Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(R\) be the region in the first quadrant below the curve \(y=x^{-2 / 3}\) and to the left of \(x=1\). (a) Show that the area of \(R\) is finite by finding its value. (b) Show that the volume of the solid generated by revolving \(R\) about the \(x\)-axis is infinite.

Step-by-Step Solution

Answer

The area is 3, and the volume is infinite.

Step 1: Understand the Problem

We need to determine the area of the region \( R \) under the curve \( y = x^{-2/3} \) from \( x = 0 \) to \( x = 1 \), then find the volume when this region is revolved about the \( x \)-axis.

Step 2: Set Up the Integral for Area

The area \( A \) of region \( R \) can be found by integrating \( y = x^{-2/3} \) with respect to \( x \) from 0 to 1. The integral is set up as follows:\[ A = \int_{0}^{1} x^{-2/3} \, dx \]

Step 3: Evaluate the Integral for Area

Calculate the integral:\[ A = \left[ \frac{x^{1/3}}{1/3} \right]_{0}^{1} = 3\left[ x^{1/3} \right]_{0}^{1} = 3(1^{1/3} - 0^{1/3}) = 3(1 - 0) = 3 \]Thus, the area of the region \( R \) is 3.

Step 4: Set Up the Integral for Volume

Using the disk method, the volume \( V \) of the solid formed by revolving the region \( R \) around the \( x \)-axis is:\[ V = \pi \int_{0}^{1} (x^{-2/3})^2 \, dx \]Simplify the integrand:\[ V = \pi \int_{0}^{1} x^{-4/3} \, dx \]

Step 5: Evaluate the Integral for Volume

Integrate to find the volume:\[ V = \pi \left[ \frac{x^{-1/3}}{-1/3} \right]_{0}^{1} = -3\pi [x^{-1/3}]_{0}^{1} \]Calculating the limits, we find:\[ V = -3\pi (1^{-1/3} - \lim_{x \to 0^{+}} x^{-1/3}) \]The term \( \lim_{x \to 0^{+}} x^{-1/3} \) approaches infinity, making the volume infinite.

Key Concepts

Integration

Integration is a fundamental concept in calculus. It allows us to find quantities like areas, volumes, and other accumulations over a specific interval. In simpler terms, integration helps us sum up small pieces to find the whole value for continuous functions.
When integrating a function, we often look for the antiderivative, which is a function whose derivative is the original function we started with.

In our example, we calculated the area under the curve \( y = x^{-2/3} \) using the definite integral from 0 to 1:

• Set up the integral: \( A = \int_{0}^{1} x^{-2/3} \, dx \)
• Find the antiderivative: \( \left( \frac{x^{1/3}}{1/3} \right) \)
• Evaluate at the boundaries: \( 3(1 - 0) = 3 \)

Thus, integration provided us with a finite area of 3 for region \( R \). This showcases how integration helps us find areas of regions under a curve.

Disk Method

The disk method is a technique used in calculus to find the volume of a solid of revolution.
When a region in the plane is revolved around a line (usually one of the axes), this method helps in calculating the resultant volume.

Here’s a brief overview of how the disk method works:

• You take thin slices (disks) of the solid perpendicular to the axis of rotation.
• Each of these slices has a thickness and a radius that depends on the distance from the axis.
• By summing up the volumes of all these disks, you get the total volume of the solid.

In our problem, we revolve region \( R \) around the \( x \)-axis. We use the integral:

• \( V = \pi \int_{0}^{1} (x^{-2/3})^2 \, dx \)
• This simplifies to \( \pi \int_{0}^{1} x^{-4/3} \, dx \).

Unfortunately, this resulted in an improper integral which led to an infinite volume, showcasing how sometimes, even bounded areas can produce unbounded volumes.

Improper Integrals

Unlike regular integrals, improper integrals deal with infinite limits or discontinuous functions within the integral’s bounds.
These integrals extend our ability to evaluate functions that approach infinity or have undefined points over a specific interval. This involves using limits to handle the infinite behavior.

In our scenario, the integral
\[ V = \pi \int_{0}^{1} x^{-4/3} \, dx \]
required us to evaluate behavior near zero, a point where the function is unbounded.

The process usually involves:

• Evaluating the integral as you approach the problematic boundary.
• Taking the limit of the evaluated expression as it nears the point of discontinuity or infinity.

In this exercise, the result of the limit was infinite. Even though the initial region had a finite area, improper integrals showed the volume to be infinite after rotation. Improper integrals highlight the fascinating complexity of calculus when it encounters discontinuities or infinite stretches.