Question

In Exercises 11-20, find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis. $$y=\sec x, \quad y=\sqrt{2}, \quad-\pi / 4 \leq x \leq \pi / 4$$

Short answer

The volume of the solid generated by revolving the region bounded by the curves is \(V=\pi (2 - \pi/2)\) cubic units

Step-by-step solution

Understand The Boundaries And Plot The Functions

Plot the functions \(y = \sec(x)\), \(y = \sqrt{2}\), and vertical lines \(x = -\pi/4\) and \(x = \pi/4\). This will give you the area bounded by these lines which will be revolved around the x-axis.

Identify The Outer And Inner Radii

The space between \(y = \sec(x)\) and \(y = \sqrt{2}\) forms a washer when the bounded area is revolved around the x-axis. The outer radius is the distance from x-axis to the curve \(y = \sec(x)\), represented as \(R(x) = \sec(x)\), and the inner radius is the distance from the x-axis to the line \(y = \sqrt{2}\), represented as \(r(x) = \sqrt{2}\).

Set Up The Integral Using Washer Method

To find the volume of revolution, we will integrate the volume of each infinitesimal washer. The infinitesimal volume is \(\pi (R^2 - r^2) dx\). We will integrate between the x-values of \(-\pi/4\) and \(\pi/4\), leading to the integral expression to calculate the volume: \(V = \int_{-\pi/4}^{\pi/4} \pi ((\sec(x))^2 - (\sqrt{2})^2) dx\)

Solve The Integral

The volume of the solid of revolution is the result of the integral, which can be found by using standard integration methods. The integral of \(\sec^2(x)\) is \(\tan(x)\) and the integral of \(\sqrt{2}^2\) is \(2x\). This leads to: \(V = \pi [\tan(x) - 2x]_{-\pi/4}^{\pi/4}\)

Evaluate The Integral

Finally, evaluate the expression at \(x=\pi/4\) and \(x=-\pi/4\) and subtract the two results to get the volume of the solid: \(V = \pi ( (\tan(\pi/4) - 2*\pi/4) - (\tan(-\pi/4) - 2*(-\pi/4)) ) = \pi (2 - \pi/2)\)