Question

Find the volume of the solid generated when \(y=\frac{1}{x}\) for \(1 \leq x \leq 5\) is rotated about the \(x\) axis.

Short answer

Answer: \(\frac{4\pi}{5}\)

Step-by-step solution

Determine the cross-sectional area of the infinitesimal disk

To find the cross-sectional area, we consider a small disk with thickness \(dx\) and radius equal to the function value \(y(x)\). The area of this disk is given by the formula: Area = \(\pi \cdot y^2\) Since \(y = \frac{1}{x}\), we get the Area as: Area = \(\pi\left(\frac{1}{x}\right)^2 = \frac{\pi}{x^2}\)

Determine the infinitesimal volume of the disk

The infinitesimal volume of the disk can be found by multiplying its cross-sectional area with the thickness \(dx\). Infinitesimal volume = Area \(\cdot dx\) Infinitesimal volume = \(\frac{\pi}{x^2}dx\)

Set up the integral to find the total volume

To find the total volume of the solid, we need to integrate the infinitesimal volume with respect to \(x\) along the interval \(1\leq x \leq 5\). This can be written as: Volume = \(\int_{1}^{5} \frac{\pi}{x^2}dx\)

Evaluate the integral

To solve the integral, we notice that the integrand represents the power function. Therefore, we can rewrite it as \(\pi x^{-2}\) and find the antiderivative. Antiderivative of \(\pi x^{-2}\) is \(-\pi x^{-1}+C\) Now apply the limits of integration: \(-\pi\left(\frac{1}{5}\right) - \left(-\pi\left(\frac{1}{1}\right)\right) = -\frac{\pi}{5} + \pi\)

Determine the total volume

Now we just need to simplify the expression to find the total volume of the solid: Volume = \(- \frac{\pi}{5} + \pi = \pi - \frac{\pi}{5} = \frac{5\pi}{5} - \frac{\pi}{5} = \frac{4\pi}{5}\) The volume of the solid generated when \(y=\frac{1}{x}\) for \(1 \leq x \leq 5\) is rotated about the \(x\) axis is \(\frac{4\pi}{5}\).

Key concepts

Integration in Mathematics

Integration is a fundamental tool in mathematics, particularly within the field of calculus. It involves the process of finding the whole from the sum of its parts—essentially determining the accumulated quantity. When we integrate a function, we usually seek the area under the curve of a graph of the function, or more broadly, the total accumulation of some quantity across a specified range.

The process of integration is often used to solve problems involving area, volume, displacement, or any scenario where summing infinite small quantities over a range or interval is required. In the exercise at hand, integration becomes a powerful technique because it allows the calculation of the volume of a solid shape generated by revolving a curve around an axis. The integral aggregates infinitesimally thin disks or washers to determine this volume, which would be extremely difficult to calculate using basic geometry.

Single Variable Calculus

Single variable calculus is the branch of calculus that deals with functions of a single variable and includes the study of limits, derivatives, integrals, and infinite series. It is often the first type of calculus students encounter. The problems and exercises in single variable calculus typically involve finding a quantity related to a one-dimensional function—such as its rate of change or the total area under its curve.

In the context of the given exercise, single variable calculus specifically addresses finding the volume of a solid of revolution. This solid is created by rotating a one-dimensional curve, in this case, \(y=\frac{1}{x}\), around a single axis (the x-axis). The main calculus tool used to find this volume is the integral, which accumulates the infinitely small volumes of individual disks along the interval from \(x=1\) to \(x=5\).

Disk Method

The disk method is a specific technique used in single variable calculus to find the volume of a solid of revolution. When a region in the plane is revolved around a line, such as an axis, the resulting volume can be calculated by imagining it as a stack of thin circular disks.

Each disk's volume is a function of its radius—which is given by the distance from the axis to the function—and its infinitesimal thickness, typically denoted as \(dx\) or \(dy\), depending on the axis of rotation. The disk method involves squaring the function that represents the radius, multiplying by \(\pi\) to find the area of the circular face of the disk, and then integrating over the specified range to find the total volume.

In our exercise, since the solid is created by rotating the function \(y=\frac{1}{x}\) around the x-axis, the radius of each disk is \(y\), and the thickness is \(dx\). Thus, the volume of each infinitesimal disk is \(\pi y^2 dx\), and the total volume is obtained by integrating this expression from \(x=1\) to \(x=5\).