/*! 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 30 Let \(R\) be the region bounded ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 30

Let \(R\) be the region bounded by the following curves. Use the washer method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=x, y=x+2, x=0, x=4$$

Short Answer

Expert verified
Question: Determine the volume of the solid generated by revolving the region R, bounded by the curves \(y=x\), \(y=x+2\), \(x=0\), and \(x=4\), around the x-axis. Answer: The volume of the solid generated when the region R is revolved about the \(x\)-axis is \(V = \frac{104\pi}{3}\).

Step by step solution

01

Understand the given functions

The given functions are \(y=x\), \(y=x+2\), \(x=0\), and \(x=4\). The first two lines are oblique lines and the last two lines are vertical lines, which will form a parallelogram shape. Determine the points of intersection of all the functions to get the range for integration.
02

Find the points of intersection

Set the two functions equal to each other and solve for x. You will get \(x = x+2 \Rightarrow x=2\). The points of intersection are \((0,0),(0,2),(4,4),\) and \((4,6)\), from which we see the range for integration is from \(x=0\) to \(x=4\).
03

Determine inner and outer radii for the washer method

As the region R is revolved around the \(x\)-axis, we need to find the radius of the inner and outer circles of the washers formed. They are equal to the distance of the curves from the x-axis. The outer radius is the distance from the \(x\)-axis to \(y=x+2\), which is just given by \(x+2\). The inner radius is the distance from the \(x\)-axis to \(y=x\), which is just given by \(x\).
04

Set up the integration formula

Using the washer method, the volume formula is given by: $$V = \pi \int_{a}^{b} [R_o^2(x) - R_i^2(x)] dx$$ Where \(R_o(x)\) is the outer radius function, \(R_i(x)\) is the inner radius function, and a and b are the limits of integration.
05

Substitute the values and integrate

For our problem, the outer radius is \((x+2)^2\) and the inner radius is \(x^2\). The volume V integral will be as follows: $$V = \pi \int_{0}^{4} [(x+2)^2 - x^2] dx$$ Integrate, $$V = \pi \Biggr[\frac{1}{3}(x+2)^3 - \frac{1}{3}x^3 \Biggr]_0^4$$
06

Calculate the volume

Finally, substitute the upper and lower limits and calculate the volume: $$V = \pi \Biggr[\frac{1}{3}(6)^3 - \frac{1}{3}(4)^3 - \frac{1}{3}(2)^3 \Biggr]$$ $$V = \frac{104\pi}{3}$$ Therefore, the volume of the solid generated when the region R is revolved about the \(x\)-axis is \(V = \frac{104\pi}{3}\).

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.

Volume of Revolution
When we talk about the volume of a solid of revolution, we are referring to the three-dimensional shape created by rotating a two-dimensional region around an axis. This concept is particularly useful in disciplines like engineering and physics, where such shapes frequently appear. To understand this, imagine taking a flat drawing of a region and spinning it around a line (the axis of revolution) -- the space that this spinning region occupies is your solid of revolution.
In our specific problem, we're revolving the region bounded by the lines \(y=x\), \(y=x+2\), \(x=0\), and \(x=4\) around the \(x\)-axis. This action creates a series of circular disks or "washers" stacked along the axis, producing a solid shape like a cylinder with a varying radius along its length. This is where the washer method comes into play, helping us determine the shape's overall volume.
Integral Calculus
Integral calculus is the branch of mathematics that deals with the summation of parts to find whole quantities. It's a tool used to calculate areas, volumes, and other concepts that can be summed into a total. When calculating the volume of a solid of revolution, integral calculus plays a critical role through techniques like the washer method.

In the washer method, the idea is to integrate across the region's boundaries, taking into account the shapes formed when the area is revolved around the axis. This method involves setting up an integral with specific limits from one bound to the other and subtracting the area of the inner disk from the area of the outer disk, both squared and expressed as functions of \(x\). The process of setting the definite integral to calculate the volume is key and relies on evaluating the integral of the difference between these outer and inner circle areas throughout the entire defined region.
Solid Geometry
Solid geometry is the study of 3D shapes, much like the ones formed through volumes of revolution. It involves understanding geometric figures in three-dimensional space鈥攕uch as spheres, cubes, and the kinds of varying shapes we encounter using the washer method.
In this exercise, solid geometry helps us visualize and comprehend the final result of revolving a plane region around a line. The resulting shape is a hybrid of familiar geometric shapes where the dimensions are described by functions such as \(y=x\) and \(y=x+2\) that define the contours of the sphere as it revolved around the \(x\)-axis. Understanding these shapes鈥� dimensions helps us effectively set up our equations in integral calculus, refining the integration process as we calculated a specific solid's volume. This interplay between equations and geometric visualizations is what makes solid geometry a vital part of solving real-world problems in mathematics and its applications.

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

Two points \(P\) and \(Q\) are chosen randomly, one on each of two adjacent sides of a unit square (see figure). What is the probability that the area of the triangle formed by the sides of the square and the line segment \(P Q\) is less than one-fourth the area of the square? Begin by showing that \(x\) and \(y\) must satisfy \(x y<\frac{1}{2}\) in order for the area condition to be met. Then argue that the required probability is \(\frac{1}{2}+\int_{1 / 2}^{1} \frac{d x}{2 x}\) and evaluate the integral.

Kelly started at noon \((t=0)\) riding a bike from Niwot to Berthoud, a distance of \(20 \mathrm{km},\) with velocity \(v(t)=15 /(t+1)^{2}\) (decreasing because of fatigue). Sandy started at noon \((t=0)\) riding a bike in the opposite direction from Berthoud to Niwot with velocity \(u(t)=20 /(t+1)^{2}\) (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours. a. Make a graph of Kelly's distance from Niwot as a function of time. b. Make a graph of Sandy's distance from Berthoud as a function of time. c. When do they meet? How far has each person traveled when they meet? d. More generally, if the riders' speeds are \(v(t)=A /(t+1)^{2}\) and \(u(t)=B /(t+1)^{2}\) and the distance between the towns is \(\vec{D},\) what conditions on \(A, B,\) and \(D\) must be met to ensure that the riders will pass each other? e. Looking ahead: With the velocity functions given in part (d), make a conjecture about the maximum distance each person can ride (given unlimited time).

Evaluate the following integrals. $$\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta$$

Use l'H么pital's Rule to evaluate the following limits. $$\lim _{x \rightarrow \infty} \frac{1-\operatorname{coth} x}{1-\tanh x}$$

A hemispherical bowl of radius 8 inches is filled to a depth of \(h\) inches, where \(0 \leq h \leq 8\). Find the volume of water in the bowl as a function of \(h\). (Check the special cases \(h=0 \text { and } h=8 .)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.