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In Exercises \(23-26,\) 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. \(y=3 x, \quad y=0, \quad x=2\) a. The \(y\) -axis b. The line \(x=4\) c. The line \(x=-1\) d. The \(x\) -axis e. The line \(y=7\) f. The line \(y=-2\)

Short Answer

Expert verified
a) \(16\pi\); b) \(32\pi\); c) \(28\pi\); d) \(0\); e) \(12\pi\); f) \(24\pi\).

Step by step solution

01

Understanding the Problem

The problem requires using the shell method to find the volumes of generated solids by revolving a region bounded by the curves around specified lines. The region is bounded by the curves \( y = 3x \), \( y = 0 \), and \( x = 2 \).
02

Setting up the Shell Method Formula

The shell method for a region bounded by vertical lines \( x = a \) to \( x = b \) is given by the formula: \[ V = 2\pi \int_{a}^{b} (x)(f(x) - g(x))\, dx \]where \( f(x) \) is the top curve and \( g(x) \) is the bottom curve. Here, \( f(x) = 3x \) and \( g(x) = 0 \), with limits from \( x = 0 \) to \( x = 2 \).
03

Volume when Revolving around the y-axis

Revolve around the \( y \)-axis: The shell method volume formula becomes \[ V = 2\pi \int_{0}^{2} (x)(3x - 0)\, dx = 2\pi \int_{0}^{2} 3x^2\, dx \].Evaluating the integral gives:\[ V = 2\pi [x^3]_{0}^{2} = 2\pi [8] = 16\pi \].
04

Volume when Revolving around x = 4

Revolve about \( x = 4 \): The radius of the shell becomes \( 4 - x \). The volume becomes: \[ V = 2\pi \int_{0}^{2} (4-x)(3x)\, dx \].This expands to \[ V = 2\pi \int_{0}^{2} (12x - 3x^2)\, dx \] which evaluates to \[ V = 2\pi ([6x^2 - x^3]_{0}^{2}) = 2\pi [24 - 8] = 32\pi \].
05

Volume when Revolving around x = -1

Revolve about \( x = -1 \): The radius of the shell is \( x - (-1) = x + 1 \). The integral becomes:\[ V = 2\pi \int_{0}^{2} (x+1)(3x)\, dx \].This expands to \[ V = 2\pi \int_{0}^{2} (3x^2 + 3x)\, dx \] which evaluates to\[ V = 2\pi [x^3 + \frac{3}{2}x^2]_{0}^{2} = 2\pi [8 + 6] = 28\pi \].
06

Volume when Revolving around the x-axis

Revolves around the \( x \)-axis: The shell method doesn't apply directly as the axis of rotation changes. However, treat this as horizontal slices (washers/cylindrical shells not typical) which evaluates to trivial (since we just have a flat region along width). Hence the volume is \( 0 \).
07

Volume when Revolving around y = 7

For rotation about \( y = 7 \): The height of cylindrical shell becomes \( 7 - 3x \). Calculate:\[ V = 2\pi \int_{0}^{2} x(7 - 3x)\, dx \].This expands to \[ V = 2\pi \int_{0}^{2} (7x - 3x^2)\, dx \] which evaluates to\[ V = 2\pi ([\frac{7x^2}{2} - x^3]_{0}^{2}) = 2\pi [14 - 8] = 12\pi \].
08

Volume when Revolving around y = -2

For rotation about \( y = -2 \): Shell height becomes \( 3x + 2 \). Calculate:\[ V = 2\pi \int_{0}^{2} x(3x + 2)\, dx \].This expands to \[ V = 2\pi \int_{0}^{2} (3x^2 + 2x)\, dx \] which evaluates to\[ V = 2\pi [x^3 + x^2]_{0}^{2} = 2\pi [8 + 4] = 24\pi \].

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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 revolution," we refer to the process of generating a three-dimensional solid by revolving a two-dimensional area around a given axis. This technique is crucial in engineering and mathematics for determining the properties of objects that have rotational symmetry.

Consider a simple flat shape, like a rectangle or a triangle. If you spin this shape around a line (an axis), it sweeps out a volume in space, creating a solid. Calculating the volume of this solid involves determining how much space it occupies, a task made easier by various calculus techniques.

