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Chapter 15: Problem 33

A swimming pool is circular with a \(40-\) ft diameter. The depth is constant along east-west lines and increases linearly from 2 ft at the south end to 7 \(\mathrm{ft}\) at the north end. Find the volume of water in the pool.

Short Answer

Expert verified
The volume of the pool is \(72000\pi\) cubic feet.

Step by step solution

01

Understand the Shape and Dimensions

The pool is circular with a diameter of 40 feet, which means it has a radius of 20 feet. The depth varies linearly from 2 feet at the south to 7 feet at the north.
02

Find the Area of the Circular Base

Since the pool is circular, use the formula for the area of a circle, \( A = \pi r^2 \). Plugging in the radius of 20 feet, \( A = \pi (20)^2 = 400\pi \) square feet.
03

Set Up the Depth Function

The depth changes linearly from 2 feet to 7 feet across a span of 40 feet (the diameter). We can describe this with the linear function \( d(x) = 2 + \frac{5}{40}x = 2 + \frac{1}{8}x \), where \( x \) is the distance from the south end in feet.
04

Determine the Volume Integral

To find the volume, integrate the area of a cross-section of the pool (which is a rectangle's area per unit width with \( width = 1 \)), from 0 to 40, \[ V = \int_0^{40} 400\pi d(x) \, dx = \int_0^{40} 400\pi (2 + \frac{1}{8}x) \, dx \].
05

Solve the Integral

Calculate the integral \[ V = 400\pi \int_0^{40} (2 + \frac{1}{8}x) \, dx = 400\pi \left[ 2x + \frac{1}{8} \frac{x^2}{2} \right]_0^{40} = 400\pi \left[ 2(40) + \frac{1}{16}(1600) \right] \].
06

Evaluate and Simplify the Expression

Evaluate the expression, \[ 400\pi (80 + 100) = 400\pi \times 180 = 72000\pi \]. This gives us the volume of the water in cubic feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Calculation
Calculating volume is like figuring out how much space something occupies. In our swimming pool example, we need to determine how much water fits inside the pool. Here's where the concept of volume calculation comes into play. Imagine you want to fill the pool with little cubes of water, each having a side length of 1-foot. The question becomes: How many of those cubes can fit into the pool?
  • Circular Base: Our pool's bottom is a circle, so we'll need the area of that circle to start. We'll multiply this area by the average depth to estimate the total volume.
  • Variable Depth: Our pool doesn鈥檛 have the same depth throughout, but fear not! We can handle it with some more sophisticated calculations.
In situations with non-uniform shapes like this pool, integrals help by summing infinitely small units, leading us seamlessly to our next concept: integral calculus.
Integral Calculus
Integral calculus is all about accumulation, like adding up little pieces to find a bigger total. Think of it as a way to total up infinite tiny slices of something to get a complete picture. When calculating a pool's volume, integral calculus comes in handy as we are summing up small volume slices across different depths.
  • Setting Up the Integral: To use integration for the pool, we need two things: the base area and how depth changes along the pool's length.
  • The Integral Formula: We set the integral to account for areas of very thin slices of the pool's water from one end to the other.
This method allows us to precisely calculate how much water the pool holds, by not just calculating averages, but considering all incremental depths.
Circular Area
The circular area is fundamental when dealing with round shapes, like the base of our swimming pool. Whenever you hear "circular," think about the shape of a pie.
  • Radius and Diameter: Our pool's diameter is 40 ft, so the radius鈥攈alf of the diameter鈥攊s 20 ft.
  • Area Calculation: The area of our circular pool can be found using the formula \( A = \pi r^2 \). This gives us a comprehensive base area of \( 400\pi \) square feet.
This area becomes a crucial part of further calculations, as it forms a constant component in our integral volume solution.
Linear Depth Function
A linear depth function helps describe how something changes smoothly in value from point to point. In our pool, the water depth increases evenly from the shallow to the deep end.
  • Form of the Function: A linear function here means that depth can be calculated by simply plugging in the distance from the shallow end. Our function is \( d(x) = 2 + \frac{1}{8}x \).
  • Understanding the Function: This tells us that for every foot we move north from the south end, the pool gets uniformly deeper by about 0.125 feet.
Understanding this function is key to setting up our integral, as it represents how much each section of the pool contributes to the overall volume.

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