To find the volume of water in the pool, we can approach this problem as follows:

1. Identify the shape of the pool: The pool is circular, with a diameter of 40 feet. This means the radius $r$ of the pool is: $r = \frac{40}{2} = 20 \text{ feet}$

2. Determine the varying depth: The depth of the pool changes linearly from 4 feet at the south end to 9 feet at the north end. This suggests the average depth $d_{\text{avg}}$ can be calculated as: $d_{\text{avg}} = \frac{4 + 9}{2} = \frac{13}{2} = 6.5 \text{ feet}$

3. Calculate the volume of the pool: Since the pool is cylindrical in shape with a circular base, we can use the volume formula for a cylinder, $V = \pi r^2 h$, where $h$ is the average depth.

   Substituting the values we found: $V = \pi \times (20)^2 \times 6.5$ $V = \pi \times 400 \times 6.5$ $V = 2600 \pi$

4. Compute the approximate volume: Using $\pi \approx 3.14159$, $V \approx 2600 \times 3.14159 \approx 8168.14 \text{ cubic feet}$

5. Round the answer: Rounding to the nearest whole number, the volume of the water in the pool is approximately: $\boxed{8168} \text{ cubic feet}$

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

Here are some related questions to explore:

1. How would the volume change if the depth increased linearly from 3 feet to 10 feet instead?
2. What would be the effect on volume if the diameter were 50 feet instead of 40 feet?
3. How would you calculate the volume if the pool had an irregular, non-linear depth change?
4. What other real-world applications might require calculating volumes with varying depths?
5. How could we calculate the water volume if the pool were an elliptical shape instead of circular?

Tip: For problems involving linear changes in dimensions, taking the average value simplifies the calculations.