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Chapter 6: Problem 35

Let \(V_{x}\) and \(V_{y}\) be the volumes of the solids that result when the region enclosed by \(y=1 / x, y=0, x=\frac{1}{2},\) and \(x=b\) \(\left(b>\frac{1}{2}\right)\) is revolved about the \(x\) -axis and \(y\) -axis, respec. tively. Is there a value of \(b\) for which \(V_{x}=V_{y} ?\)

Short Answer

Expert verified
Yes, \(b = 1\) makes \(V_x = V_y\).

Step by step solution

01

Understanding the Problem

We are given a region enclosed by the curve \(y = \frac{1}{x}\), \(y = 0\), \(x = \frac{1}{2}\), and \(x = b\). The problem asks us to find if there is a value of \(b\) for which the volume of the solid when this region is rotated about the x-axis is equal to the volume when it is rotated about the y-axis.
02

Calculate Volume of Solid Rotated about x-axis

The volume of the solid \(V_x\) when the region is revolved around the x-axis is given by the integral using the disk method: \[ V_{x} = \pi \int_{\frac{1}{2}}^{b} \left(\frac{1}{x}\right)^2 \, dx = \pi \left[ -\frac{1}{x} \right]_{\frac{1}{2}}^{b} = \pi \left(-\frac{1}{b} + 2\right). \]
03

Calculate Volume of Solid Rotated about y-axis

The volume of the solid \(V_y\) when the region is revolved around the y-axis is given using the shell method: \[ V_{y} = 2\pi \int_{\frac{1}{2}}^{b} \left(x \cdot \frac{1}{x}\right) \, dx = 2\pi \int_{\frac{1}{2}}^{b} \, dx = 2\pi (b - \frac{1}{2}). \]
04

Set the Volumes Equal and Solve for b

To find the value of \(b\) where \(V_x = V_y\), set the expressions equal: \[ \pi \left(-\frac{1}{b} + 2\right) = 2\pi (b - \frac{1}{2}). \]Divide both sides by \(\pi\) and simplify: \[ -\frac{1}{b} + 2 = 2b - 1. \]Combine and rearrange terms to solve for \(b\):\[ 0 = 2b + \frac{1}{b} - 3 \].
05

Solve the Equation for b

The equation \(2b + \frac{1}{b} - 3 = 0\) can be solved by multiplying through by \(b\) to clear the fraction: \[ 2b^2 - 3b + 1 = 0. \]Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \(A=2, B=-3, C=1\), solve for \(b\):\[ b = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 2 \times 1}}{2 \times 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}. \]This gives \(b = 1\) or \(b = \frac{1}{2}\). Given \(b > \frac{1}{2}\), the solution is \(b = 1\).

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Key Concepts

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

Volume of Solids
In calculus, finding the volume of solids involves integrating across a certain interval to determine the three-dimensional space that an object occupies. When a region enclosed by curves is rotated around an axis, it creates a solid of revolution. The volume of such a solid is determined by integrating the cross-sectional area perpendicular to the axis of rotation.
  • The region is enclosed by boundaries which in this case are the curve defined by the equation, the x-axis, and vertical lines at defined points.
  • The goal is to find the volume of the solid when this region is revolved around a given axis.
  • The integration limits are determined by the intersection points or set boundaries of the enclosed region.
Calculating volumes accurately for these shapes largely relies on derivations through methods like the Disk Method and Shell Method, depending on the axis of rotation.
Disk Method
The Disk Method provides a way to calculate the volume of a solid of revolution when rotating around the x-axis. Imagine slicing the solid into thin disks perpendicular to the axis. The volume of each disk has thickness \( \Delta x \) and resembles a cylinder:
  • The radius of each disk is determined by the function value at that slice, typically \( y \) for a curve \( y = f(x) \).
  • The volume of each tiny disk is thus \( \pi \, (radius)^2 \, \Delta x \).
To find the total volume, integrate this expression over the interval of the x-axis relevant to the region.
For instance, the volume \( V_x \) when the region is rotated around the x-axis is calculated by:\[V_x = \pi \int_{a}^{b} [f(x)]^2 \, dx\]In our problem, this was applied with \( f(x) = \frac{1}{x} \) to get the volume increment across the interval from \( x = \frac{1}{2} \) to \( x = b \).
Shell Method
The Shell Method is another powerful tool for finding the volume of a solid of revolution, especially useful for rotations around the y-axis. In this method, think about building the solid from shells (cylinders) instead of disks:
  • Each shell is a hollow cylinder with thickness \( \Delta x \) parallel to the axis of rotation, and the height is determined by the function describing the region \( y = f(x) \).
  • The circumferential length of each shell is \( 2\pi x \), representing the distance around the shell.
The incremental volume of each shell is given by:\[2\pi \int x \, [f(x)] \, dx\]By integrating from \( x = a \) to \( x = b \), the volume \( V_y \) about the y-axis was calculated in our scenario, effectively summing up all the shells' volumes over the specified interval. This technique suits problems where revolving the region horizontally to the axis doesn't lend itself to easy calculation via Disk or Washer methods.

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