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 π cubic units.

Step 1: Graph the region

Plot the given curves y=sec(x), y=0, x=0, and x=π/4 on a coordinate plane. Notice that the region R is bounded by these curves and lies in the first quadrant.

Step 2: Set up the disk method integral

The disk method states that the volume of the solid generated when a region is revolved about the x-axis is given by the integral $$ V = \pi \int_a^b [f(x)]^2\,dx, $$ where f(x) is the height of the region from the x-axis and [a, b] is the interval over which the region is defined. In this case, f(x) = sec(x), because that's the function we're revolving around the x-axis. The interval [a, b] is [0, π/4]. So, the integral representing the volume of the solid generated when R is revolved about the x-axis is $$ V = \pi \int_{0}^{\pi/4} \sec^2(x)\, dx. $$

Step 3: Evaluate the integral

Now, we need to evaluate the integral: $$ V = \pi \int_{0}^{\pi/4} \sec^2(x)\, dx. $$ To find the antiderivative of sec^2(x), notice that the derivative of tan(x) is sec^2(x). Therefore, the antiderivative of sec^2(x) is tan(x). Now, we can evaluate the integral: $$ V = \pi [\tan(x)]_{0}^{\pi/4} = \pi(\tan(\pi/4) - \tan(0)) = \pi(1-0) = \pi. $$ So, the volume of the solid generated when R is revolved about the x-axis is π cubic units.

Key Concepts

Volume of Revolution

When discussing the volume of revolution, we're often referring to the 3D shape formed by rotating a 2D region around a specified axis. This method is widely used in calculus and is crucial for calculating volumes of solids that might otherwise be tricky to determine.
In this problem, the region defined by \(y = \sec x\), \(y = 0\), \(x = 0\), and \(x = \frac{\pi}{4}\) is revolved around the x-axis using the disk method. The disk method involves slicing the solid perpendicular to the axis of revolution, resulting in circular disks.

• The formula for volume in terms of disks is \(V = \pi \int_a^b [f(x)]^2 \, dx\).
• The function \(f(x)\) represents the radius of the disks, originating at the axis of revolution and stretching to the curve.
• Integrating over the interval \([a, b]\) gives the complete volume of the solid.

Through integration, the cumulative volume of all these disks is calculated, producing the volume of the solid. This approach provides an intuitive visualization of how a flat region becomes a 3D object.

Integration Techniques

Integration is a fundamental concept in calculus used to find areas under curves and volumes of solids, among other things. The disk method itself employs integration to calculate these volumes, and it's vital to grasp the underlying techniques.
In the given problem, we perform integration with the function \(\sec^2(x)\) over the interval from 0 to \({\pi}/4\). The goal is to find the antiderivative of this function, which requires knowledge of basic antiderivative rules.
The antiderivative of \(\sec^2(x)\) is known to be \(\tan(x)\), as the derivative of \(\tan(x)\) yields \(\sec^2(x)\). This direct relationship simplifies the integration process.

• First, identify the function to integrate: \(\sec^2(x)\).
• Recognize that its antiderivative is \(\tan(x)\).
• Substitute the limits into the antiderivative: \[V = \pi[\tan(x)]_{0}^{\pi/4}\].
• Calculate the result for the upper and lower limits to find the total volume.

This process emphasizes how understanding the function's derivative can lead directly to solving integration problems efficiently.

Secant Function

The secant function, \(\sec(x)\), is a trigonometric function and is the reciprocal of the cosine function. Understanding this function is essential in problems involving volumes of revolution, especially when it serves as a boundary for regions being revolved.
The secant function has unique properties and behavior due to its relationship with the cosine function. It is defined as \(\sec(x) = \frac{1}{\cos(x)}\), which implies certain domains and ranges:

• It's undefined where \(\cos(x) = 0\) (for example, \(x = \frac{\pi}{2}, \frac{3\pi}{2},...\)).
• This also means \(\sec(x)\) has vertical asymptotes at these points.
• The function ranges from 1 to positive or negative infinity, based on \(\cos(x)\) values.

In the context of our problem, \(\sec(x)\) is used as a boundary in the first quadrant, from \(0\) to \(\frac{\pi}{4}\). It's crucial to visualize this graphically as the region stretching above the x-axis, interacting in the volume calculation process by determining the radius of the disks in the disk method.