To solve this problem, we need to calculate the volume of a circular pool with a varying depth.

Given Information:

1. Diameter of the pool: 20 ft.
2. Depth:
   • At the south end: 5 ft.
   • At the north end: 10 ft.
   • Depth increases linearly from 5 ft to 10 ft along the pool.

Steps to Solve

1. Calculate the radius of the circular pool: $\text{radius} = \frac{\text{diameter}}{2} = \frac{20 \, \text{ft}}{2} = 10 \, \text{ft}$

2. Model the depth function: Since the depth increases linearly from 5 ft to 10 ft, we can represent it as a linear function $d(x)$ along the north-south axis, where $x$ goes from $-10$ ft (south end) to $10$ ft (north end): $d(x) = 5 + \frac{10 - 5}{20} \times (x + 10) = 5 + \frac{5}{20} \times (x + 10) = 5 + 0.25(x + 10)$ Simplifying this, we get: $d(x) = 5 + 0.25x + 2.5 = 7.5 + 0.25x$

3. Set up the volume integral: The volume $V$ can be calculated by integrating the area of circular cross-sections along the depth: [ V = \int_{-10}^{10} \pi \cdot (10)^2 \cdot d(x) , dx