/*! 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 36 Shell and washer methods Let \(R... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Shell and washer methods Let \(R\) be the region bounded by the following curves. Use both the shell method and the washer method to find the volume of the solid generated when \(R\) is revolved about the indicated axis. $$y=\frac{1}{x+1}, y=1-\frac{x}{3} ; \text { about the } x \text { -axis }$$

Short Answer

Expert verified
Answer: The volume of the solid is \(\frac{8\pi}{3} - 2\pi \ln(4)\).

Step by step solution

01

Intersection Points

To find the points of intersection, we need to solve the following system of equations: $$\frac{1}{x+1} = 1-\frac{x}{3}$$ Solving the system: $$3 = (x+1)(3-x)$$ $$x^2-2x+3=0$$ By applying the quadratic formula, we get \(x=1\) and \(x=3\). So, the points of intersection are \((1,\frac{1}{2})\) and \((3,0)\), and this will give us the limits of integration. Step 2: Shell Method
02

Shell Method Setup

To set up the shell method integral, we consider that the height of the shell is the difference between the two functions, and the circumference of the shell is \(2\pi y\). Since our shape is bounded between \(x=1\) and \(x=3\), our integral will look like: $$V_{shell} = 2\pi\int_{1}^3 y\left( (1-\frac{x}{3}) - \frac{1}{x+1} \right) dx$$ Step 3: Shell Method Evaluation
03

Shell Method Evaluation

Now, we need to evaluate the integral to find the volume: $$V_{shell} = 2\pi\int_{1}^3 y\left( 1-\frac{x}{3} - \frac{1}{x+1} \right) dx= 2\pi\int_{1}^3 \left(\frac{x}{3} - \frac{1}{x+1}\right) dx$$ $$V_{shell} = 2\pi\left[\frac{x^2}{6} - \ln(x+1) \right]_{1}^{3} = 2\pi\left(\frac{8}{6} - \ln(4)\right) = \frac{8\pi}{3} - 2\pi \ln(4)$$ Step 4: Washer Method
04

Washer Method Setup

Similarly, we'll set up the integral for the washer method, considering the outer and inner radii formed by revolving the region around the x-axis: $$V_{washer} = \pi\int_{1}^3 [(1 - \frac{x}{3})^2 - (\frac{1}{x+1})^2] dx$$ Step 5: Washer Method Evaluation
05

Washer Method Evaluation

We'll now evaluate this integral: $$V_{washer} = \pi\int_{1}^3 [(1 - \frac{x}{3})^2 - (\frac{1}{x+1})^2] dx = \pi\left[\frac{x^3}{9} - \frac{7x}{6} + x\ln(x+1) \right]_{1}^{3}$$ $$V_{washer} = \pi\left(\frac{8}{9} - \frac{15}{6} + 3\ln(4) \right) = \frac{8\pi}{3} - 2\pi \ln(4)$$ From both methods, we can conclude that the volume is the same: $$V_{shell} = V_{washer} = \frac{8\pi}{3} - 2\pi \ln(4)$$

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 91Ó°ÊÓ!

Key Concepts

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

Shell Method
The shell method is a powerful technique for finding the volume of a solid of revolution, particularly when the axis of rotation is parallel to the axis of integration. This approach visualizes the solid as a series of cylindrical shells. Each shell’s surface area is 2\(\text{π} \) times its radius times its height, and the volume is then this surface area times the thickness of the shell, which is an infinitely small change in the radius, denoted as \( dx \) or \( dy \), depending on the axis of rotation.

For a solid revolved around the x-axis, as in our exercise, the radius of each shell is a function of y, and the height is the difference between the outer and inner function values at a particular y. The integral for the volume of the shell method is set up as \[V_{shell} = 2\text{Ï€}\int y \cdot h(y) \text{d}x\], where \( h(y) \) is the height of the shell. By integrating from the lower to the upper limit of x, we can obtain the total volume.

In the example provided, the exercise improvement advice would suggest explicitly stating the reasoning behind choosing the shell method over other methods: Shell method is often favored when functions are easier to work with respect to one axis, and it avoids the inconvenience of finding and working with the outer and inner radii in the washer method.
Washer Method
Alternatively, the washer method approaches the volume of a solid of revolution by slicing the solid perpendicularly to the axis of rotation. These slices are essentially washers (disks with holes), and each washer's volume can be calculated by taking the area of the larger disk (outer radius) and subtracting the smaller disk (inner radius).

