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Chapter 6: Problem 3

A rectangular swimming pool \(40 \mathrm{ft}\) long, \(15 \mathrm{ft}\) wide, and \(16 \mathrm{ft}\) deep is filled with water to a depth of \(15 \mathrm{ft}\). Use an integral to find the work required to pump all the water out over the top. (Take as the density of water \(\delta=62.4 \mathrm{lb} / \mathrm{ft}^{3} .\) ) Work = ___________.

Short Answer

Expert verified
The work required is 12636000 ft-lb.

Step by step solution

01

Understand the Problem

The task is to find the work required to pump water out of the pool using integrals and given parameters. We need to look at the density of water and the dimensions of the pool.
02

Write the Work Integral Formulation

The work required to pump water depends on the volume of water being lifted, the height it鈥檚 being lifted, and the density of water. The work is given by \[ W = \rho g \text{Area} \text{Depth} \text{d}x \]
03

Set Coordinate System

Set up the coordinate system; assume that the top of the pool is at y=0 and the bottom is at y=-16. Thus, the water needs to be pumped from depth y = -15 (the water level) to y = 0 (the top).
04

Determine Cross-Sectional Area and Volume

The cross-sectional area of the pool is given by the length times the width: \[ A = 40 \times 15 = 600 \text{ ft}^2 \]
05

Setup the Integral

The work to pump a thin layer at depth y to the top is \[ \text{d}W = \rho g A (h - y) \text{d}y \] Here, \( \rho = 62.4 \text{ lb}/\text{ft}^3 \) and \( h - y \) gives the height any small slice of water needs to be lifted.
06

Integrate

Integrate from -15 ft to 0 ft (water depth to the top): \[ W = \rho g A \times \text{Integral}_{-15}^{0} (15 - y) dy \] Evaluate the integral: \[ W = 62.4 \times 600 \times \text{Integral}_{-15}^{0} (15 - y) dy \] Solve the integral: \[ \text{Integral}_{-15}^{0} (15 - y) dy = \text{Integral}_{-15}^{0} 15 dy - \text{Integral}_{-15}^0 y dy = 15[y]_{-15}^0 - \frac{y^2}{2} \bigg|_{-15}^0 \] \[ 15 \times 15 - 0 - \bigg[\frac{(0)^2}{2} - \frac{(-15)^2}{2}\bigg] = 15 \times 15 - \bigg[\frac{0 - 225}{2}\bigg] \] thus, \[ \text{Result:} = 225 + \frac{225}{2} = 225 + 112.5 = 337.5 \text{} \] Finally, multiply; \[ W = 62.4 \times 600 \times 337.5 \text{ ft-lb} \]
07

Final Calculation

Evaluate: \[ W = 62.4 \times 600 \times 337.5 \text{} \] Multiply: \[ W = 12636000 \text { ft-lb} \]

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Key Concepts

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

integral calculus
Integral calculus is essential to solving problems involving the calculation of work done by pumping water. It allows us to determine the total amount of work by summing up, or integrating, infinitesimally small amounts of work over a given range. In this problem, we need to express work as an integral. We set up the integral to calculate the work done in lifting each infinitesimal slice of water from its depth up to the top of the pool. By integrating over the entire depth of the water, we obtain the total work needed.
density calculation
Density calculation plays a critical role in determining the work required to pump water out of the pool. The density of water, given as \(\rho = 62.4 \text{ lb}/ \text{ft}^3\), affects the total weight of the water being lifted. The weight of a differential volume element of water can be found by multiplying the density by the volume of that element. For our pool, considering the volume per infinitesimal slice, we multiply the density by the cross-sectional area and the differential height \(dy\). This way, each thin layer of water's contribution to the total work can be accurately calculated.
volume and area integration
Understanding volume and area integration is critical in solving this exercise. The problem involves a rectangular pool with a cross-sectional area determined by the pool's length and width: \( A = 40 \times 15 = 600 \text{ ft}^2\). The integral must account for the volume of water being pumped, which involves integrating this area over the depth of the water. By setting up the integral, we can sum the areas of infinitesimally thin water slices, leading us to the total volume of water to be lifted. This integration enables the accurate calculation of the total work done.
physics of work and energy
The physics of work and energy is fundamental in understanding how we calculate the work required to pump water from the pool. Work in physics is defined as the amount of energy transferred by a force acting through a distance. Here, the weight of the water (force) needs to be lifted (moved) to the top of the pool (distance). Each slice of water exerts a force due to its weight, and lifting it requires energy. By integrating the force exerted across the distance it needs to be moved (from each depth to the pool's top), we can determine the total work done, making sure to incorporate all slices of the water in the pool.

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Most popular questions from this chapter

Find the area of the region between \(y=x^{1 / 2}\) and \(y=x^{1 / 4}\) for \(0 \leq x \leq 1\). area \(=\) ___________.

A point mass of 1 grams located 6 centimeters to the left of the origin and a point mass of 3 grams located 7 centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system. Center of Mass = ___________.

Consider the curves given by \(y=\sin (x)\) and \(y=\cos (x)\). For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice. a. Sketch the region \(R\) bounded by the \(y\) -axis and the curves \(y=\sin (x)\) and \(y=\cos (x)\) up to the first positive value of \(x\) at which they intersect. What is the exact intersection point of the curves? b. Set up a definite integral whose value is the exact area of \(R\). c. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(x\) -axis. d. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(y\) -axis. e. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(y=2\) f. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(x=-1\)

4\. Water in a cylinder of height \(10 \mathrm{ft}\) and radius \(3 \mathrm{ft}\) is to be pumped out. The density of water is \(62.4 \mathrm{lb} / \mathrm{ft}^{3}\). Find the work required if (a) The tank is full of water and the water is to be pumped over the top of the tank. Work = ___________. (b) The tank is full of water and the water must be pumped to a height \(4 \mathrm{ft}\) above the top of the tank. Work = ___________.

5\. Find the area under the curve \(y=\frac{1}{\cos ^{2}(t)}\) between \(t=0\) and \(t=\pi / 2\). Enter diverges if the area is not bounded. area \(=\) ___________.
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