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Chapter 6: Problem 24

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. \(y=x^{3}, \quad y=8, \quad x=0\) a. The \(y\) -axis b. The line \(x=3\) c. The line \(x=-2\) d. The \(x\) -axis e. The line \(y=8\) f. The line \(y=-1\)

Short Answer

Expert verified
a) \(\frac{256\pi}{5}\), b) \(\frac{164\pi}{5}\), c) \(67.2\pi\). Other revolutions require different methods (e.g., washer).

Step by step solution

01

Understand the Problem

We need to find the volume of the solid formed by revolving the region bounded by the curve \(y = x^3\), the line \(y = 8\), and \(x = 0\) around various axes and lines using the shell method.
02

Shell Method Formula

The shell method formula for the volume is \( V = 2\pi \int_{a}^{b} p(x)h(x) \, dx \), where \( p(x) \) is the radius of the shell and \( h(x) \) is the height of the shell.
03

Identify the Region

The region to be revolved is bounded by \(x = 0\), \(y = x^3\), and \(y = 8\). Therefore, the intersection point of \(y = x^3\) and \(y = 8\) needs to be determined. The solutions of \(x^3 = 8\) is \(x=2\), giving an interval of \([0, 2]\).
04

Step 4a: Revolving about the y-axis

For revolving about the y-axis, \( p(x) = x \) and \( h(x) = 8 - x^3 \). The volume is computed as \( V = 2\pi \int_{0}^{2} x(8-x^3) \, dx = 2\pi \int_{0}^{2} (8x - x^4) \, dx \).
05

Step 5a: Integrate for y-axis

Calculate \( V = 2\pi [4x^2 - \frac{x^5}{5}]_{0}^{2} = 2\pi [4(4) - \frac{32}{5}] = 2\pi [\frac{160}{5} - \frac{32}{5}] \).
06

Step 6a: Simplify Result for y-axis

\( V = 2\pi \times \frac{128}{5} = \frac{256\pi}{5} \).
07

Step 4b: Revolving about the line x=3

For revolving about \(x=3\), \( p(x) = 3 - x \) and \( h(x) = 8 - x^3 \). Calculate the volume: \( V = 2\pi \int_{0}^{2} (3-x)(8-x^3) \, dx \).
08

Step 5b: Integrate for line x=3

Simplify to \( V = 2\pi \int_{0}^{2} [24 - 8x^3 - 3x + x^4] \, dx \). Evaluate: \( V = 2\pi [24x - 8\frac{x^4}{4} - \frac{3x^2}{2} + \frac{x^5}{5}]_{0}^{2} \).
09

Step 6b: Solve Integral for line x=3

Calculate \( V = 2\pi [48 - 32 - 6 + \frac{32}{5}] = 2\pi [10 + \frac{32}{5}] \).
10

Step 7b: Simplify Result for line x=3

\( V = 2\pi \times \frac{82}{5} = \frac{164\pi}{5} \).
11

Step 4c: Revolving about the line x=-2

For revolving about \(x=-2\), \( p(x) = x + 2 \) and \( h(x) = 8 - x^3 \). Calculate the volume: \( V = 2\pi \int_{0}^{2} (x+2)(8-x^3) \, dx \).
12

Step 5c: Integrate for line x=-2

Simplify to \( V = 2\pi \int_{0}^{2} [8x + 16 - x^4 - 2x^3] \, dx \). Evaluate: \( V = 2\pi [4x^2 + 16x - \frac{x^5}{5} - \frac{2x^4}{4}]_{0}^{2} \).
13

Step 6c: Calculate Result for line x=-2

\( V = 2\pi [16 + 32 - 6.4 - 8 ] = 2\pi [33.6] = 67.2\pi \).
14

Step 4d: Revolving about the x-axis

Revolving about the x-axis is not possible with the shell method because it requires rotation around a vertical line. For x-axis, use the method of washers or disks.
15

Step 4e/f: Revolving about lines y=8, y=-1

As with the x-axis, neither \(y=8\) nor \(y=-1\) can be addressed by the shell method around vertical lines. Use appropriate methods for horizontal axes (such as washers or slices).

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Key Concepts

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

Volume of Solids of Revolution
When we discuss the volume of solids of revolution, we're talking about a fascinating way calculus is used to determine the space filled by an object when it's rotated around a particular axis. This is a crucial concept in calculus that combines geometry and integration.
Simply put, how would you calculate the volume of a shape created by revolving a curve or region around a line?
There are several methods for doing this: two popular ones are the Disk/Washer method and the Shell method.
These methods help describe a solid generated by rotating a two-dimensional area around an axis.
  • The Disk/Washer method slices the area into disks or washers perpendicular to the axis.
  • The Shell method, on the other hand, uses cylindrical shells and is handy when dealing with vertical axes of rotation.
The choice of method often depends on the axis of rotation and the function's alignment. Understanding these techniques provides powerful insights into volume calculations for complex solids.
Place special attention on understanding the axis of rotation - it defines how you set up your integral in solving the problem.
Integration Techniques
Integration techniques are fundamental to solving calculus problems, especially when it comes to finding volumes, areas, or solving differential equations.
These techniques are extensions of basic integration rules but can handle more complicated functions and situations.
  • The power of integration shines when dealing with continuous functions that describe regions or solids.
  • In the context of the Shell Method, you often need to set up an integral that includes understanding limits and finding appropriate functions for calculation.
  • Generally, integration may involve manipulating algebraic expressions, factoring, or applying substitution to simplify the integral.
For the shell method, identifying the radius and height of the shell is key.
This involves algebraic manipulation and simplification before integrating.
Once set up, integration becomes the tool that computes the actual volume.
Calculus Problem Solving
Solving calculus problems effectively often involves combining multiple concepts and methods.
Understanding the framework of the problem and breaking it into smaller, more manageable tasks is crucial.
When faced with a volume-related problem using the Shell Method, it's all about carefully pulling together the relevant pieces of information.
Initially, identify the function or functions representing the bounded region.
From there, determine how to set up the problem:
  • Define the radius and height of the cylindrical shell.
  • Find intersection points or boundaries to set your limits of integration.
  • Use geometry knowledge to simplify or rearrange equations as necessary.
Each step builds upon the previous one, requiring both precision and attention to detail.
Ultimately, calculus problem-solving is about applying known techniques to reach a desired solution. Recognizing patterns and common problem types enhances proficiency in navigation through these challenges.

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Most popular questions from this chapter

a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface too. c. Use your utility's integral evaluator to find the surface's area numerically. \(y=\int_{1}^{x} \sqrt{t^{2}-1} d t, \quad 1 \leq x \leq \sqrt{5} ; \quad x\) -axis

Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=x^{3}\) from \(x=0\) to \(x=1\).

Find the centroid of the thin plate bounded by the graphs of the given functions. Use Equations (6) and (7) with \(\delta=1\) and \(M=\) area of the region covered by the plate. $$\begin{aligned}&g(x)=0, \quad f(x)=2+\sin x, \quad x=0, \quad \text { and } \quad x=2 \pi\\\&\left(\text {Hint}: \int x \sin x \, d x=\sin x-x \cos x+C .\right)\end{aligned}$$

Use a CAS to perform the following steps for the given graph of the function over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(6.22 .)\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2.4 .8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$f(x)=x^{1 / 3}+x^{2 / 3}, \quad 0 \leq x \leq 2$$

Find the volume of the solid generated by revolving the region bounded by the parabola \(y=x^{2}\) and the line \(y=1\) about a. the line \(y=1\) b. the line \(y=2\) c. the line \(y=-1\)
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