Problem 41 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\)-axis. \(y=\sec x, \quad y=\sqrt{2}, \quad-\pi / 4 \leq x \leq \pi / 4\)

Step-by-Step Solution

Answer

The volume is \(\pi^2 - 2\pi\).

Step 1: Understand the Problem

We need to find the volume of the solid generated by revolving the region bounded by the curves around the x-axis. The curves given are \(y = \sec x\) and \(y = \sqrt{2}\) over the interval \(-\pi/4 \leq x \leq \pi/4\).

Step 2: Setup the Integral for the Volume

The volume of the solid formed by revolving a region around the x-axis can be found using the disk method. The formula is given by \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \), where \(f(x)\) represents the function revolving around the x-axis. Here, \(f(x) = \sqrt{2} - \sec x\).

Step 3: Establish the Limits and Integrand

Since we revolve between \(x = -\pi/4\) and \(x = \pi/4\), and the outer function \(f(x) = \sqrt{2}\) is higher than the inner \(g(x) = \sec x\), our volume integral setup becomes: \[ V = \pi \int_{-\pi/4}^{\pi/4} [\sqrt{2}^2 - \sec^2 x] \, dx \].

Step 4: Simplify the Integrand

First calculate \(\sqrt{2}^2 = 2\). Hence, the integrand inside of the volume calculation becomes \(2 - \sec^2 x\). Thus, our integral expression is:\[ V = \pi \int_{-\pi/4}^{\pi/4} (2 - \sec^2 x) \, dx \].

Step 5: Integrate

Compute the integral:\[ \pi \int_{-\pi/4}^{\pi/4} (2 - \sec^2 x) \, dx = \pi \left[ 2x - \tan x \right]_{-\pi/4}^{\pi/4} \].Evaluate from \(-\pi/4\) to \(\pi/4\).

Step 6: Evaluate the Integral

Evaluate \( \left[ 2x - \tan x \right]_{-\pi/4}^{\pi/4} \):- At \(x = \pi/4\), \(2(\pi/4) - \tan(\pi/4) = \pi/2 - 1\).- At \(x = -\pi/4\), \(2(-\pi/4) - \tan(-\pi/4) = -\pi/2 + 1\).Subtract results to find the total: \[ (\pi/2 - 1) - (-\pi/2 + 1) = \pi - 2 \].Thus, the total volume is: \[ V = \pi(\pi - 2) \].

Step 7: Conclusion

The volume of the solid generated by revolving the region around the x-axis is \( \pi^2 - 2\pi \).

Key Concepts

Disk Method

The Disk Method is a crucial concept in finding volumes of solids of revolution. Imagine slicing a solid into thin, disk-like shapes. Each disk has a small thickness and a radius defined by the curve that is being revolved around an axis. The formula for the volume using the Disk Method is given by:

• \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]

where:

• \(V\) is the volume of the solid.
• \(f(x)\) is the function representing the radius of the disk at a point \(x\).
• \(a\) and \(b\) are the limits of integration, representing the interval over which the solid is being revolved.

For the problem at hand, when revolving the region about the x-axis between \(-\pi/4\) and \(\pi/4\), we used \(f(x) = \sqrt{2} - \sec x\) to calculate the volume. This ensures that we only include the area between the two curves, preventing any overlap.

Calculus Integration

In calculus, integration allows us to calculate the accumulation of quantities, such as areas under curves or volumes of solids. When dealing with volumes of revolution, integration is used to sum up the volumes of infinitely many disks or washers.
The integral:

• \[ \int_{a}^{b} [f(x)] \, dx \]

represents the area under the curve \(f(x)\) from \(x = a\) to \(x = b\). When extended to find volumes, the function \(f(x)\) might represent the radius of rotation, as in the disk method.
In this exercise, we use integration to sum up the thin disk volumes from \(-\pi/4\) to \(\pi/4\), giving us the complete volume. This process involves setting up the integral, simplifying the function inside, and computing.

Definite Integral

The concept of a definite integral is fundamental in finding exact values for areas, volumes, and other quantities in calculus. A definite integral has specific limits, \(a\) and \(b\), and is represented as:

• \[ \int_{a}^{b} f(x) \, dx \]

These limits define the region over which the integration is performed, allowing us to calculate the precise value of the accumulated quantity.
In our volume problem, the definite integral is used to accumulate the disk volumes over the interval \(-\pi/4\) to \(\pi/4\). The setup \[V = \pi \int_{-\pi/4}^{\pi/4} (2 - \sec^2 x) \, dx\] indicates that we compute between these bounds. Evaluating this integral involves applying antiderivatives and determining the precise volume for this range.

Trigonometric Functions

Trigonometric functions like \(\sec(x)\) play an essential role in calculus, especially in problems involving geometric shapes and rotations. The secant function, \(\sec(x)\), is the reciprocal of the cosine function, defined as:

• \(\sec(x) = \frac{1}{\cos(x)}\)

In the exercise, we dealt with \(y = \sec(x)\) as part of the boundary defining the solid of revolution. Remember that:

• \(\sec(x)\) is undefined where \(\cos(x) = 0\), emphasizing the importance of considering appropriate intervals to avoid singularities in calculations.
• When integrated, trigonometric functions often lead to expressions involving other trigonometric identities, like tangent, which were used in our problem-solving steps.

Understanding these functions allows for correctly setting up and simplifying integrals during volume calculations.