Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(8\pi^3 + 4\pi^2\).

Step 1: Understand the Problem

We need to find the volume of the solid obtained by revolving the region bounded by the curve \(y = 2 + \sin x\), the lines \(y = 0\), \(x = 0\), and \(x = 2\pi\) about the \(y\)-axis.

Step 2: Set Up the Method of Cylindrical Shells

To find the volume of the solid of revolution, we apply the method of cylindrical shells. The formula for the volume is:\[V = \int_{a}^{b} 2\pi x (f(x) - g(x)) \, dx\]where \(f(x) = 2 + \sin x\) and \(g(x) = 0\) over the interval \([0, 2\pi]\).

Step 3: Simplify the Integral Expression

Substitute the given functions into the volume formula:\[V = \int_{0}^{2\pi} 2\pi x (2 + \sin x) \, dx\]Simplifying further, we get:\[V = 2\pi \int_{0}^{2\pi} (2x + x \sin x) \, dx\]This can be split into two separate integrals:\[V = 2\pi \left( \int_{0}^{2\pi} 2x \, dx + \int_{0}^{2\pi} x \sin x \, dx \right)\]

Step 4: Evaluate \(\int_{0}^{2\pi} 2x \, dx\)

Compute the first integral:\[\int_{0}^{2\pi} 2x \, dx = \left[ x^2 \right]_{0}^{2\pi} = (2\pi)^2 - 0^2 = 4\pi^2\]

Step 5: Evaluate \(\int_{0}^{2\pi} x \sin x \, dx\) Using the Hint

Using the hint provided:\[\int x \sin x \, dx = \sin x - x \cos x + C\]Evaluate from 0 to 2\pi:\[\left[ \sin x - x \cos x \right]_{0}^{2\pi} = (0 - 2\pi(-1)) - (0 - 0) = 2\pi\]

Step 6: Compute the Total Volume

Combine the results from Steps 4 and 5 into the volume expression:\[V = 2\pi \left( 4\pi^2 + 2\pi \right)\]Simplify:\[V = 2\pi (4\pi^2 + 2\pi) = 8\pi^3 + 4\pi^2\]

Key Concepts

Cylindrical Shells Method

The Cylindrical Shells Method is an elegant technique in calculus used for finding the volume of solids of revolution. Imagine a sheet of paper rolled into a cylinder; that's what we deal with when using this method. This approach is particularly useful when the solid is revolved about an axis, especially the vertical one, like the y-axis.

In this method, each 'shell' is a cylindrical slice of the solid. The volume of the entire shape can be calculated by adding up the volumes of these infinitesimally thin shells. To achieve this, we use integration. The formula for the volume using cylindrical shells when revolving around the y-axis is given by:

\[V = \int_{a}^{b} 2\pi x (f(x) - g(x)) \, dx\]
where:

• \(x\) is the distance of the shell from the axis of rotation.
• \(f(x)\) and \(g(x)\) are the top and bottom functions defining the region.
• \([a, b]\) specifies the interval of integration on the x-axis.

In the given problem, this method allows us to transform the complex problem of finding the volume into manageable integrals.

Volume Integration

Volume integration is a powerful technique used to find the volumes of three-dimensional shapes by integrating cross-sectional areas. There are various forms of volume integration, but when dealing with solids of revolution, two primary methods shine: the Disk Method and the Cylindrical Shells Method.

The choice between these methods often depends on the axis of rotation compared to the function's orientation. The Cylindrical Shells Method is often simpler when the axis of rotation is parallel to the height of the solid, like in this problem about rotating a region around the y-axis.

In volume integration using the shells method, we essentially calculate the volume of a singular, infinitesimally thin cylindrical shell and then sum these volumes over the interval \([a, b]\). The integration process breaks down the solid into tiny pieces, evaluates each piece's volume, and then aggregates them to form the total volume.

Sine Function

The sine function, denoted as \(\sin(x)\), is a fundamental trigonometric function crucial for modeling periodic phenomena. It oscillates between -1 and 1 and exhibits a regular wave pattern over its domain.

In the given problem, the region involved includes the sine function translated upwards by 2 units, represented by \(y = 2 + \sin(x)\). This translation does not affect the periodicity or the graph's wave pattern but shifts it vertically, meaning all points are 2 units higher than the basic sine function.

This transformation impacts how we set up integrals for volume calculations. The integral must account for both the constant shift and the wave-like behavior of the sine function. The challenge here, which is simplified by integration, is to address these oscillations in the calculation process, ensuring they accurately contribute to the total volume of the solid formed by the revolution.