Question

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\left(x^{2}+1\right)^{-1 / 2}\) and the \(x\) -axis on the interval \([2, \infty)\) is revolved about the \(x\) -axis.

Short answer

Answer: The volume of the solid is \(\pi^2 \Big[ \frac{1}{2} - \frac{1}{\pi} arctan(2) \Big]\).

Step-by-step solution

Set up the integral for the volume

Using disk method, we want to find the volume of the solid formed by revolving the area between the curve \(f(x) = (x^{2}+1)^{-\frac{1}{2}}\) and the \(x\)-axis around the \(x\)-axis. The volume is given by the integral, $$V = \pi \int_{2}^{\infty} [f(x)]^2 dx = \pi\int_{2}^{\infty} [(\frac{1}{\sqrt{x^2+1}})^2] dx$$

Simplify the integrand

Simplify the integrand inside the integral to make it easier to integrate: $$ V = \pi\int_{2}^{\infty} [\frac{1}{x^2+1}] dx $$

Integrate the function and find the volume

Now we need to find the indefinite integral of the function inside and then evaluate at the limits of integration, \(2\) and \(\infty\). This function's antiderivative is \(arctan(x)\): $$ V = \pi\Big[ arctan(x)|_{2}^{\infty} \Big] $$ Evaluate the antiderivative at the limits: $$ V = \pi\Big[ \lim_{x\to\infty} arctan(x) - arctan(2) \Big] $$ Since \(\lim_{x\to\infty} arctan(x) = \frac{\pi}{2}\), we have: $$ V = \pi\Big[ \frac{\pi}{2} - arctan(2) \Big] $$

Simplify the volume expression

Simplify the expression for the volume: $$ V = \pi^2 \Big[ \frac{1}{2} - \frac{1}{\pi} arctan(2) \Big] $$ Finally, the volume of the described solid of revolution is given by: $$ V = \pi^2 \Big[ \frac{1}{2} - \frac{1}{\pi} arctan(2) \Big] $$

Key concepts

Volume of revolution

When we talk about finding the "volume of revolution," we're dealing with the process of creating a 3D solid by rotating a 2D shape around an axis. Imagine a flat region or curve on a graph being spun around a line. The result is a 3D object, and calculus helps us find its volume. This method is important in real-world applications like designing barrels, vases, or any object with circular symmetry.

To calculate such volumes, we typically set up an integral. The integral sums up infinitely small volume slices of the disk or washer formed by rotation. Each slice has a tiny but measurable volume. The sum of these disks gives the total volume of the solid. In this context, the "volume of revolution" provides a framework for these calculations.

Disk method

The disk method is a straightforward technique in calculus used to find the volume of a solid of revolution. It involves slicing the solid into thin disks, perpendicular to the axis of rotation. Picture a stack of coins; each coin is a disk slice. The thickness of each slice is an infinitesimally small distance, represented in calculus as \(dx\), and the radius is dictated by the function we are revolving.

To calculate the volume of each disk, use the formula for the volume of a cylinder, \(V = \pi r^2h\). Here, \(r\) is the radius of the disk, given by the function value at that point, and \(h\), the height, is the small width \(dx\). The total volume is the integral of these disks over the range of rotation.

Improper integral

Improper integrals are a fascinating part of calculus that help us handle infinite boundaries or unbounded functions. In our case, since the region of revolution extends from \(x = 2\) to \( ightarrow \infty\), we deal with an improper integral. These integrals require special care. Often, they involve limits to ensure the calculations are mathematically sound.

A common scenario involves setting up the definite integral under the assumption of finite bounds, then taking the limit as one of these bounds moves towards infinity. For example, we might initially consider the integral from 2 to some arbitrary upper bound \(b\), then compute the limit of the integral as \(b ightarrow \infty\). This way, we can evaluate the volume even when extending infinitely in one direction.

Arctan function

The Arctan function, or inverse tangent, is essential in the integration process for many problems. Found as the antiderivative of \(1/(x^2 + 1)\), it helps tackle specific integral types effectively. The arctangent curve itself represents the angle whose tangent is a given number. A characteristic of the Arctan function is that it has horizontal asymptotes at \(-\pi/2\) and \(+\pi/2\). This behavior makes it perfect for evaluating limits in improper integrals.

When resolving integrals leading to arctan, remember that \( ext{arctan}(x)\) gradually approaches \( \pi/2 \) as \(x\) goes to infinity. Similarly, this approach can be used for calculating volumes or areas related to curves asymptotic to certain lines.