Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the region \(R\) bounded by \(y=1 / x^{3}, x=1, x=3\), and \(y=0\). Set up (but do not evaluate) integrals for each of the following. (a) Area of \(R\) (b) Volume of the solid obtained when \(R\) is revolved about the \(y\)-axis (c) Volume of the solid obtained when \(R\) is revolved about \(y=-1\) (d) Volume of the solid obtained when \(R\) is revolved about \(x=4\)

Step-by-Step Solution

Answer

Setup of integrals for area and volumes around specified axes are provided.

Step 1: Identify Region Boundaries

First, identify the boundaries of the region \( R \). The curve is given by \( y = \frac{1}{x^3} \), and it is bounded between \( x = 1 \) and \( x = 3 \). The region is above the x-axis, so \( y = 0 \) is also a boundary.

Step 2: Sketch the Region

To sketch the region \( R \), plot the curve \( y = \frac{1}{x^3} \) between \( x = 1 \) and \( x = 3 \). Draw vertical lines at \( x = 1 \) and \( x = 3 \), and a horizontal line along \( y = 0 \). This outlines region \( R \).

Step 3: Set Up Area Integral

The area \( A \) of region \( R \) can be found by integrating \( y = \frac{1}{x^3} \) with respect to \( x \) from \( 1 \) to \( 3 \). The integral is: \[ A = \int_{1}^{3} \frac{1}{x^3} \, dx \]

Step 4: Volume About y-axis

To find the volume of the solid formed by revolving region \( R \) about the \( y \)-axis, use the shell method. The formula is: \[ V = 2 \pi \int_{1}^{3} x \cdot \frac{1}{x^3} \, dx \] Simplified, this is: \[ V = 2 \pi \int_{1}^{3} \frac{1}{x^2} \, dx \]

Step 5: Volume About y = -1

To set up the integral for the volume when revolving \( R \) about \( y = -1 \), apply the shell method. Each shell's radius is \( x \), and the height is \( 1 + \frac{1}{x^3} \). The volume integral is: \[ V = 2 \pi \int_{1}^{3} x \left(1 + \frac{1}{x^3}\right) \, dx \] Simplified as: \[ V = 2 \pi \int_{1}^{3} \left( x + \frac{1}{x^2} \right) \, dx \]

Step 6: Volume About x = 4

To find the volume when revolving \( R \) about \( x = 4 \), use the washer method. The outer and inner radii of each washer are \( 4 - 1 = 3 \) and \( 4 - x \), respectively. The volume integral is: \[ V = \pi \int_{1}^{3} [(3)^2 - (4-x)^2] \, dy \] Simplified further: \[ V = \pi \int_{1}^{3} [9 - (16 - 8x + x^2)] \, dy \]

Key Concepts

definite integrals

A definite integral is a fundamental concept in calculus that represents the area under a curve within a given interval. When you see an expression like \[ \int_{a}^{b} f(x) \, dx \] it calculates the area under the function \( f(x) \) from \( x = a \) to \( x = b \). This helps us in quantifying the exact area enclosed between a curve, the x-axis, and the vertical lines \( x = a \) and \( x = b \).
For example, in the problem given, the definite integral \( \int_{1}^{3} \frac{1}{x^3} \, dx \) helps us find the area of the region \( R \) between the curve \( y = \frac{1}{x^3} \) and the lines \( x = 1 \) and \( x = 3 \). This represents the total area underneath the curve from \( x = 1 \) to \( x = 3 \), above the x-axis.

Definite integrals not only help with areas but can also be applied in various scenarios where accumulation or total value within an interval is essential.

volume of revolution

The volume of revolution is a method used in calculus to determine the volume of a 3D object obtained when a region or a function is revolved around a specified axis. Imagine taking a shape and rotating it around a line; this creates a 3-dimensional solid, such as a cylinder or doughnut.
To calculate this volume, we employ techniques like the disk method, washer method, or shell method. The disk method involves slicing the solid into thin disks perpendicular to the axis of rotation, while the washer method extends this idea to objects with cavities, and the shell method is typically used for solids of revolution about vertical lines.
For instance, in the exercise, when region \( R \) is revolved around the \( y \)-axis, the shell method is used to calculate the volume, resulting in an integral in which cylindrical shells are conceptually "stacked" to fill the volume.

area under a curve

Calculating the area under a curve is a common task in calculus, particularly when evaluating definite integrals. This calculation involves determining the space between the curve of a function and the x-axis over a specified interval
.For the provided problem, we calculate the area under the curve \( y= \frac{1}{x^3} \) from \( x=1 \) to \( x=3 \). This area represents the complete space enclosed by the curve, the x-axis, and the vertical boundaries of the interval.
This concept is fundamental because it allows for determining integral values, which can relate to real-world applications like calculating distance, probability distributions, and even economic indicators. Understanding how to find the area under a curve is crucial for students aiming to grasp complex applications of calculus.

shell method

The shell method is a fascinating and effective technique within calculus to find the volume of a solid generated by revolving a region around an axis. It's especially useful when the axis of rotation is parallel to the axis of the function (vertical or horizontal lines).
In this method, imagine dividing the solid into cylindrical shells, then summing their volumes via integration. Each shell's volume is derived from its height (determined by the function), thickness (infinitesimally small), and circumference (depending on the radius from the axis of rotation).
For instance, in the problem, when revolving the region \( R \) around the \( y \)-axis, the shell method involves \[ V = 2 \pi \int_{1}^{3} x \cdot \frac{1}{x^3} \, dx \], simplifying to \[ V = 2 \pi \int_{1}^{3} \frac{1}{x^2} \, dx \]. This method helps visualize and compute volumes efficiently in problems where other methods might be cumbersome. Mastering the shell method provides advantage in understanding not only volumes but the symmetry and properties of solids of revolution.