Problem 75 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(0

Step-by-Step Solution

Answer

For what values of \(p\) is the volume of the solid generated by revolving the region \(R\) about the \(y\)-axis finite? Answer: The volume of the solid generated by revolving the region \(R\) about the \(x\)-axis is finite for \(p<\frac{1}{2}\). The volume of the solid generated by revolving the region \(R\) about the \(y\)-axis is finite for \(p>2\).

Step 1: Set up the integral for volume by the disk method

The disk method for solids of revolution about the x-axis requires us to set up the integral \(\displaystyle \pi \int_a^b [f(x)]^2 dx\), where \(f(x)\) is the function describing the curve, and \(a\) and \(b\) are the limits of integration. Here, \(f(x)=x^{-p}\), and the limits are \(0\) (for x) and \(1\). Thus, the integral becomes: \(\displaystyle V = \pi \int_0^1 x^{-2p} dx\).

Step 2: Calculate the integral

In order to find the volume, we need to evaluate the integral. \(\displaystyle V = \pi \int_0^1 x^{-2p} dx = \pi \left[ \frac{x^{-2p+1}}{-2p+1} \right]_0^1 = \pi \left( \frac{1^{-2p+1}}{-2p+1} - \frac{0^{-2p+1}}{-2p+1} \right)\). However, the term \(0^{-2p+1}\) is not well-defined when \(-2p+1 \leq 0\). To avoid this issue, we will take the limit as \(x\) approaches \(0\) instead of evaluating the integral directly at \(0\).

Step 3: Determine when the integral converges

To find the values of \(p\) for which the volume is finite, we need to examine the limit of the expression as \(x\) approaches \(0\). \(\displaystyle V = \pi \lim_{x \to 0} \frac{x^{-2p+1}}{-2p+1}\). The limit converges if and only if \(-2p+1>0\), or when \(p<\frac{1}{2}\). Thus, for \(p<\frac{1}{2}\), the volume of the solid generated by revolving the region \(R\) about the \(x\)-axis is finite. b. Solid generated when \(R\) is revolved about the \(y\)-axis

Step 1: Rewrite the function in terms of y

To use the disk method for solids of revolution about the y-axis, we need to rewrite the function in terms of y. \(f(x)=x^{-p}\) can be rearranged as \(x = y^{-\frac{1}{p}}\).

Step 2: Set up the integral for volume by the disk method

Using the disk method for solids of revolution about the y-axis, we set up the integral \(\displaystyle \pi \int_c^d [g(y)]^2 dy\), where \(g(y)\) is the function in terms of y, and \(c\) and \(d\) are the limits of integration. Here, \(g(y)=y^{-\frac{1}{p}}\), and the limits are \(0\) (for y) and \(1\). Thus, the integral becomes: \(\displaystyle V = \pi \int_0^1 y^{-\frac{2}{p}} dy\).

Step 3: Calculate the integral

We evaluate the integral to find the volume. \(\displaystyle V = \pi \int_0^1 y^{-\frac{2}{p}} dy = \pi \left[ \frac{y^{-\frac{2}{p}+1}}{-\frac{2}{p}+1} \right]_0^1 = \pi \left( \frac{1^{-\frac{2}{p}+1}}{-\frac{2}{p}+1} - \frac{0^{-\frac{2}{p}+1}}{-\frac{2}{p}+1} \right)\). As in the previous case, the term \(0^{-\frac{2}{p}+1}\) is not well-defined when \(-\frac{2}{p}+1 \leq 0\). We will also use the limit as \(y\) approaches \(0\).

Step 4: Determine when the integral converges

To find the values of \(p\) for which the volume is finite, we need to examine the limit of the expression as \(y\) approaches \(0\). \(\displaystyle V = \pi \lim_{y \to 0} \frac{y^{-\frac{2}{p}+1}}{-\frac{2}{p}+1}\). The limit converges if and only if \(-\frac{2}{p}+1>0\), or when \(p > 2\). Thus, for \(p > 2\), the volume of the solid generated by revolving the region \(R\) about the \(y\)-axis is finite.

Key Concepts

Disk Method

The disk method is a technique used in calculus to find the volume of solids of revolution. This method involves slicing the solid perpendicular to the axis of rotation into infinitesimally thin disks. Think of these disks as a stack of tiny pancakes—it helps visualize how the entire solid is constructed. Each disk has a small thickness \(dx\) or \(dy\) depending on the axis of rotation.
The formula for the volume using the disk method, when revolving around the x-axis is \(V = \pi \int_a^b [f(x)]^2 dx\). This involves:

• Defining the function \(f(x)\) which describes the curve you are rotating.
• Setting the limits of integration \(a\) to \(b\), representing the region's boundaries in terms of x.
• Squaring the function since you’re calculating an area, then multiplying by \(+pi\) for the circular disks.

When revolving around the y-axis, you need to re-define your function as \(g(y)\) and adjust the calculation accordingly. For this exercise, express \(f(x)\) in terms of \(y\) using algebra, and then proceed with a similar setup.

Integral Calculus

Integral calculus is fundamental in finding areas, volumes, and other values accumulated over a region. In these problems, integrals help calculate the volume of a solid by summing up infinitesimal slices—hence, the solid's continuous nature.
The process involves setting up an integral that represents the volume using a function that describes a boundary of the region. Start with identifying the curve and setting integration limits based on your region's domain.
For example:

• If your curve is \(f(x) = x^{-p}\) and it’s rotated about the x-axis, the integral becomes \(\pi \int_0^1 x^{-2p} dx\).
• This integrates the squared function from 0 to 1, summing up slices formed as disks or washers.

In solving, you calculate the antiderivative to find exact volumes, adjusting based on variable limits and convergence requirements.

Volume Calculation

Volume calculation using integral calculus, specifically for solids of revolution, involves the methodical application of the disk method or the washer method. You first establish the solid by revolving a region around an axis. The next step is an integral setup based on the geometry formed.
For \(f(x) = x^{-p}\), deriving its volume when revolved:

• Identify it rotates around the x-axis, using the limits from 0 to 1. The function squared forms our disk's radius.
• The integral \(\pi \int_0^1 x^{-2p} dx\) gets evaluated to find volume, employing rules of integration.
• For the y-axis rotation, convert the function in terms of \(y\): \(x = y^{-\frac{1}{p}}\).

This systematic process gives you the volume by integrating over the axis-perpendicular disks, making these elements tangible and calculable.

Limits and Convergence

Understanding limits and convergence is vital, particularly when dealing with improper integrals like those in volume calculation for solids of revolution. When evaluating an integral that meets problematic boundaries, limits help avoid undefined scenarios.
In the case of \(f(x) = x^{-p}\), when revolving around the x-axis, consider:

• Some terms, e.g., \(0^{-2p+1}\), may be undefined for certain values, especially when the exponent expression equals or exceeds zero. This can lead to divergence or an infinite volume.
• The volume converges and is finite only when such terms are resolved, typically through limits approaching 0, verifying conditions like \(-2p+1 > 0 \).

Ensuring convergence requires confirming these results. For the y-axis rotation, similarly, the limit \(\pi \lim_{y \to 0} \) plays a role in determining which conditions (e.g., \(p > 2\)) the volume is finite, ensuring you understand convergence conditions clearly.