Problem 74 - Thomas Calculus

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 \(\pi \ln 2\).

Step 1: Identify the Problem

We need to find the volume of the solid formed by revolving the region between the curve \(y=\sqrt{\cot x}\) and the \(x\)-axis from \(x=\frac{\pi}{6}\) to \(x=\frac{\pi}{2}\) around the \(x\)-axis.

Step 2: Setup the Integral

To find the volume of the solid of revolution about the \(x\)-axis, we use the disk method. This involves integrating the area of circular disks. The formula is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). Here, \(f(x) = \sqrt{\cot x}\), \(a = \frac{\pi}{6}\), and \(b = \frac{\pi}{2}\). Thus, the integral becomes: \[ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sqrt{\cot x})^2 \, dx \].

Step 3: Simplify the Integral

Simplify the integrand \((\sqrt{\cot x})^2 = \cot x\), making the integral: \[ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cot x \, dx \].

Step 4: Integrate \(\cot x\)

The integral of \(\cot x\) is \(\ln|\sin x|\). Thus, compute the integral: \[ \int \cot x \, dx = \ln|\sin x| \]. Therefore, \[ V = \pi [\ln|\sin x|]_{\frac{\pi}{6}}^{\frac{\pi}{2}} \].

Step 5: Evaluate at Upper and Lower Limits

Substitute the limits to find the volume: \[ V = \pi (\ln|\sin(\frac{\pi}{2})| - \ln|\sin(\frac{\pi}{6})|) \]. Calculate the values of \(\sin(\frac{\pi}{2}) = 1\) and \(\sin(\frac{\pi}{6}) = \frac{1}{2}\), which gives: \[ V = \pi (\ln 1 - \ln \frac{1}{2}) \].

Step 6: Simplify and Compute Final Answer

Since \(\ln 1 = 0\), the expression simplifies to \(-\ln \frac{1}{2}\). Using the property \(\ln a - \ln b = \ln \frac{a}{b}\), this becomes \(\ln 2\). Thus, the volume is: \[ V = \pi \ln 2 \].

Key Concepts

Disk Method

The Disk Method is an essential concept in calculus that helps us find the volume of a solid of revolution. When dealing with a region in the plane that is revolved around an axis, we can use this method to sum up a series of infinitesimally thin disks (or circles) along the axis of rotation.
Each disk has a very small thickness \(dx\) and a radius \(f(x)\), which is basically the distance from the curve to the axis of rotation.
Key steps involved when using the Disk Method include:

• Identifying the function and the bounds of rotation. For instance, if the region is bound by \(y=f(x)\) from \(x=a\) to \(x=b\), these become the limits of integration.
• Setting up the integral based on the formula \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\). This formula represents the volume as a sum of the areas of each disk, where \(\pi [f(x)]^2\) gives the area of a circular disk.
• Integrating to find the volume and evaluating at the boundaries.

This method can apply to a variety of regions and functions, making it a powerful tool in solving volume-related problems in calculus.

Integral Calculus

Integral Calculus provides us with the tools to accumulate quantities and find the total size or value such as area, volume, displacement, etc.
In our context, it allows us to sum up infinitesimally small elements to find the volume of a solid.
When working with functions and curves:

• The integral sign \(\int\) represents the "summation" of an infinite number of small quantities.
• The limits of integration \(a\) and \(b\) tell us over which interval we are summing. They define where we start and stop accumulating the small elements.
• The function inside the integral, like \[f(x)=\sqrt{\cot x}\] in this case, describes the quantity we want to accumulate or the process by which it modifies other elements (like height, radius, etc).

The use of integral calculus as a summation tool is fundamental in solving complex geometry problems like calculating the volume of solids of revolution.

Trigonometric Functions

The study of Trigonometric Functions often involves dealing with angles and their relationships through functions like sine, cosine, tangent, and their reciprocals.
In calculus, these functions are essential since many curves can be expressed in terms of trigonometric functions.
Specific insights about our problem include:

• The function \(y=\sqrt{\cot x}\) involves the cotangent, which is a reciprocal of the tangent function. \(\cot x = \frac{1}{\tan x}\).
• Understanding these fundamental trigonometric identities and transformations can simplify the integration process. For instance, simplifying \( (\sqrt{\cot x})^2 \) to \(\cot x\) was crucial.
• Using identities and transformations, such as the integral of \(\cot x\) being \(\ln|\sin x|\), assists in seamlessly solving integral calculus problems involving trigonometric functions.

By familiarizing ourselves with these functions, we can unlock new pathways to solving intricate calculus problems.