/*! 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 8 Use the shell method to find the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines about the \(y\) -axis. $$y=2 x, \quad y=x / 2, \quad x=1$$

Short Answer

Expert verified
The volume of the solid is \(\pi\).

Step by step solution

01

Identify the Region to Revolve

First, identify the region bounded by the curves \(y=2x\), \(y=\frac{x}{2}\), and the line \(x=1\). The region lies between the two curves and is bounded by \(x=0\) and \(x=1\). The upper boundary is \(y=2x\) and the lower boundary is \(y=\frac{x}{2}\).
02

Set Up the Shell Method Formula

The shell method involves using cylindrical shells to approximate the volume. The formula for the shell method when rotating around the \(y\)-axis is: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \]Here, the 'radius' is the distance from the shell to the \(y\)-axis, which is \(x\), and the 'height' is the difference between the curves: \((2x - \frac{x}{2})\).
03

Simplify the Functions

The height function, which is the difference between \(y=2x\) and \(y=\frac{x}{2}\), simplifies as follows:\[ \text{Height} = (2x) - \left(\frac{x}{2}\right) = \frac{4x}{2} - \frac{x}{2} = \frac{3x}{2} \]
04

Set Up the Integral for Volume

Substitute the expressions for 'radius' and 'height' into the shell method formula. The integral becomes:\[ V = 2\pi \int_{0}^{1} x \left(\frac{3x}{2}\right) \, dx \]Simplify to:\[ V = 2\pi \int_{0}^{1} \frac{3x^2}{2} \, dx \]
05

Evaluate the Integral

Evaluate the integral: \[ \int_{0}^{1} \frac{3}{2} x^2 \, dx = \frac{3}{2} \int_{0}^{1} x^2 \, dx \]Solve:\[ \frac{3}{2} \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{3}{2} \left( \frac{1}{3} \right) = \frac{1}{2} \]
06

Multiply by 2\(\pi\) to Find the Volume

Now multiply the result from the integral by \(2\pi\) to find the final volume:\[ V = 2\pi \times \frac{1}{2} = \pi \]
07

Conclusion

The volume of the solid generated by revolving the given region about the \(y\)-axis is \(\pi\).

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.

Volume of Solids
Finding the volume of solids is an essential concept in calculus, especially when dealing with three-dimensional shapes. This becomes particularly interesting when the solids are formed by revolving a two-dimensional region around an axis. These types of problems often arise in calculus and require a deeper understanding of geometric transformations. When we talk about these volumes formed by rotation, we are looking at the solid of revolution. This is a process of taking a region defined by curves and revolving it to create a three-dimensional solid. For instance, the region between the curves can form unique shapes like cylinders or complex structures depending on the boundaries and axis of rotation.
Revolving Regions
Revolving regions is the method used to create a solid by rotating a region around an axis. In the given problem, the region is revolved around the y-axis. This concept is essential because it allows us to transition from understanding basic areas to the more complex notion of volume.
The region in question is determined by the curves and lines in the xy-plane. Here, the curves are given by the equations \(y=2x\) and \(y=\frac{x}{2}\). The line \(x=1\) completes the boundary of the region. When these boundaries are revolved around the y-axis, they form a cylindrical shape. By revolving the region, we create concentric cylindrical shells, like the layers of an onion, that contribute to the overall volume of the solid.
  • Understand the curves or lines forming the boundaries of the region.
  • Identify the axis of rotation, which in this case is the y-axis.
  • Visualize the three-dimensional shape resulting from this rotation.
Calculus Problem Solving
Solving calculus problems, particularly those involving volumes of revolution, requires a series of methodical steps. It's crucial to break down the problem into manageable parts and understand each step clearly.
Firstly, identify the region to be revolved by the curves and boundaries given. Determine the limits of integration, typically where the curves intersect or the bounds are explicitly given. Then, choose a method suitable for volume calculation, such as the shell method, which is well-suited for rotation about the y-axis in this scenario.
By setting up the shell method's integral, each component of the solid is considered systematically:
  • "Radius" defined by the distance from the rotation axis.
  • "Height" as the difference between the functions of the curves.
  • Integration over the defined range to accumulate the volume of shells.

Approaching the problem in these organized steps ensures no part of the problem is overlooked, offering a clear pathway from setup to solution.
Integral Calculation
Integral calculation is the backbone of determining volumes when revolving regions. Through integration, we can efficiently find the accumulated sum of an infinite number of infinitesimally small parts—in this case, the cylindrical shells.
For the given region, the shell method's integral is formulated as \[ V = 2\pi \int_{0}^{1} x \left(\frac{3x}{2}\right) \, dx \]. Solving this integral requires understanding both the setup and the technique of integration itself.
The steps involved include:
  • Substitute and simplify expressions into the integral (e.g., height = \(\frac{3x}{2}\) and radius = \(x\)).
  • Integrate the expression \(\frac{3}{2} x^2\) from the lower bound of 0 to the upper bound of 1.
  • Solve to find the antiderivative and evaluate the definite integral.
  • Multiply the result by \(2\pi\) to account for the revolution.

This step-by-step integration provides the precise volume of the solid, illustrating how the abstract notion of integration solidifies in tangible calculations.

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

It takes a force of 96,000 N to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of \(20 \mathrm{cm}\) to its fully compressed height of \(12 \mathrm{cm}\). a. What is the assembly's force constant? b. How much work does it take to compress the assembly the first centimeter? the second centimeter? Answer to the nearest joule.

Find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region in the first quadrant bounded by the curve \(x=y-y^{3}\) and the \(y\) -axis about a. the \(x\) -axis b. the line \(y=1\)

Find the lateral (side) surface area of the cone generated by revolving the line segment \(y=x / 2,0 \leq x \leq 4,\) about the \(x\) -axis. Check your answer with the geometry formula Lateral surface area \(=\frac{1}{2} \times\) base circumference \(\times\) slant height.

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$x=2 \sqrt{4-y}, \quad 0 \leq y \leq 15 / 4 ; \quad y-\text { axis }$$

The volume of a torus The disk \(x^{2}+y^{2} \leq a^{2}\) is revolved about the line \(x=b(b>a)\) to generate a solid shaped like a doughnut and called a torus. Find its volume. (Hint: \(\int_{-a}^{a} \sqrt{a^{2}-y^{2}} d y=\pi a^{2} / 2,\) since it is the area of a semicircle of radius \(a .)\)

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.