Problem 79
Question
Volume A region in the first quadrant is bounded above by the curve \(y=\cosh x,\) below by the curve \(y=\sinh x,\) and on the left and right by the \(y\) -axis and the line \(x=2,\) respectively. Find the volume of the solid generated by revolving the region about the \(x\) -axis.
Step-by-Step Solution
Verified Answer
The volume of the solid is \(2\pi\).
1Step 1: Set Up the Problem
We are looking to find the volume of the solid generated by rotating the region defined by \(y = \cosh x\) and \(y = \sinh x\) about the \(x\)-axis from \(x = 0\) to \(x = 2\). This can be done using the method of washers (or disks) in calculus.
2Step 2: Express the Volume Element
The volume of a thin washer at a given \(x\) is the area of the washer times its thickness, \(dx\). The outer radius of the washer is given by \(\cosh x\) and the inner radius by \(\sinh x\). The area of the washer is \(\pi((\cosh x)^2 - (\sinh x)^2)\).
3Step 3: Simplify the Washer Area
Using the hyperbolic identity, \((\cosh x)^2 - (\sinh x)^2 = 1\), we see that the area simplifies to \(\pi(1) = \pi\). Thus, the volume element becomes \(\pi \cdot dx\).
4Step 4: Integrate to Find the Volume
Integrate the volume element from \(x = 0\) to \(x = 2\): \[V = \int_{0}^{2} \pi \cdot dx = \pi [x]_{0}^{2} = \pi (2 - 0) = 2\pi.\]
5Step 5: State the Result
The volume of the solid generated by revolving the given region about the \(x\)-axis is \(2\pi\).
Key Concepts
Method of WashersHyperbolic FunctionsIntegral Calculus
Method of Washers
The method of washers is a powerful tool in integral calculus used for finding the volume of a solid of revolution. This method involves revolving a region around an axis to form a solid and then calculating its volume.
- The basic idea is to slice the solid perpendicular to the axis of revolution.
- Each slice forms a washer or disk - essentially a flat cylinder.
- The volume of each thin washer is computed by finding the area of the washer face and multiplying by its negligible thickness, denoted as \(dx\) or \(dy\).
- Outer radius: the distance from the axis to the outer edge, given by the function farther from the axis.
- Inner radius: the distance from the axis to the inner edge, represented by the function closer to the axis.
Hyperbolic Functions
Hyperbolic functions serve as analogs to trigonometric functions, yet they arise from hyperbolas rather than circles.
In this exercise, this identity allows us to easily simplify the washer method's area calculation. By identifying that the area equation \(\pi((\cosh x)^2 - (\sinh x)^2)\) simplifies to \(\pi\), the problem becomes computationally straightforward.
- Definitions:
- Hyperbolic sine, \(\sinh x = \frac{e^x - e^{-x}}{2}\).
- Hyperbolic cosine, \(\cosh x = \frac{e^x + e^{-x}}{2}\).
- These functions are useful in scenarios where growth models, physics equations, or calculus problems extend beyond circular arcs.
In this exercise, this identity allows us to easily simplify the washer method's area calculation. By identifying that the area equation \(\pi((\cosh x)^2 - (\sinh x)^2)\) simplifies to \(\pi\), the problem becomes computationally straightforward.
Integral Calculus
Integral calculus is fundamental for finding areas, volumes, and other quantities under curves or surfaces. By summarizing infinitesimal slices, integration helps calculate continuous processes.
Integration, especially in the context of solids of revolution, is a step-by-step summing reality, piecing together infinitesimally small slices into a coherent, complete volume measure.
- Definite integrals accumulate quantities over an interval \([a, b]\), providing results like total volume or area.
- In this problem, the integral \(\int_{0}^{2} \pi \, dx\) calculates the total volume of the solid between \(x = 0\) and \(x = 2\).
Integration, especially in the context of solids of revolution, is a step-by-step summing reality, piecing together infinitesimally small slices into a coherent, complete volume measure.
Other exercises in this chapter
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