To calculate the amount of work required to pump all of the Kool-Aid contaminated water out of the swimming pool, we need to use the physical definition of work in this context. Specifically, work $W$ is defined as:

$W = \int_{a}^{b} F(x) \, dx$

Where:

• $F(x)$ is the force required to lift a small slice of water at height $x$,
• $a$ and $b$ are the limits of integration, representing the range of heights from which the water is pumped,
• $x$ is the height above the bottom of the pool, and
• $dx$ represents an infinitesimally small slice of water at height $x$.

Step 1: Geometry of the Pool

The pool is cylindrical with:

• Radius $r = 13.5 \, \text{ft}$,
• Height of water $h = 5.5 \, \text{ft}$ (only this amount of water is in the pool),
• The total height of the pool is 10 feet, but we're only concerned with the height of the water.

Step 2: Volume of a Slice of Water at Height $x$

A thin slice of water at height $x$ has a volume:

$\Delta V = \pi r^2 \, dx = \pi (13.5)^2 \, dx$

The radius of the pool remains constant, so we can calculate:

$\pi (13.5)^2 = \pi \times 182.25 = 573.64 \, \text{ft}^2$

Thus, the volume of a thin slice of water at height $x$ is:

$\Delta V = 573.64 \, dx \, \text{ft}^3$

Step 3: Force Required to Lift the Slice of Water

The force required to lift a slice of water depends on its weight. The weight of the slice is:

$\text{Weight} = \sigma \times \Delta V = 65.3 \times 573.64 \, dx = 37453.892 \, dx \, \text{lbs}$

Since the slice of water at height $x$ needs to be lifted a distance $10 - x$ (because the water needs to be pumped over the side of the pool, which is 10 feet high), the force needed is:

$F(x) = 37453.892 \times (10 - x)$

Step 4: Work Calculation

The work required to lift a slice of water from height $x$ to the top is given by:

$dW = F(x) \, dx = 37453.892 (10 - x) \, dx$

To find the total work, integrate this expression from $x = 0$ to $x = 5.5$ (since the water only fills up to 5.5 feet):

$W = \int_0^{5.5} 37453.892 (10 - x) \, dx$

Step 5: Solving the Integral

We can break this integral into two parts:

$W = 37453.892 \int_0^{5.5} (10 - x) \, dx = 37453.892 \left( \int_0^{5.5} 10 \, dx - \int_0^{5.5} x \, dx \right)$

Now, compute each integral:

$\int_0^{5.5} 10 \, dx = 10x \bigg|_0^{5.5} = 10(5.5) = 55$ $\int_0^{5.5} x \, dx = \frac{x^2}{2} \bigg|_0^{5.5} = \frac{(5.5)^2}{2} = \frac{30.25}{2} = 15.125$

Thus, the total work is:

$W = 37453.892 \left( 55 - 15.125 \right) = 37453.892 \times 39.875$

Step 6: Final Calculation

Now, multiply the two terms to find the total work:

$W = 37453.892 \times 39.875 = 1493108.672 \, \text{ft-lbs}$

Conclusion:

The total work required to pump all of the Kool-Aid contaminated water over the side of the pool is approximately:

$\boxed{1,493,108.67 \, \text{ft-lbs}}$

Would you like more details on any part of this solution?

---

Related Questions:

1. How does the height of the pool affect the amount of work required to pump water out?
2. How would the work change if the water level were increased to 8 feet?
3. What if the density of the water were different? How would that impact the work calculation?
4. How does the radius of the pool influence the amount of work?
5. Could this problem be solved using different units (like SI units)?

Tip: Work is directly related to both the volume and the height the water must be lifted. Increasing the water level significantly increases the work required!