Problem 29 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the washer method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=e^{x / 2}, y=e^{-x / 2}, x=\ln 2, x=\ln 3$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(V=\pi\frac{7}{6}\).

Step 1: Sketch the region

To have a better understanding of the problem, we start by sketching the region R bounded by the given curves. We should plot the functions \(y=e^{x/2}\), \(y=e^{-x/2}\), the vertical lines \(x=\ln 2\) and \(x=\ln 3\), and observe the enclosed region.

Step 2: Identify inner and outer radii

For the washer method, we must identify the inner and outer radii of each washer generated when the region R is revolved about the x-axis. The outer radius is defined by the curve \(y=e^{x/2}\), and the inner radius is defined by the curve \(y=e^{-x/2}\). Since we are revolving around the x-axis, we should express the outer and inner radii in terms of x as follows: \[ Outer\,radius = e^{x/2}, \, Inner\,radius = e^{-x/2} \]

Step 3: Set up the integral

We will now set up the integral for the washer method, which will give us the volume of the solid. The formula for the washer method is: \[ V = \pi \int_{a}^{b} (R_{outer}^2 -R_{inner}^2) \, dx \]where \(V\) is the volume of the solid, \(R_{outer}\) and \(R_{inner}\) are the outer and inner radii, and \(a\) and \(b\) are the limits of integration. We have already found the outer and inner radii in step 2. The limits of integration are given by the vertical lines \(x=\ln 2\) and \(x=\ln 3\). Thus, we have the integral: \[ V = \pi \int_{\ln 2}^{\ln 3} \left[(e^{x/2})^2 - (e^{-x/2})^2\right] \, dx \]

Step 4: Simplify and integrate

We can simplify the expressions inside the integral as follows: \[ (e^{x/2})^2 = e^{x},\, (e^{-x/2})^2 = e^{-x} \]Now we can rewrite the integral as: \[ V = \pi \int_{\ln 2}^{\ln 3} (e^x - e^{-x}) \, dx \]Integrating each term separately, we get: \[ V = \pi \left[\int_{\ln 2}^{\ln 3} e^x \, dx - \int_{\ln 2}^{\ln 3} e^{-x} \, dx\right] \]And after integrating, we have: \[ V = \pi \left[e^x - (-e^{-x})\right]_{\ln 2}^{\ln 3} \]

Step 5: Evaluate the integral

Now it is time to evaluate the integral at the limits of integration: \[ V = \pi \left[\left(e^{\ln 3} - (-e^{-\ln 3})\right) - \left(e^{\ln 2} - (-e^{-\ln 2})\right)\right] \]We can use the properties of logarithms to simplify: \[ V = \pi \left[(3 - \frac{1}{3}) - (2 - \frac{1}{2})\right] \]Computing the final expression, we find the volume: \[ V = \pi \left(\frac{8}{3} - \frac{3}{2}\right) = \pi \frac{7}{6} \]Thus, the volume of the solid generated when region R is revolved about the x-axis is \(V=\pi\frac{7}{6}\).

Key Concepts

Volume of Solid of Revolution

The concept of the volume of a solid of revolution is fascinating and vital in calculus. It involves taking a two-dimensional region and rotating it around an axis, forming a three-dimensional shape, known as a solid of revolution. This is often used to find volumes of objects that have circular symmetry.
The Washer Method, which is specifically used in this context, is applied when the solid has a hole in the middle, resembling a washer. This method allows us to calculate the volume by subtracting the volume of the inner solid from the outer solid. The difference between the areas of two circular disks (outer and inner radii) gives the volume of each washer.

• Outer radius defines the boundary of the solid that is furthest from the axis.
• Inner radius defines the boundary closest to the axis of rotation.

For this exercise, these concepts help in determining the volume when the region bounded by given curves is revolved around the x-axis, emphasizing the need to understand both inner and outer radii.

Integration

Integration is a core mathematical operation used extensively in calculus, and it is essential for calculating areas, volumes, and quantities in general. While differentiation deals with the rate of change, integration accumulates or adds up quantities. In the calculus of solids of revolution, integration becomes crucial when using methods like the Washer Method.
In this problem, after determining the expressions for the outer and inner radii in terms of the variable 'x,' integration is applicable to sum up these expressions over a specific interval. This gives the accumulated volume for the solid generated by the revolution.

• Integration is used here to combine an infinite number of infinitesimally small washers into a finite volume.
• The integral of a function over an interval provides the total value or accumulation of the function over that interval.

These principles of integration are applied to evaluate the volume from the region defined by the curves.

Definite Integral

The definite integral is a specific application of integration, where we evaluate the integral of a function within a given range or interval. It is particularly useful in real-world applications where we need definite quantities, like area or volume over a particular region.
The definite integral essentially "locks in" the limits of integration providing a numeric result, unlike indefinite integrals which provide general forms of functions without specific boundaries. In the Washer Method, setting up the definite integral involves:

• Determining the boundaries of the region (from x = ln 2 to x = ln 3 in our example).
• Using the integral to calculate the precise volume between these bounds.

The evaluated result, such as shown in the problem, gives the volume of the solid as a specific value, ensuring precise calculations necessary for understanding areas and volumes derived from functions.

Exponential Functions

Exponential functions are a type of mathematical function characterized by their rapid growth or decay. They are represented in the form of \( y = a^x \), where 'a' is a positive constant. These functions appear frequently in modeling natural phenomena like population growth, radioactive decay, and more.
In the given problem, the regions are bounded by two exponential functions: \( y = e^{x/2} \) and \( y = e^{-x/2} \). Exponential functions feature prominently in integration problems due to their distinct properties and the fact that their derivatives and integrals are relatively straightforward.

• When differentiated or integrated, exponential functions tend to preserve their forms, making calculations simpler.
• Exponential growth (e.g., \( e^{x/2} \)) results in rapidly increasing values while exponential decay (e.g., \( e^{-x/2} \)) indicates values decreasing at a rate proportional to their current size.

Understanding these functions provides insight into their behavior and aids in setting up and solving integrals for problems involving solids of revolution.