Question

The loudspeaker created by revolving \(y=1 / x\) from \(x=1\) to \(x=4\) around the \(x\) -axis.

Short answer

The volume is \(\frac{3\pi}{4}.\)

Step-by-step solution

Understand the Problem

We need to find the volume of the solid formed by revolving the curve given by the function \(y = \frac{1}{x}\) between \(x = 1\) and \(x = 4\) around the \(x\)-axis.

Set Up the Integral for Volume

The formula for the volume \(V\) of a solid of revolution around the \(x\)-axis using the disk method is \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \(f(x)\) is the function being revolved. Here, \(f(x) = \frac{1}{x}\), \(a = 1\), and \(b = 4\).

Write the Integral Expression

Substitute the function \(f(x) = \frac{1}{x}\) into the volume formula: \[ V = \pi \int_{1}^{4} \left(\frac{1}{x}\right)^2 \, dx \] This simplifies to \[ V = \pi \int_{1}^{4} \frac{1}{x^2} \, dx \]

Evaluate the Integral

The integral \(\int \frac{1}{x^2} \, dx\) can be evaluated by rewriting it as \(\int x^{-2} \, dx\). This integrates to \(-x^{-1}\), or \(-\frac{1}{x}\). So the definite integral becomes: \[ V = \pi \left[-\frac{1}{x}\right]_{1}^{4} \]

Compute the Definite Integral

Apply the limits of integration: \[ V = \pi \left(-\frac{1}{4} - \left(-\frac{1}{1}\right)\right) = \pi \left(-\frac{1}{4} + 1\right) \] Simplify the expression: \(\left(-\frac{1}{4} + 1 \right) = \frac{3}{4}\). Thus, \[ V = \pi \times \frac{3}{4} = \frac{3\pi}{4} \]

Conclusion

The volume of the solid created by revolving \(y = \frac{1}{x}\) from \(x = 1\) to \(x = 4\) around the \(x\)-axis is \(\frac{3\pi}{4}.\)

Key concepts

Volume of Solid

The volume of a solid, especially those generated by revolving a curve around an axis, is a fascinating concept in calculus. It involves determining the three-dimensional space occupied by such a shape. When a curve is revolved around an axis (like the x-axis), it sweeps out a solid. Calculating this volume can be crucial in various fields such as engineering and physics. In this context, it helps one understand the real-world dimensions of objects with circular symmetries.
For example, imagine taking the curve described by the equation \(y = \frac{1}{x}\) from \(x=1\) to \(x=4\) and spinning it around the x-axis. This action creates a solid object in 3D space, and our task is to find out how much space this object occupies---also known as its volume.
Knowing how to compute the volume of such solids of revolution enables us to determine key properties of objects and systems based on their geometric shapes.

Disk Method

The disk method is a powerful technique used in integral calculus to find the volume of a solid of revolution. This method involves slicing the solid perpendicular to the axis of revolution, resulting in thin circular disks. By adding up the volume of these disks, you can determine the entire volume of the solid.
The fundamental idea behind the disk method is to consider each disk having a small thickness \(\Delta x\), a radius equal to the value of the function at a particular x-value \(f(x)\), and thus an area \(\pi[f(x)]^2\). When these areas are summed up and the limit taken as the thickness approaches zero, it translates into a definite integral:

• Volume \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)
• In our problem, \(f(x) = \frac{1}{x}\) with limits \(a=1\) and \(b=4\).

By substituting \(f(x)\) in the integral, you can compute the desired volume of the solid formed by the curve.

Definite Integral

The definite integral plays a central role in calculating the volume of a solid of revolution. It allows us to sum an infinite number of infinitesimally small pieces (like the circular disks we considered) to find a total area or volume.
A definite integral has upper and lower limits, which specify the interval over which the integration is performed. For instance, in the problem concerning \(y = \frac{1}{x}\), the definite integral is evaluated between \(x=1\) and \(x=4\). It is written as:
\[ \int_{1}^{4} \frac{1}{x^2} \, dx \]
This integral essentially sums up the infinitesimal volumes of disks formed by revolving the segment of the curve about the x-axis.
Evaluating the integral results in the exact volume of the solid, thereby bridging the concrete geometric interpretation with the abstract calculus computation.

Integral Calculus

Integral calculus is the branch of mathematics concerned with accumulation of quantities and area computation. It's the reverse process of differentiation and is fundamentally about finding functions when their derivatives are known.
In the context of solids of revolution, integral calculus provides the tools necessary to accumulate infinite numbers of infinitesimally small parts—such as the slices or disks we've discussed. This makes it possible to explore and analyze complex geometric constructions.
The key components include:

• Understanding the concept of an antiderivative: In this problem, the integral \( \int x^{-2} \, dx \) leads to an antiderivative \(-\frac{1}{x}\).
• Working with definite integrals for precise calculations between limits.

By mastering integral calculus, you gain the ability to solve diverse mathematical and real-world problems, including computing the volume of solids.