Problem 24 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=\sqrt{\sin ^{-1} x}, y=\sqrt{\pi / 2}, \text { and } x=0$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\pi\) cubic units.

Step 1: Sketching the Region R

Begin by sketching the curves on the xy-plane. Once you sketch the functions, you can shade the bounded area formed by the intersecting curves. This will help visualize the region we need to revolve around the x-axis.

Step 2: Set up the Disk Method Integral

Typically, the equation for the disk method can be written as follows: \(V = \pi \int_{a}^{b} [f(x)]^2 dx\). In this case, we will use the secant function and the given bounds for the region. Thus, the integral will be set up like this: \(V = \pi \int_{0}^{\frac{\pi}{4}} [\sec(x)]^2 dx\).

Step 3: Evaluate the Integral

First, let's rewrite the integral as follows: \(V = \pi \int_{0}^{\frac{\pi}{4}} [\sec^2(x)] dx\). Notice that the derivative of \(\tan(x)\) is \(\sec^2(x)\). So, we can integrate this function and find the antiderivative: \(\int \sec^2(x) dx = \tan(x) + C\). Now, let's find the volume by evaluating the antiderivative at the given bounds: \(V = \pi \left[\tan\left(\frac{\pi}{4}\right) - \tan(0)\right]\).

Step 4: Calculate the Volume

Finally, let's plug in the values of the bounds into the equation we derived in the previous step and find the volume of the solid: \(V = \pi \left[\tan\left(\frac{\pi}{4}\right) - \tan(0)\right] = \pi [1 - 0] = \pi\). The volume of the solid generated by revolving region R about the x-axis is \(\pi\) cubic units.

Key Concepts

Volume of Revolution

Understanding the concept of the volume of revolution is essential when you're learning about three-dimensional shapes created through calculus. Imagine you have a two-dimensional region on a graph, and you rotate this region around a line (usually an axis). The shape that's formed by this rotation is what we call a solid of revolution, and we can find its volume using various techniques, one of them being the disk method. For instance, in the given exercise, we generate a solid by revolving the region bound by the secant function and the x-axis around the x-axis itself. By applying the disk method, we can transform an abstract two-dimensional area into a tangible three-dimensional object, and calculate the exact volume it would contain.

Integral Calculus

Integral calculus is a branch of mathematics that is concerned with the accumulation of quantities and the area under curves. When faced with the task of finding the volume of a solid of revolution, integral calculus is the tool we reach for. It allows us to sum up infinite circular slices (disks) of the solid to determine its overall volume.

By setting up an integral for the volume, such as \(V = \pi \int_{a}^{b} [f(x)]^2 dx\), we're essentially stacking up an infinite number of infinitesimally thin disks from the interval \(a\) to \(b\) along the x-axis. The area of each disk is \(\pi[f(x)]^2\), representing the circle's area with radius \(f(x)\). Integrating this area over the specified interval gives us the total volume of the solid.

Secant Function

The secant function, denoted as \(\sec(x)\), is one of the six fundamental trigonometric functions, reciprocal to the cosine function, so \(\sec(x) = \frac{1}{\cos(x)}\). It has its idiosyncrasies, such as being undefined for certain values where \(\cos(x) = 0\), causing vertical asymptotes on its graph.

In the exercise, the region bounded by the secant function up to \(x = \frac{\pi}{4}\) and the x-axis is being used to form a solid of revolution. Understanding how the function behaves and what its graph looks like is crucial in setting up the integral for the disk method.

Antiderivative

An antiderivative of a function is another function that reverses the process of differentiation. In other words, if the derivative of function \(F(x)\) is \(f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). This concept is central to solving problems in integral calculus, as finding the antiderivative allows us to evaluate definite integrals, which in turn gives us area, volume, and other accumulation values.

In our exercise, the antiderivative of \(\sec^2(x)\) is needed to find the volume using the disk method. Luckily, since \(\sec^2(x)\) is a well-known derivative of \(\tan(x)\), we can quickly determine the antiderivative and evaluate the integral to find the solid's volume. Remember, don't forget to apply the limits of integration to get the final volume!