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Chapter 5: Problem 8

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \(y=2 x^{2}, \quad y=0, \quad x=2\) (a) the \(y\) -axis (b) the \(x\) -axis (c) the line \(y=8\) (d) the line \(x=2\)

Short Answer

Expert verified
The volumes of the solids generated by revolving the region around the y-axis, the x-axis, the line \(y = 8\), and the line \(x = 2\), are respectively \(\frac{4\pi}{3}\), \(\frac{32\pi}{5}\), \(\frac{192\pi}{5}\), and \(8\pi\).

Step by step solution

01

Sketch the region

First, sketch the parabola \(y = 2x^2\) from \(x = 0\) to \(x = 2\) and the x-axis.
02

Volume around the y-axis

Next, use the disk method to calculate the volume of the solid generated by revolving the region around the y-axis. The volume is given by \(V = \pi \int_{a}^{b}[g(x)]^2 dx\). Here, \(g(x) = \sqrt{y/2}\), \(a = 0\) and \(b = 8\). So, \(V = \pi \int_{0}^{8}(\sqrt{y/2})^2 dy = \frac{4\pi}{3}\).
03

Volume around the x-axis

Now, use the disk method to calculate the volume of the solid when the region is revolved about the x-axis. In this case, \(g(x) = 2x^2\), \(a = 0\) and \(b = 2\). So, \(V = \pi \int_{0}^{2}(2x^2)^2 dx = \frac{32\pi}{5}\).
04

Volume around the line y=8

To find the volume of the solid when the region is revolved about the line \(y = 8\), use the washer method. The volume is \(V = \pi \int_{a}^{b}([R(x)]^2-[r(x)]^2) dx\), where \(R(x) = \sqrt{y/2}\) and \(r(x) = 2x^2\). Here, \(R(x) = 8\) and \(r(x) = 2x^2\), and \(a = 0\) and \(b = 2\). The volume is \(V = \pi \int_{0}^{2}[(8)^2 - (2x^2)^2] dx = \frac{192\pi}{5}\).
05

Volume around the line x=2

Finally, use the washer method for the solid obtained by rotating the region about the line \(x = 2\). Now, \(R(x) = 2 - 0 = 2\), and \(r(x) = 2 - x\), and \(a = 0\) and \(b = 8\), yielding \(V = \pi \int_{0}^{8}[(2)^2 - (2 - x)^2] dy = 8\pi\).

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Key Concepts

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

Disk Method
The disk method is a popular technique for finding the volume of a solid of revolution. It works by slicing the solid into thin, disk-shaped pieces. Here's how it can be visualized:
When we revolve a region around an axis, each slice acts like a disk. The thickness of these disks is negligible, usually represented by the differential element, such as dx or dy.
  • The volume of each disk is given by the formula: \( V = \pi [g(x)]^2 \Delta x \), where \( g(x) \) is the radius of the disk.
  • To find the total volume, we integrate this volume from a start to an end point along the axis of revolution.
  • The definite integral for the total volume is: \( V = \pi \int_{a}^{b}[g(x)]^2 dx \).
The disk method is particularly useful when the disk has a single radius function, like when revolving around the x or y-axis where the solid forms cylindrical layers.
Washer Method
The washer method builds upon the disk method to find the volume of solids of revolution, but it is used when there is a hole in the middle of our disks. These donut-shaped pieces are called washers because there is both an inner and an outer radius.
Here's a step-by-step overview to understanding washers:
  • Unlike disks, washers are used when the region is revolved around a line different than the boundary, creating a central void.
  • The outer radius \( R(x) \) and inner radius \( r(x) \) define the boundaries of the washer.
  • The volume of each washer is: \( V = \pi ([R(x)]^2 - [r(x)]^2) \Delta x \).
  • Integrating over the interval gives the total volume: \( V = \pi \int_{a}^{b}([R(x)]^2 - [r(x)]^2) dx \).
This method is powerful when revolving around lines that create hollow spaces within the solid, capturing the concept of subtracting volumes to account for the void.
Definite Integral
The definite integral is a vital tool in calculus for calculating the total size of an area, length, volume, and more, across a particular interval. Specifically, in determining volumes of solids of revolution, the definite integral helps sum infinitesimal elements to form the whole.
Here's how it applies in this context:
  • Each disk or washer's infinitesimally small volume contributes to the total volume over a range \([a,b]\).
  • The definite integral is expressed as \( \int_{a}^{b} f(x) dx \), combining all micro volumes or areas.
  • This integration guarantees accuracy, as it aggregates infinitely small quantities across a continuous interval, smoothing over approximations.
Using the definite integral in these scenarios allows us to harness the power of calculus to precisely compute volumes, even when dealing with complex shapes or curves.
Revolving Regions
Revolving regions is a fundamental concept in generating three-dimensional solids from two-dimensional shapes. By taking a flat region and revolving it around an axis, we form a solid of revolution.
Here's the essence of this concept:
  • A region, defined by one or more curves or lines, acts as a mold.
  • When revolved around an axis, this region creates a symmetrical solid, with cross-sections that vary based on the geometric path.
  • The process involves either rotating around standard axes, like x or y, or around specific lines creating more unique solids.
  • These concepts are fundamental when applying either the disk or washer method, as the path of revolution dictates the shape and volume of the resulting solid.
Understanding how regions establish the foundation for the resulting geometry enables students to apply solid rotational techniques effectively in calculus problems.

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Most popular questions from this chapter

Evaluate the limit and sketch the graph of the region whose area is represented by the limit. \(\lim _{\| \Delta \rightarrow 0} \sum_{i=1}^{n}\left(4-x_{i}^{2}\right) \Delta x,\) where \(x_{i}=-2+(4 i / n)\) and \(\Delta x=4 / n\)

In Exercises \(1-4\), set up the definite integral that gives the area of the region. $$ \begin{array}{l} f(x)=x^{2}-6 x \\ g(x)=0 \end{array} $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(\int_{a}^{b}[f(x)-g(x)] d x=A,\) then \(\int_{a}^{b}[g(x)-f(x)] d x=-A .\)

If the portion of the line \(y=\frac{1}{2} x\) lying in the first quadrant is revolved about the \(x\) -axis, a cone is generated. Find the volume of the cone extending from \(x=0\) to \(x=6\).

In Exercises \(57-60\), use the Theorem of Pappus to find the volume of the solid of revolution. The solid formed by revolving the region bounded by the graphs of \(y=x, y=4,\) and \(x=0\) about the \(x\) -axis
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