/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find the volumes of the solids g... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 21

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises \(17-22\) about the \(x\) -axis. $$ y=\sqrt{\cos x}, \quad 0 \leq x \leq \pi / 2, \quad y=0, \quad x=0 $$

Short Answer

Expert verified
The volume of the solid is \( \pi \).

Step by step solution

01

Identify the Boundary Function and Region

The given boundary function is \( y = \sqrt{\cos x} \) and the limits are \( 0 \leq x \leq \pi/2 \). We are revolving this area about the \( x\)-axis to find the volume of the solid.
02

Use the Disk Method Formula

To find the volume of the solid of revolution, we use the disk method, which is given by:\[V = \int_{a}^{b} \pi [f(x)]^2 \,dx\]where \( f(x) = \sqrt{\cos x} \), \( a = 0 \), and \( b = \pi/2 \).
03

Set Up the Integral

Substitute \( f(x) = \sqrt{\cos x} \) into the formula:\[V = \int_{0}^{\pi/2} \pi \left(\sqrt{\cos x}\right)^2 \, dx = \int_{0}^{\pi/2} \pi \cos x \, dx\]
04

Evaluate the Integral

Compute the integral:\[V = \pi \int_{0}^{\pi/2} \cos x \, dx\]The antiderivative of \( \cos x \) is \( \sin x \). Therefore, evaluate the bounds:\[V = \pi [ \sin x ]_{0}^{\pi/2} = \pi (\sin(\pi/2) - \sin(0)) = \pi (1 - 0) = \pi\]
05

Summarize the Result

The volume of the solid generated by revolving the region bounded by \( y = \sqrt{\cos x} \), \( y = 0 \), and \( x = 0 \) around the \( x\)-axis from \( x = 0 \) to \( x = \pi/2 \) is \( \pi \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Disk Method
When it comes to finding the volume of solids formed by revolving a region around an axis, the disk method is an efficient technique. It's particularly useful when the solid is generated by rotating a region bounded by a function around the x-axis. Here鈥檚 how it works:
  • Visualize slicing the solid into thin, circular disks perpendicular to the x-axis.
  • The thickness of each disk corresponds to a small change in the x-value, denoted as \( dx \).
  • The radius of each disk is determined by the value of the function at that particular slice, \( f(x) \).
The disk method formula is \( V = \int_{a}^{b} \pi [f(x)]^2 \,dx \), where \( [f(x)]^2 \) gives the area of the circular face of each disk. By integrating from \( a \) to \( b \), you sum up all the infinitesimally thin disks to find the total volume.
Integral Calculus
Integral calculus is a fundamental part of calculus concerned with the concept of integration. In the context of volume of solids of revolution:
  • An integral can be understood as the limit of a sum, which adds up an infinite number of infinitesimally small quantities.
  • Integration helps to calculate not only areas under curves but also volumes of solids formed by those curves.
  • In our scenario, integrating \( \pi \left(\sqrt{\cos x}\right)^2 \) over a specific interval allows us to determine the volume of the entire revolved region.
Integral calculus makes it possible to generalize solutions for a variety of problems involving curves and their properties over an interval. It effectively transforms geometric problems into calculations manageable by algebraic means.
Definite Integral Evaluation
The process of evaluating a definite integral involves solving an integral over a specified interval. In the exercise, we need to find the definite integral from 0 to \( \pi/2 \) of the function \( \pi \cos x \). This is the final step in applying the disk method.Here's a step-by-step breakdown:
  • Identify the function to be integrated: \( \pi \cos x \).
  • Find the antiderivative. The antiderivative of \( \cos x \) is \( \sin x \), so the antiderivative of \( \pi \cos x \) is \( \pi \sin x \).
  • Evaluate this antiderivative at the two ends of the interval (0 and \( \pi/2 \)), and calculate the difference: \( \pi [ \sin(\pi/2) - \sin(0) ] \).
This results in a volume of \( \pi \), which exemplifies the elegance and utility of using definite integrals to solve real-world geometric problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(39-42\) , find the volume of the solid generated by revolving each region about the \(y\) -axis. The region in the first quadrant bounded above by the parabola \(y=x^{2},\) below by the \(x\) -axis, and on the right by the line \(x=2\)

Bathroom scale \(A\) bathroom scale is compressed 1\(/ 16\) in. when a 150 -lb person stands on it. Assuming that the scale be- haves like a spring that obeys Hooke's Law, how much does someone who compresses the scale 1\(/ 8\) in. weigh? How much work is done compressing the scale 1\(/ 8\) in.?

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-22\) about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. \(x=y^{3} / 3, \quad 0 \leq y \leq 1 ; \quad y\) -axis

Compute the volume of the solid generated by revolving the triangular region bounded by the lines \(2 y=x+4, y=x,\) and \(x=0\) about a. the \(x\) -axis using the washer method. b. the \(y\) -axis using the shell method. c. the line \(x=4\) using the shell method. d. the line \(y=8\) using the washer method.

Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild restrictions preventing doubling back mentioned earlier), calculate the length of the semicircle \(y=\sqrt{1-x^{2}}\) with these two different parametrizations. $$ \begin{array}{l}{\text { a. } x=\cos 2 t, \quad y=\sin 2 t, \quad 0 \leq t \leq \pi / 2} \\ {\text { b. } x=\sin \pi t, \quad y=\cos \pi t, \quad-1 / 2 \leq t \leq 1 / 2}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.