Problem 74 - Thomas Calculus in SI Units

Question

The region between the curve \(y=\sqrt{\cot x}\) and the \(x\) -axis from \(x=\pi / 6\) to \(x=\pi / 2\) is revolved about the \(x\) -axis to generate a solid. Find the volume of the solid.

Step-by-Step Solution

Answer

The volume of the solid is \(\pi \log 2\).

Step 1: Understand the Problem

We need to find the volume of a solid generated by revolving the curve \(y = \sqrt{\cot x}\) around the \(x\)-axis from \(x=\pi / 6\) to \(x=\pi / 2\). This involves using the disk method for calculating volumes of revolution.

Step 2: Set up the Integral

The formula for the volume of a solid of revolution using the disk method is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx\). For this problem, \( f(x) = \sqrt{\cot x} \), \( a = \pi / 6 \), and \( b = \pi / 2 \). Thus the volume is:\[V = \pi \int_{\pi / 6}^{\pi / 2} (\cot x) \, dx\]

Step 3: Integrate the Function

To evaluate the integral \( \int_{\pi / 6}^{\pi / 2} \cot x \, dx \), remember that the derivative of \(\log|\sin x|\) is \(\cot x\). So,\[\int \cot x \, dx = \log|\sin x|\].Therefore, evaluate the definite integral:\[\int_{\pi / 6}^{\pi / 2} \cot x \, dx = [\log|\sin x|]_{\pi / 6}^{\pi / 2}\].

Step 4: Evaluate the Definite Integral

Substitute the limits into the integrated function:\[= \log|\sin(\pi/2)| - \log|\sin(\pi/6)|\]\[= \log 1 - \log (1/2) = 0 - (-\log 2) = \log 2\].

Step 5: Calculate the Volume

Plug the value of the evaluated integral back into the volume equation:\[V = \pi \times \log 2\].Therefore, the volume of the solid is \(\pi \log 2\).

Key Concepts

Volumes of Revolution

To understand volumes of revolution, picture a region in the plane that is rotated around a line to create a three-dimensional object. This is like spinning a flat region around an axis. The solid shape formed can be visualized similarly to a potter spinning clay into a bowl. This method often involves rotating a curve around an axis, like the x-axis, to form a symmetrical three-dimensional shape.

In this exercise, the region under the curve of the function is revolved around the x-axis. The disk method is typically used here. This involves slicing the solid into thin disks, calculating each disk's volume, and then combining them to find the overall volume.

• A disk is simply a cylinder with a very small height.
• The volume of each disk is based on its radius, which equals the function's value at a particular point along the x-axis.
• The sum of these disk volumes, represented by an integral, helps us determine the whole volume.

Thus, by understanding volumes of revolution, you can find the volume of complex shapes formed by rotating a function around an axis.

Integral Calculus

Integral calculus is a fundamental concept that helps us with various computations involving areas and volumes. It allows us to find the antiderivative or the original function from which a derivative is derived. In simple terms, integration is the process of finding the whole from the parts.

In this context, integral calculus plays a critical role in determining the volume of a solid of revolution. We use the integral to sum up an infinite number of infinitesimally thin disks that make up the solid. Each disk's radius is the value of the function at any given point, and the thickness is an infinitely small change in x.

• The definite integral represents the summing up of these infinite slices along a specific interval.
• Integral calculus also provides us the tools to evaluate areas under curves, allowing for precise volume calculations.

By mastering the basics of integral calculus, you will be able to solve complex volume related problems effectively.

Definite Integral Calculation

Definite integrals are used in this problem to find the precise volume of the solid created when a region is rotated. These integrals have limits which define the interval over which you integrate, giving us a specific value rather than a general function.

In this exercise, the limits of integration are from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{2}\). The goal is to calculate the area under the curve \(y = \sqrt{\cot x}\) within these bounds, but revolved about the x-axis.

• The integral of the function gives us a new function, which when evaluated at these bounds provides an actual number.
• This number, when multiplied by \(\pi\), represents the volume of the solid.

This method of using definite integrals lets you find exact measurements for volumes of revolution, illustrating the power and precision of calculus.