The volume integral using the washer method is \[V_{washer} = \text{Ï€}\int [R(x)^2 - r(x)^2] \text{d}x\], where \( R(x) \) and \( r(x) \) represent the outer and inner radii respectively. This method is often more straightforward when the solid can be thought of as a stack of washers, or when the axis of rotation is perpendicular to the axis of integration.

In the context of the given exercise, the advice for enhancing student understanding is to highlight the necessity of squaring the functions representing radii. It's crucial to point out that squaring turns distance functions (linear) into area functions (quadratic), reflecting the two-dimensional nature of the washers' cross-sectional areas.
Integration Techniques
Integration techniques are mathematical tools used to calculate the area under a curve or the accumulation of quantities, which is essential in determining the volume of solids of revolution. Understanding how to correctly set up and evaluate integrals is critical to successfully employ both the shell and washer methods.

Key techniques for calculating integrals include knowing how to expand expressions, using substitution, and integrating by parts. Sometimes, simplifying the integrand can make the integral more approachable. For example, transforming products into sums or differences might enable easier integration. Additionally, recognizing that logarithmic integration is effective for integrating expressions in the form of \(1/u\) where \(u\) is a linear function of \(x\) can be particularly helpful as seen in the step-by-step solution provided for the shell method.

Improvements for student comprehension might include focusing on pre-integration steps, such as simplifying integrands and using algebraic identities, which could make the calculus part more straightforward. Breaking down complex integrals into simpler parts, perhaps by using partial fraction decomposition when necessary, is another useful strategy.
Solid of Revolution
A solid of revolution refers to a three-dimensional object obtained by rotating a two-dimensional shape around an axis. Understanding the concepts behind solids of revolution is foundational for learning methods like the shell and washer methods.

To imagine a solid of revolution, picture a line or curve on a graph being spun 360 degrees around an axis (either the x-axis or the y-axis). The resulting shape is very much like spinning a clay figure on a potter’s wheel: the original 2D shape extrudes into a symmetrical 3D object. The formula to calculate the volume of a solid of revolution entails integrals, which mathematically accumulate the infinitely many cross-sections generated by this rotation.

For educational purposes, drawing diagrams that visually represent the generation of a solid can be extremely useful. Encouraging students to visualize the rotating graph will aid their understanding of how a certain interval on an axis translates into the three-dimensional volume they are calculating. The step-by-step solution arguably becomes easier to grasp when this visualization is clearly established in the learner's mind.

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

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _______. $$v(t)=2 t+6,\( for \)0 \leq t \leq 8$$

Mass of two bars Two bars of length \(L\) have densities \(\rho_{1}(x)=4 e^{-x}\) and \(\rho_{2}(x)=6 e^{-2 x},\) for \(0 \leq x \leq L\) a. For what values of \(L\) is bar 1 heavier than bar \(2 ?\) b. As the lengths of the bars increase, do their masses increase without bound? Explain.

Where do they meet? Kelly started at noon \((t=0)\) riding a bike from Niwot to Berthoud, a distance of \(20 \mathrm{km}\), with velocity \(v(t)=\frac{15}{(t+1)^{2}}\) (decreasing because of fatigue). Sandy started at noon \((t=0)\) riding a bike in the opposite direction from Berthoud to Niwot with velocity \(u(t)=\frac{20}{(t+1)^{2}}\) (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours. 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 \(v(t)=\frac{A}{(t+1)^{2}}\) and \(u(t)=\frac{B}{(t+1)^{2}}\) and the distance between the towns is \(D\) what conditions on \(A, B,\) and \(D\) must be met to ensure that the riders will pass each other? Looking ahead: With the velocity functions given in part (d). make a conjecture about the maximum distance each person can ride (given unlimited time).

Suppose \(R\) is the region bounded by \(y=f(x)\) and \(y=g(x)\) on the interval \([a, b],\) where \(f(x) \geq g(x)\) a. Show that if \(R\) is revolved about the vertical line \(x=x_{0},\) where \(x_{0}b ?\)

Work function A spring has a restoring force given by \(F(x)=25 x .\) Let \(W(x)\) be the work required to stretch the spring from its equilibrium position \((x=0)\) to a variable distance \(x .\) Find and graph the work function. Compare the work required to stretch the spring \(x\) units from equilibrium to the work required to compress the spring \(x\) units from equilibrium.

See all solutions

Recommended explanations on Math Textbooks

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.