/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Show how to approximate the requ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 16

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? (Use the fact that water weighs 62.5 \(\mathrm{lb} / \mathrm{ft}^{3} . )\)

Short Answer

Expert verified
The work required is \(108000 \pi\) foot-pounds.

Step by step solution

01

Understand the Problem

We have a circular swimming pool with a diameter of 24 ft and a side height of 5 ft, filled with water to a depth of 4 ft. We need to find the work required to pump the entire volume of water over the side. The work is essentially moving the water up to a height of 5 ft.
02

Establish the Geometry

The pool is circular with a radius of 12 ft (since diameter = 24 ft). The volume of water in the pool is determined by the product of the area of the base and the depth: \[ V = \pi \times (12)^2 \times 4 \text{ ft}^3 \]
03

Determine the Weight of Water

Water weighs 62.5 lb/ft鲁, so the total weight of the water is:\[ \text{Total weight} = V \times 62.5 \text{ lb/ft}^3 \]Substitute the volume from Step 2.
04

Setup the Riemann Sum

To approximate the work using a Riemann sum, imagine slicing the pool into thin horizontal layers of thickness \( \Delta y \). The work to move each slice is weight \( \times \) height lifted, which is:\[ \Delta W = 62.5 \times \pi \times (12)^2 \times \Delta y \times (5 - y) \]
05

Express Work as an Integral

Converting the Riemann sum to an integral, the total work is:\[ W = \int_0^4 62.5 \times \pi \times (12)^2 \times (5 - y) \, dy \]
06

Evaluate the Integral

Compute the integral:\[ W = 62.5 \times \pi \times 144 \times \int_0^4 (5 - y) \, dy \]Breaking it into simpler integrals:\[ \int_0^4 (5 - y) \, dy = \left[ 5y - \frac{y^2}{2} \right]_0^4 \]Substituting the limits:\[ = \left(5 \times 4 - \frac{4^2}{2}\right) - \left(5 \times 0 - \frac{0^2}{2}\right) = 20 - 8 = 12 \]
07

Compute the Total Work

Combine the results from Step 6 and calculate:\[ W = 62.5 \times \pi \times 144 \times 12 = 108000 \times \pi \, \text{foot-pounds} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Integral Calculus
Integral calculus is essential for calculating quantities that are continuously changing. It helps us find accumulated totals, like areas under curves or total work done. In the case of a swimming pool, integrating allows us to find the total work needed to pump all water over the side.
We used a Riemann sum to approximate the integral representing this work. A Riemann sum breaks the pool into small slices and adds up the work for each slice.
By converting this sum into an integral, \[ W = \int_0^4 62.5 \times \pi \times (12)^2 \times (5 - y) \, dy \]students can find an exact total using calculus. Integrals, thus, offer precise results that Riemann sums approximate step by step.
Work in Physics
The concept of work in physics relates to applying force over a distance. For the swimming pool problem, work is done when moving water from the pool to the top of its side. The formula for calculating work is:\[ \text{Work} = \text{Force} \times \text{Distance} \]
Here, force is the weight of each water slice, and distance is the height the water is lifted. For ease:
  • Weight of each slice (force) = 62.5 lb/ft鲁 脳 volume of the slice.
  • Distance lifted = difference in height from the water surface to pool's top, \, (5 - y) \, ft.
This mix of influence from both core physics principles and calculus creates a way to calculate work in real-world scenarios, like the energy needed to pump a swimming pool.
Volume Calculation
Calculating volume is vital when estimating work in such exercises. The swimming pool's volume is calculated using its geometric shape: a cylinder based on a circular area and depth.
The formula to find the volume of the pool is:\[ V = \pi \times (12)^2 \times 4 \, \text{ft}^3 \]
where:
  • \( \pi \) represents the base area of the circle (\( \pi \times (radius)^2 \)).
  • '12 ft' is the radius since the diameter was given as 24 ft.
  • '4 ft' is the water depth.
By knowing how to calculate this accurately, students can apply this to various scenarios involving liquid containers.
Weight of Water
Weight plays a central role in these calculations as it forms the force to perform work. Water has a specific weight of 62.5 lb/ft鲁. This is a constant in calculations your physics problems most often use when analyzing liquids in similar conditions.
By multiplying the pool's volume by this weight per unit volume:\[ \text{Total weight} = V \times 62.5 \, \text{lb/ft}^3 \]
students derive the total weight of the water in the pool.
This weight directly affects how much work is necessary to lift water against gravity out and over the pool's side. Thus, understanding and calculating the weight of water is key in solving physics problems involving fluids.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(29-32=\) Each integral represents the volume of a solid. Describe the solid. $$2\pi \int_{0}^{2} \frac{y}{1+y^{2}} d y$$

\(29-32=\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=1-x y, \quad(0,0)$$

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A heavy rope, 50 ft long, weighs 0.5 \(\mathrm{lb} / \mathrm{ft}\) and hangs over the edge of a building 120 \(\mathrm{ft}\) high. (a) How much work is done in pulling the rope to the top of the building? (b) How much work is done in pulling half the rope to the top of the building?

(a) A model for the shape of a bird's egg is obtained by rotating about the \(x\) -axis the region under the graph of $$ f(x)=\left(a x^{3}+b x^{2}+c x+d\right) \sqrt{1-x^{2}} $$ Use a CAS to find the volume of such an egg. (b) For a Red-throated Loon, \(a=-0.06, b=0.04\) \(c=0.1,\) and \(d=0.54 .\) Graph \(f\) and find the volume of an egg of this bird.

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is $$F=\frac{m g R^{2}}{(x+R)^{2}}$$ where \(x=x(t)\) is the object's distance above the surface at time \(t, R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and \(s o\) $$m \frac{d v}{d t}=-\frac{m g R^{2}}{(x+R)^{2}}$$ (a) Suppose a rocket is fired vertically upward with an initial velocity \(v_{0 .}\) Let \(h\) be the maximum height above the surface reached by the object. Show that $$v_{0}=\sqrt{\frac{2 g R h}{R+h}}$$ $$ [\text { Hint: By the Chain Rule, } m(d v / d t)=m v(d v / d x) .]$$ (b) Calculate \(v_{e}=\lim _{h \rightarrow \infty} v_{0 .}\) This limit is called the escape velocity for the earth. (c) Use \(R=3960 \mathrm{mi}\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) to calculate \(v_{e}\) in feet per second and in miles per second.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.