Problem 60 - Calculus Early Transcendentals

Question

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{-x}, y=0, x=0,\) and \(x=p>0\) revolved about the \(x\) -axis (Is the volume bounded as \(p \rightarrow \infty ?\))

Step-by-Step Solution

Answer

Question: Determine if the volume of a solid of revolution formed by revolving the region bounded by the functions \(y=e^{-x}, y=0, x=0, x=p>0\) around the \(x\)-axis is bounded as \(p\) approaches infinity. Answer: Yes, the volume is bounded as \(p\) approaches infinity and its limit is \(\frac{\pi}{2}\).

Step 1: Sketch the region

Define the functions \(y=e^{-x}\) and \(y=0\), and depict the region bounded by these functions and the axes \(x=0\) and \(x=p\). Sketch the region by plotting these functions on a graph.

Step 2: Set up the integral

To find the volume of the solid of revolution using the disk method, set up the integral as follows: \(V = \pi \int_{0}^{p} (e^{-x})^2\,dx\) where \(V\) is the volume of the solid of revolution, and we square \(e^{-x}\) to obtain the area of each infinitesimal disk formed by revolving the region around the \(x\)-axis.

Step 3: Integrate the function

Integrate the function with respect to \(x\): \(V = \pi \int_{0}^{p} e^{-2x}\,dx\) To integrate, we will use a substitution method. Let \(u=-2x\), then \(du=-2dx\). Now, the new integral becomes: \(V = \pi \int_{0}^{p} -\frac{1}{2}e^u\,du\) Evaluating this integral, we get: \(V = -\frac{\pi}{2}[e^u]_0^p = -\frac{\pi}{2} (e^{-2p} - e^0)\) Simplify the expression: \(V = \frac{\pi}{2}(1 - e^{-2p})\)

Step 4: Evaluate the limit

We will now determine if the volume is bounded as \(p \rightarrow \infty\) by evaluating the limit: \(\lim_{p\rightarrow\infty} V = \lim_{p\rightarrow\infty} \left(\frac{\pi}{2}(1 - e^{-2p})\right)\) As \(p\) approaches infinity, \(e^{-2p}\) approaches 0: \(\lim_{p\rightarrow\infty} V = \frac{\pi}{2}(1 - 0)\) So, the volume is bounded as \(p \rightarrow \infty\) and its limit is: \(V = \frac{\pi}{2}\)

Key Concepts

The Disk Method

The disk method is a powerful technique used to find the volume of a solid of revolution. This method involves slicing the solid into a series of disk-shaped cross-sections. Each disk is imagined as being perpendicular to the axis of rotation. In this exercise, the region between the curve and the x-axis was revolved around the x-axis, creating solid disks. The volume of each thin disk is calculated, and then these volumes are summed up using integration.
The formula for the volume of a single disk is determined by its radius and thickness. The radius of the disk is given by the function value, in this case, \(y = e^{-x}\). When the function is squared, \((e^{-x})^2\), it represents the area of the cross-section of a disk. The thickness of each disk is an infinitesimal change in \(x\), denoted as \(dx\).
Thus, the integral

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

is used to accumulate these disk volumes from one edge of the region (from \(x = 0\) to \(x = p\)). This results in the total volume of the solid created by the revolution.

Understanding Improper Integrals

Improper integrals are an important concept, especially when calculating the volume of some solids of revolution. These occur when either the interval of integration is infinite or when the function being integrated has an infinite discontinuity within the interval. In our problem, the limit as \(p \to \infty\) was considered.
Improper integrals require evaluating limits to determine if they converge to a finite value or diverge to infinity. In the given exercise, analyzing the behavior of the exponential decay \(e^{-2p}\) was crucial. As \(p\) becomes very large, \(e^{-2p}\) tends rapidly towards zero, making the overall integral approach a bounded number:

• \(\lim_{p\rightarrow\infty} \left(\frac{\pi}{2}(1 - e^{-2p})\right)\)
• Evaluating this limit showed the volume was finite and equaled \(\frac{\pi}{2}\).

This ensures comprehension of when a volume is considered bounded, particularly when dealing with infinite limits in integrals.

Exploring Exponential Functions

Exponential functions involve expressions where the variable appears in the exponent. They naturally arise in many real-world scenarios, such as in growth and decay processes. In mathematics, they are powerful and exhibit unique traits due to their continuous growth and decay patterns.
In this exercise, the exponential function \(y = e^{-x}\) represents an exponential decay. As \(x\) increases, the value of \(y\) decreases, tending towards zero but never quite reaching it. This property was pivotal when revolving the region to create a volume. The resulting figure highlighted the diminishing contribution of each cross-sectional disk as the distance along the x-axis increased.
Some key properties of exponential functions include:

• Exponential decay, where \(f(x) = e^{-x}\) decreases rapidly as \(x\) increases.
• Finding solutions often involves manipulating the expression, such as squaring \(e^{-x}\) to fit the disk method formula.
• The behavior at infinity plays a central role, evident in analyzing the limits of the integral when \(p\) approaches infinity.

Through understanding these properties, students can effectively solve problems involving volumes, rates, and changes in growth or decay.