 \) ft.
- The height of the pool is 6 ft, but the depth of the water is only 5 ft.
- The weight density of water is given as \( 62.5 \) lb/ft³.

The total work required to pump the water out is the sum of all the small amounts of work needed to pump each layer of water to the top of the pool.

### 2. Formula for Work
The general formula for work is:
\[
W = \int \rho A(y) D(y) dy
\]
Where:
- \( \rho = 62.5 \) lb/ft³ (weight density of water),
- \( A(y) = \pi r^2 \) is the cross-sectional area of the pool at height \( y \),
- \( D(y) \) is the distance the water needs to be pumped,
- \( y \) is the height of the water being pumped from.

In this case, the pool is circular, so the area \( A(y) = \pi \times 10^2 = 100\pi \) ft², which remains constant.

### 3. Setup the Integral
Each slice of water at height \( y \) is lifted a distance \( D(y) \) to the top of the pool, which is 6 ft high. Therefore, the distance the water at height \( y \) must be pumped is \( D(y) = 6 - y \).

The volume of a thin slice of water at height \( y \) with thickness \( dy \) is \( A(y) dy = 100\pi dy \), and the weight of this slice is \( \rho \cdot A(y) dy = 62.5 \times 100\pi dy \).

So, the work to pump this slice is:
\[
dW = \text{weight} \times \text{distance} = (62.5 \times 100\pi) (6 - y) dy
\]

Now, we integrate this expression from \( y = 0 \) to \( y = 5 \) (since the depth of the water is 5 ft):

\[
W = \int_0^5 (62.5 \times 100\pi) (6 - y) dy
\]

### 4. Solve the Integral
Factor out constants:
\[
W = 6250\pi \int_0^5 (6 - y) dy
\]

Now, integrate:
\[
\int_0^5 (6 - y) dy = \left[ 6y - \frac{y^2}{2} \right]_0^5
\]
Evaluating this:
\[
= \left(6(5) - \frac{5^2}{2}\right) - \left(6(0) - \frac{0^2}{2}\right)
= (30 - 12.5) - 0
= 17.5
\]

Thus, the total work is:
\[
W = 6250\pi \times 17.5
= 109375\pi \, \text{ft-lb}
\]

### 5. Final Answer
Using \( \pi \approx 3.1416 \), the work required is approximately:
\[
W \approx 109375 \times 3.1416 = 343774.6 \, \text{ft-lb}
\]

### Conclusion
The work required to pump all the water out of the pool is approximately **343,774.6 ft-lb**.

Would you like any more details, or have any questions?

### Related Questions:
1. How would the work change if the water depth was 6 ft (full pool)?
2. How does the weight density of water affect the total work required?
3. How much work would be required if the pool had a different radius?
4. How does the height of the pool walls affect the amount of work needed?
5. What is the significance of integrating over the depth in calculating work?

### Tip:
When calculating work to move fluids, always integrate over the depth since each layer of fluid requires a different amount of work based on its distance from the target location.

A circular swimming pool has a diameter of 20 ft, the sides are 6 ft high, and the depth of the water is 5 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/ft^3)

Learn how to calculate the work required to pump water out of a circular pool with a diameter of 20 ft and a water depth of 5 ft. This solution uses calculus to integrate over the water's depth, applying principles of fluid mechanics and energy. Suitable for college students studying calculus and physics.

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