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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
   =

A swimming pool is circular with a 20-ft diameter. The depth is constant along east-west lines and increases linearly from 5 ft at the south end to 10 ft at the north end. Find the volume of water (in ft^3) in the pool. (Round your answer to the nearest whole number.)

Learn how to calculate the volume of water in a circular swimming pool with a 20-ft diameter and varying depth, increasing linearly from 5 ft to 10 ft. This solution involves calculus concepts and volume integration, suitable for college-level students.

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