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Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the \(y\) -axis. $$y=1 / x, y=0, x=1, x=3$$

Short Answer

Expert verified
The volume is \(4\pi\).

Step by step solution

01

Identify the Region

Identify the region bounded by the curves: \(y = \frac{1}{x}\), \(y = 0\), \(x = 1\), and \(x = 3\). This region is a part of the first quadrant under the curve \(y = \frac{1}{x}\) from \(x = 1\) to \(x = 3\).
02

Setup the Shell Method Formula

For a solid of revolution about the \(y\)-axis, the shell method formula is: \[ V = 2\pi \int_{a}^{b} x [f(x)] \, dx \] where \(f(x)\) is the height of the shell and \(x\) is the radius. For this problem, \(f(x) = \frac{1}{x}\), \(a = 1\), and \(b = 3\).
03

Integrate to Find the Volume

Substitute \(f(x)\) into the shell method formula:\[ V = 2\pi \int_{1}^{3} x \left( \frac{1}{x} \right) \, dx = 2\pi \int_{1}^{3} 1 \, dx \]Evaluate the integral:\[ V = 2\pi [x]_{1}^{3} = 2\pi [3 - 1] = 2\pi \times 2 \]
04

Calculate the Volume

Multiply the result of the integral by \(2\pi\):\[ V = 4\pi \]
05

Conclusion

The volume of the solid generated by revolving the region around the \(y\)-axis is \(4\pi\).

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Key Concepts

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

Volume of Solids
When we talk about the volume of solids, especially in calculus, we're interested in figuring out how much space a three-dimensional object occupies. Imagine filling a bucket with water and considering how much water it can hold. That's a measure of volume. In calculus, we calculate this precisely using methods such as integration.

To find the volume of solids generated by revolving a region, we need to consider the shape and dimensions of the original region. We can use various methods like the disk method and the shell method, each with specific advantages depending on the problem at hand. Understanding which method to use requires recognizing the symmetry of the solid and its axis of rotation.

By mastering integration, you can find the volume of very complex shapes just as easily as simple ones. For instance, the volume of our object formed in this exercise is calculated through a systematic approach using the shell method. The result is a neat value of \(4\pi\), indicative of the solid's rotational symmetry and inherent geometric properties.
Shell Method
The shell method is a powerful technique to calculate the volume of solids of revolution. It's particularly useful when the solid is revolved around a vertical line, like the y-axis. Instead of slicing the solid into disks or washers, we consider it as a series of cylindrical shells.

In this method:
  • Each cylindrical shell has a small thickness \(dx\).
  • The radius of these shells is the distance from the axis of rotation to the shell, represented by \(x\) in our integral formula.
  • The height of the shell is given by the function of \(x\), \(f(x)\).

The volume of each small shell is approximately the circumference of the circle (given by \(2\pi x\)), multiplied by the height \(f(x)\), and the thickness \(dx\). Integrating this from the lower to upper bounds \(a\) to \(b\) gives the total volume of the solid.

The shell method formula:\[ V = 2\pi \int_{a}^{b} x [f(x)] \, dx \]is straightforward yet versatile, allowing easy handling of complex problems. In our example, using \(f(x) = \frac{1}{x}\) resulted in a simple integral to calculate, leading directly to the answer.
Revolution about the y-axis
Revolution about the y-axis refers to rotating a two-dimensional region around a vertical line, creating a three-dimensional object. This is a common technique in calculus to derive volumes of complex solids using symmetry.

Imagine a flat sheet of paper shaped like the region. If you were to rotate this paper around the y-axis, it sweeps out a volume in space, forming a solid that has rotational symmetry about the y-axis.

In practice:
  • You identify the region of interest based on the curves given.
  • For this exercise, the region is bounded by the curve \(y = \frac{1}{x}\), between \(x = 1\) and \(x = 3\).
  • When this area is revolved around the y-axis, it forms a band-like volume.

This method allows you to harness mathematics to explore how flat shapes in the x-y plane can form intriguing volumes when rotated. It simplifies the calculation by mathematically capturing the essence of rotation in integral form, leading us to correctly find the volume of the solid, \(4\pi\), for the region in consideration.

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Most popular questions from this chapter

In the International System of Units, work and kinetic energy have the same units.

Suppose that a hollow tube rotates with a constant angular velocity of \(\omega\) rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0,\) and assume that the object is at rest and \(r=0\) when \(t=0 .\) It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)]$$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) (a) Graph \(r\) versus \(t\) for \(0 \leq t \leq 1\) (b) Assuming that the tube has a length of \(1 \mathrm{m},\) approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate \(d r / d t\) at the instant that the object reaches the end of the tube. (GRAPH CAN'T COPY).

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Prove that the centroid of a triangle is the point of intersection of the three medians of the triangle. [Hint: Choose coordinates so that the vertices of the triangle are located at \((0,-a),(0, a), \text { and }(b, c) .]\)

A cone-shaped water reservoir is \(20 \mathrm{ft}\) in diameter across the top and \(15 \mathrm{ft}\) deep. If the reservoir is filled to a depth of \(10 \mathrm{ft},\) how much work is required to pump all the water to the top of the reservoir?

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