Problem 70 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid obtained by revolving the region under the graph of \(y=\sin (x / 2)\) from \(x=0\) to \(x=2 \pi\) about the \(y\)-axis.

Step-by-Step Solution

Answer

The volume is \( 4\pi^{2} \).

Step 1: Identify the problem type

We are dealing with a problem of finding the volume of a solid of revolution. Since the problem involves revolving a region about the y-axis, we need to use the method of cylindrical shells.

Step 2: Set up the cylindrical shell formula

To find the volume using cylindrical shells, the formula is \[ V = 2\pi \int_{a}^{b} x \, f(x) \, dx \]In this case, \( f(x) = \sin(x/2) \), \( a = 0 \), and \( b = 2\pi \).

Step 3: Integrate the function

Substitute \( f(x) = \sin(x/2) \) into the formula: \[ V = 2\pi \int_{0}^{2\pi} x \sin(x/2) \, dx \]To solve this integral, we perform substitution. Let \( u = x/2 \), then \( du = 1/2 dx \) or \( dx = 2du \). Thus, \( x = 2u \). Change the limits of integration: When \( x = 0 \), \( u = 0 \) and when \( x = 2\pi \), \( u = \pi \). The integral becomes:\[ V = 2\pi \, \int_{0}^{\pi} 2u \, \sin(u) \, 2du \]

Step 4: Simplify and solve the integral

Simplify the integral to:\[ V = 4\pi \, \int_{0}^{\pi} u \sin(u) \, du \]To solve this, use integration by parts where \( v = u \) and \( dw = \sin(u) \, du \). Then, \( dv = du \) and \( w = -\cos(u) \). Thus:\[ V = 4\pi \left.-u \cos(u)\right|_{0}^{\pi} + 4\pi \int_{0}^{\pi} \cos(u) \, du \]

Step 5: Evaluate the integral and find the volume

First, evaluate the boundary term:\[ -u \cos(u)\bigg|_{0}^{\pi} = -(\pi)(\cos(\pi)) + (0)(\cos(0)) = \pi \]Next, solve \( 4\pi \int_{0}^{\pi} \cos(u) \, du \):\[ 4\pi \left. \sin(u) \right|_{0}^{\pi} = 4\pi(0 - 0) = 0 \]Hence, the volume is:\[ V = 4\pi \cdot \pi = 4\pi^{2} \]

Step 6: Conclusion

Thus, the volume of the solid obtained by revolving the given region about the y-axis is \( 4\pi^{2} \).

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a technique used to find the volume of a solid of revolution. This method is advantageous when revolving around the y-axis, as seen in this exercise. The concept is based on wrapping a rectangular strip, parallel to the axis of rotation, around it to form a cylindrical shell.
When you revolve a small element of width around an axis, it forms a cylindrical shell. The total volume is calculated by integrating these infinitesimally thin shells. The formula used is \[ V = 2\pi \int_{a}^{b} x f(x) \, dx \]where:

• \( x \) is the distance from the axis of rotation to the shell.
• \( f(x) \) represents the height of the shell.
• \( a \) and \( b \) are the limits of integration.

Applying this to the function \( y = \sin(x/2) \) under rotation about the y-axis, we find the shells by setting up the integral with the right limits and function substitutions.

Integration by Parts

Integration by parts is a crucial tool in integral calculus used to integrate the product of two functions. It is particularly helpful in this exercise, where calculating the volume requires integrating \( u \sin(u) \).
The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]For the exercise in question:

• Set \( u = u \), which implies that \( du = du \).
• Identify \( dv = \sin(u) \, du \), leading to \( v = -\cos(u) \).

By substituting these terms, you can solve the integral, simplifying by evaluating \(-u \cos(u)\) at the boundaries and integrating \( \cos(u) \). Integration by parts helps tackle complex integrals that are products of simple functions.

Calculus Application

This exercise showcases the calculus application in determining the volume of a solid formed by revolution, emphasizing practical scenarios where calculus shines. By understanding how functions behave and utilizing methods such as the cylindrical shells, you can solve real-world volume problems effectively.
Calculus enables us to:

• Analyze and interpret the behavior of functions through integration and differentiation.
• Solve physical problems by setting them up as mathematical equations.

In learning these methods, students can navigate between theory and real-world applications, bridging the gap between abstract concepts and tangible solutions.

Definite Integrals

Definite integrals are fundamental in calculus for calculating the area under curves and solving various types of practical problems, including volume. In this problem, the definite integral computes the volume of the solid of revolution formed from the curves.
The setup requires:

• Determining the correct limits of integration \( a \) and \( b \), based on the given range for \( x \) which is from 0 to \( 2\pi \).
• Substituting the function into the cylindrical shell formula, ensuring the integral represents the actual volume.

Definite integrals take the abstraction out of calculus by providing a bounded estimate over a given interval, leading to precise, calculable values as seen in the solution \( V = 4\pi^2 \).