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Magazine Chapter 6: Problem 32
Let \(V\) be the volume of the solid that results when the region enclosed by
\(y=1 / x, y=0, x=2\), and \(x=b(0
Short Answer
Expert verified
The value of \(b\) is \(\frac{2\pi}{3 + \pi}\).
Step by step solution
01
Identify the Volume Formula
The region bounded by the curves is to be revolved around the x-axis. We use the disk method to find the volume of the solid. For a given region rotated about the x-axis, the volume formula is:\[ V = \pi \int_{a}^{b} (f(x))^2 \, dx \]where \(f(x) = \frac{1}{x}\) in this case, and the limits of integration are determined by \(x = 2\) and \(x = b\) (with \(0 < b < 2\)).
02
Set Up the Integral
Set up the definite integral for the volume:\[ V = \pi \int_{b}^{2} \left(\frac{1}{x}\right)^2 \, dx = \pi \int_{b}^{2} \frac{1}{x^2} \, dx \]
03
Integrate the Function
Integrate \(\int \frac{1}{x^2} \, dx\), which results in:\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C \]Applying the limits \(b\) to 2, we evaluate:\[ V = \pi \left[ -\frac{1}{x} \right]_{b}^{2} = \pi \left( -\frac{1}{2} + \frac{1}{b} \right) \]
04
Solve for b Given V = 3
Set the expression for volume equal to 3 and solve for \(b\):\[ \pi \left( \frac{1}{b} - \frac{1}{2} \right) = 3 \]Divide through by \(\pi\):\[ \frac{1}{b} - \frac{1}{2} = \frac{3}{\pi} \]Rearrange to solve for \(b\):\[ \frac{1}{b} = \frac{3}{\pi} + \frac{1}{2} \]\[ b = \frac{1}{\frac{3}{\pi} + \frac{1}{2}} \]
05
Simplify the Expression for b
Simplify the expression:\[ b = \frac{1}{\frac{3}{\pi} + \frac{1}{2}} = \frac{1}{\frac{3 + \pi}{2\pi}} = \frac{2\pi}{3 + \pi} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Volume of Revolution
The concept of the volume of revolution involves creating a 3D figure by rotating a 2D region around an axis. In our exercise, we are rotating the region beneath the curve of the function \( y = \frac{1}{x} \) between \( x = 2 \) and \( x = b \). This region will form a 'solid of revolution' once revolved around the x-axis. To calculate the volume of this solid, the disk method is typically used.
- The disk method involves slicing the solid into thin disks perpendicular to the axis of rotation.
- Each disk has a small thickness \( \Delta x \) and a radius equal to the function \( f(x) \).
- The volume of each thin disk is approximated by the formula \( \pi (f(x))^2 \Delta x \).
Definite Integral
Definite integrals play a crucial role in calculating areas and volumes, particularly in finding the volume of solids formed by revolving a region around an axis. In our scenario, the definite integral \( \pi \int_{b}^{2} \frac{1}{x^2} \, dx \) is used to compute the volume of the solid formed when the area under \( y = \frac{1}{x} \) is revolved around the x-axis.
- A definite integral is denoted with upper and lower limits, which represent the region's bounds.
- The integral calculates the accumulated total area under the curve \( y = \frac{1}{x} \) from \( x = b \) to \( x = 2 \).
- When evaluating the integral, we first find the indefinite integral or antiderivative and then apply the limits to find the definite value.
Solid of Revolution
A solid of revolution is a three-dimensional object created by rotating a two-dimensional shape around an axis. In the given exercise, we consider the region enclosed by \( y = \frac{1}{x} \), \( y = 0 \), and two vertical lines at \( x = 2 \) and \( x = b \). When this region is revolved around the x-axis, it forms a shape known as a solid of revolution.
- Any two-dimensional curve can produce a solid of revolution through rotation.
- This solid's shape and volume can be determined using calculus techniques such as the disk method.
- The axis of rotation is critical, as it influences the geometry of the resulting solid.
