To solve this problem, we need to follow the steps outlined in the questions:

Part (a): Express the volume of the water in the pool as a function of height $h$.

The cross-section of the swimming pool shows that the depth varies, which means that the volume of water depends on the height of the water at any given point. We need to consider different sections of the pool and treat them piecewise.

We have two main sections of the pool:

1. The first section is a trapezoid that extends from the shallow end (3 feet deep) to the midpoint of the pool where the depth reaches 9 feet.
2. The second section is a constant rectangular prism, where the depth remains constant at 9 feet.

We will derive the volume of water for each section separately as a function of the water height, $h$, and then combine the results into a piecewise function.

Section 1: The sloped part of the pool (from 3 feet to 9 feet depth)

For the first part of the pool, the cross-section is a trapezoid. The width of the pool is constant at 20 ft, and the depth varies linearly from 3 feet to 9 feet over a distance of 34 feet. We can break this into:

• From $h = 3$ to $h = 9$, the width changes linearly.

Using linear interpolation, the width of the pool at any depth $h$ can be expressed as: $W(h) = \frac{6}{6}(h - 3) + 6 = h - 3 + 6 = h + 3 \quad \text{(for} \ 3 \leq h \leq 9\text{)}.$ The volume of this section is: $V_1(h) = \text{area of trapezoid} \times \text{width of pool}.$ We'll continue this process and provide a solution for both parts.

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