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Chapter 6: Problem 28

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves The region between the curve \(y=\sqrt{\cot x}\) and the \(x\) -axis from \(x=\pi / 6\) to \(x=\pi / 2\)

Short Answer

Expert verified
The volume is \( \pi \ln 2 \).

Step by step solution

01

Understand the Problem

We need to find the volume of the solid formed by revolving the region bounded by the curve \(y = \sqrt{\cot x}\) and the \(x\)-axis, from \(x = \pi/6\) to \(x = \pi/2\), around the \(x\)-axis.
02

Set Up the Integral for Volume

The volume \(V\) of the solid of revolution can be calculated using the disk method. The formula is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \), where \( f(x) = \sqrt{\cot x} \), \(a = \pi/6\), and \(b = \pi/2\).
03

Substitute the Function into the Formula

Substitute the function into the volume formula: \[ V = \pi \int_{\pi/6}^{\pi/2} \left(\sqrt{\cot x}\right)^2 \, dx = \pi \int_{\pi/6}^{\pi/2} \cot x \, dx. \]
04

Integrate the Function

Integrate \( \cot x \): \[ \int \cot x \, dx = \ln|\sin x| + C. \] Apply the bounds \( \pi/6 \) and \( \pi/2 \) to this expression.
05

Evaluate the Definite Integral

Compute the definite integral:\[ V = \pi \left[ \ln|\sin(x)| \right]_{\pi/6}^{\pi/2} = \pi \left( \ln|\sin(\pi/2)| - \ln|\sin(\pi/6)| \right). \]
06

Simplify the Expression

Since \( \sin(\pi/2) = 1 \) and \( \sin(\pi/6) = \frac{1}{2} \), the expression simplifies to:\[ V = \pi ( \ln 1 - \ln \frac{1}{2} ) = \pi (0 + \ln 2). \]
07

Write the Final Answer

The volume of the solid is \( V = \pi \ln 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Disk Method
The Disk Method is a powerful technique used in calculus to find the volume of a solid of revolution. When you revolve a region around an axis, specifically the x-axis in this case, you can envision slicing the solid into thin disks perpendicular to the axis. Each disk has a small thickness, which we'll denote as \( \Delta x \), and a certain radius that depends on the curve.
  • The radius of each disk is determined by the function \( f(x) \).
  • The volume of one disk can be calculated as \( \pi [f(x)]^2 \Delta x \).
  • To find the total volume, you integrate these disk volumes over the interval from \( a \) to \( b \).
This results in the integral formula for the Disk Method: \[V = \pi \int_{a}^{b} [f(x)]^2 \, dx.\]The Disk Method is especially useful when the solid has circular cross-sections parallel to the axis of rotation.
Definite Integral
The Definite Integral plays a crucial role in calculating the volume of solids of revolution. It allows us to sum up infinitely many infinitesimally small quantities, giving an exact measure of a total volume.
  • The Definite Integral is expressed in the form \( \int_{a}^{b} f(x) \, dx \), with limits of integration \( a \) and \( b \).
  • The integral calculates the net area under the curve \( f(x) \) between the two limits.
  • The "definite" aspect means it results in a numeric value, representing a physical quantity like volume or area.
In the context of our problem, we used the Definite Integral to find the volume of a solid that results from revolving a specific region bounded by the curve \( y = \sqrt{\cot x} \) around the axis.
Trigonometric Functions
Trigonometric Functions are fundamental in calculus, especially when dealing with problems involving periodic or rotational scenarios. In this exercise, the function \( y = \sqrt{\cot x} \) plays a key role.
  • The function \( \cot x \) is the reciprocal of the tangent function, \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \).
  • When we revolve the region under \( y = \sqrt{\cot x} \), the computation involves trigonometric identities and integral properties.
  • Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) often require derivatives and integrations familiar in advanced calculus topics.
In solving these integrals, understanding the properties of these trigonometric functions ensures precise calculation and simplification of results.
Calculus Problem Solving
Problem-solving in calculus often involves a blend of techniques and concepts. Each exercise can be seen as a puzzle, where you apply known methods and adapt them to a unique scenario.
  • Understanding the problem is the first step, such as recognizing what kind of solid is being generated.
  • Identify the right method or formula, such as using the Disk Method for solids of revolution.
  • Carefully follow each step of integration, ensuring proper substitution and simplification.
  • Verification is key; reviewing and double-checking the computations helps verify the accuracy of the solution.
Engaging with exercises like these helps improve mathematical intuition, provides a deeper understanding of calculus concepts, and builds confidence in solving complex problems.

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Most popular questions from this chapter

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