To set up the integral representing the work required to pump all the water out of the swimming pool, we need to consider the following factors:

1. Dimensions of the pool:

   • Diameter of the pool = 8 m, so the radius $r = 4$ m.
   • Height of the pool = 3 m.
   • Depth of the water = 2 m (so the water does not completely fill the pool).

2. Work formula: The work $W$ required to pump water out is given by the formula: $W = \int (\text{force}) \cdot (\text{distance}) \, dy$ where $y$ is the vertical position of the water slice, the force is the weight of the slice of water, and the distance is the height the water needs to be lifted.

3. Volume of a slice of water: Consider a thin horizontal slice of water at a height $y$, with thickness $dy$. The area of the slice is the area of the circular pool, which is: $A = \pi r^2 = \pi (4^2) = 16\pi \, \text{m}^2$ The volume of the slice is then: $dV = A \, dy = 16\pi \, dy$

4. Weight of the water slice: The mass of the slice is the volume $dV$ multiplied by the density of water $\rho = 1000 \, \text{kg/m}^3$: $dm = \rho \, dV = 1000 \times 16\pi \, dy$ The weight of the slice is $dm \times g$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity: $\text{Weight} = 1000 \times 16\pi \times 9.8 \, dy = 156800\pi \, dy$

5. Distance the water must be lifted: The water at height $y$ (measured from the bottom of the pool) must be lifted a distance of $(3 - y)$ meters to be pumped out, since the sides of the pool are 3 meters high.

6. Integral for work: The total work is obtained by integrating from $y = 0$ (the bottom of the pool) to $y = 2$ (the top of the water): $W = \int_0^2 156800\pi (3 - y) \, dy$

Thus, the integral that represents the work required to pump all the water out over the side of the pool is:

$W = 156800\pi \int_0^2 (3 - y) \, dy$

Would you like any details on how to evaluate this integral or clarification on any step?

Here are 5 related questions for further exploration:

1. How does changing the water density affect the work needed to pump the water out?
2. What happens if the depth of the water reaches the full 3 meters?
3. How would the work change if the pool were half the size (radius of 2 m)?
4. What if gravity on another planet is different? How would this affect the result?
5. What additional work would be required to lift water over an obstacle higher than 3 m?

Tip: Always ensure units are consistent when calculating physical quantities like work—especially when using integrals.