Using calculus, specifically the integral calculus, we can compute volumes of complex shapes that are created when curves define the boundaries of regions to be revolved. This is where methods like the shell method or disk/washer methods come into play, allowing us to compute these volumes accurately.
Cylindrical Shells
Cylindrical shells provide a unique approach to finding the volume of revolution by considering thin, hollow tubes stacked together to fill the space of the solid. Imagine a can of soda: it's essentially a cylindrical shell revolving around the center. Each slice or shell around the axis makes up part of the total volume.

The shell method formula, \[ V = 2\pi \int_{a}^{b} (x)(f(x) - g(x))\, dx \]gives us a powerful way to sum up these slices. The formula integrates across the specified bounds and considers the cylindrical shell's height and radial distance from the axis of rotation. Here, the integration limits are determined by the bounds of the region being revolved, and the functions \(f(x)\) and \(g(x)\) represent the curves forming the region's upper and lower boundaries.

By using this method effectively, we can analyze how each small part (or shell) contributes to the overall volume, making it especially useful when dealing with complicated regions and asymmetric objects.
Integral Calculus
Integral calculus is the branch of mathematics focused on the concept of integration, which generally deals with the accumulation of quantities and the areas under and between curves. In the context of volumes of revolution, integral calculus helps us sum up an infinite number of infinitesimally small pieces to find the total volume.

Integration, as used in the shell method, involves calculating an integral to add up all the shells spanning from the start to the end of the region, with each shell's contribution being part of the whole. It’s like adding up slices of bread to make a full loaf.

The definite integral, in terms of volume calculation, evaluates the total area under the curve when revolved about the axis. Mathematically, this is achieved by defining the integral on specific bounds, informed by the functions involved. This analysis allows accurate creation of models from theoretical shapes, leading directly to applications in physics and engineering.
Revolving Solids
The concept of revolving solids involves generating three-dimensional objects by rotating a two-dimensional shape around a predefined axis. These solids are known as "solids of revolution." The challenge here is understanding how a flat shape like a curve or a line behaves when spun around, and how this results in a solid with volume.

Each revolution changes a simple curve or area into a more complex, round shape, like turning a simple line segment into a cone by revolving around its base. This revolution creates objects whose properties can then be analyzed to understand more about their physical capacities, like mass, center of gravity, or material strength.

Calculating the volumes of such solids requires applying mathematical methods that consider not just the shape but also its symmetry and other geometric properties. Understanding these concepts deeply is key in disciplines that need precise measurement and modeling, such as computer-aided design (CAD), manufacturing, and architectural engineering.

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Most popular questions from this chapter

Volume of a hemisphere Derive the formula \(V=(2 / 3) \pi R^{3}\) for the volume of a hemisphere of radius \(R\) by comparing its cross-sections with the cross- sections of a solid right circular cylinder of radius \(R\) and height \(R\) from which a solid right circular cone of base radius \(R\) and height \(R\) has been removed, as suggested by the accompanying figure.

Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3L if you make it 9 \(\mathrm{cm}\) deep and give the sphere a radius of 16 \(\mathrm{cm}\) . To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (1 \(=1000 \mathrm{cm}^{3}\) )

For some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the \(y\) -axis, for example, and washers are used, we must integrate with respect to \(y .\) It may not be possible, however, to express the integrand in terms of \(y .\) In such a case, the shell method allows us to integrate with respect to \(x\) instead. Exercises 29 and 30 provide some insight. $$ \begin{array}{l}{\text { Compute the volume of the solid generated by revolving the triangular }} \\ {\text { region bounded by the lines } 2 y=x+4, y=x, \text { and } x=0} \\ {\text { about }}\end{array} $$ $$ \begin{array}{l}{\text { a. the } x \text { -axis using the washer method. }} \\\ {\text { b. the } y \text { -axis using the shell method. }} \\ {\text { c. the line } x=4 \text { using the shell method. }} \\ {\text { d. the line } y=8 \text { using the washer method. }}\end{array} $$

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(y\)-axis. The region enclosed by \(x=\sqrt{5} y^{2}, \quad x=0, \quad y=-1, \quad y=1\)

By integration, find the volume of the solid generated by revolving the triangular region with vertices \((0,0),(b, 0),(0, h)\) about a. the \(x\)-axis. \(\quad\) b. the \(y\)-axis.

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