Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

22\. The region bounded by \(y=2+\sin x, y=0, x=0\), and \(x=2 \pi\) is revolved about the \(y\) -axis. Find the volume that results. Hint: \(\int x \sin x d x=\sin x-x \cos x+C\).

Step-by-Step Solution

Answer

The volume is \(8\pi^3\).

Step 1: Understand the Solid's Structure

The problem involves rotating the area bounded by the curve \(y=2+\sin x\), the x-axis, and the vertical lines \(x=0\) and \(x=2\pi\) around the y-axis. This will create a three-dimensional solid, and we need to find its volume.

Step 2: Visualize the Integral for Volume

The volume of the solid can be found using the method of cylindrical shells. The formula for volume using cylindrical shells when rotating around the y-axis is \( V = \int_{a}^{b} 2\pi x f(x) \, dx \). In this case, \(f(x) = 2+\sin x\), \(a=0\), and \(b=2\pi\).

Step 3: Set Up the Integral

The integral for volume becomes: \[ V = \int_{0}^{2\pi} 2\pi x (2 + \sin x) \, dx = 2\pi \int_{0}^{2\pi} (2x + x \sin x) \, dx \].Break the integral into two parts: \( \int 2x \, dx \) and \( \int x \sin x \, dx \).

Step 4: Calculate the First Integral

Calculate \(\int 2x \, dx\) from 0 to \(2\pi\). This simplifies as follows:\[ \int 2x \, dx = [x^2]_{0}^{2\pi} = (2\pi)^2 - 0^2 = 4\pi^2 \].

Step 5: Calculate the Second Integral Using the Given Hint

For \(\int x \sin x \, dx\), use the given hint: \[ \int x \sin x \, dx = \sin x - x \cos x + C \].Evaluate from 0 to \(2\pi\): \[ [(-x \cos x + \sin x)]_{0}^{2\pi} = [(-2\pi \cdot \cos 2\pi + \sin 2\pi) - (0 \cdot \cos 0 + \sin 0)] = 0 \].

Step 6: Combine the Results

Now, sum up the results from Steps 4 and 5: \[ V = 2\pi \left( 4\pi^2 + 0 \right) = 2\pi \times 4\pi^2 = 8\pi^3 \].

Step 7: Conclude the Calculation

Thus, the volume of the solid formed by revolving the given region about the y-axis is \( 8\pi^3 \).

Key Concepts

Cylindrical Shells Method

The Cylindrical Shells Method is a powerful technique used in integral calculus to find the volume of a solid of revolution, especially when the solid is rotated about a vertical axis. Unlike the Disk and Washer methods, which involve slicing the solid into disks or washers parallel to the axis of revolution, the Cylindrical Shells Method involves wrapping the solid into cylindrical shells perpendicular to the axis.

This method is particularly useful when the curves involved are functions of x, and the region is rotated around the y-axis. The formula for finding the volume using the Cylindrical Shells Method is:

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

In this formula, \( x \) represents the radius of the cylindrical shell, \( f(x) \) is the height, and \( 2\pi x \) represents the circumference.

This approach efficiently handles cases where the axis of rotation is not a boundary of the region. By integrating over the interval \( [a, b] \), the method calculates the total volume by summing up the volumes of all infinitesimally thin cylindrical shells.

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the concept of integration. It is used to accumulate quantities, such as areas under curves, total mass, or volume—the primary focus in this exercise.

In the context of finding volumes of solids of revolution, integral calculus plays a critical role. By using definite integrals, we can calculate the exact volume of complex shapes and regions when they are revolved around an axis.
For example, when using the Cylindrical Shells Method, we construct an integral that sums up the infinitesimal contributions of each shell to the total volume. In this way, integral calculus transforms problems that appear geometrically complex into solvable mathematical forms.

• Definite integrals accumulate all contributions between two boundaries \( a \) and \( b \).
• The process of integration involves finding an antiderivative and evaluating it over the given bounds.
• In applied contexts, integrals can help determine real-world measurements, like volumes and areas, otherwise difficult to compute.

Understanding integral calculus allows for solving problems like this exercise, where the integration process is used to derive the volume of a solid created by rotation.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are foundational in mathematics and appear frequently in integral calculus, especially when dealing with functions and curves in rotational volume problems.

In this exercise, the function \( y = 2 + \sin x \) plays a central role, as it determines the height of the cylindrical shells when the region is revolved around the y-axis. The sine function is a periodic trigonometric function characterized by:

• Amplitude: The height variation of the wave. Here \( +2 \) shifts the sine graph upwards, impacting the shells' heights.
• Period: The interval length over which the pattern repeats, significant when setting the bounds for integration.
• Phase Shift: Here, the sine begins at 2, rather than starting from zero, adjusting the curve's starting point in the context of shell height.

Trigonometric identities and properties often aid in simplifying integrations involving these functions. For example, when integrating \( \int x \sin x \, dx \), using the given identity helps break down the complexity, making it possible to evaluate the integral effectively.

Recognizing patterns and using properties of trigonometric functions can simplify calculations and lead to more straightforward solutions in problems involving volumes of solids of revolution.