/*! 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 52 Find the volume of the described... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 52

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\left(x^{2}-1\right)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(y\) -axis.

Short Answer

Expert verified
Answer: The volume of the solid of revolution is \(V = \frac{2\pi}{3}\).

Step by step solution

01

Set up the integral for the sum of the disk volumes

Since we're revolving around the y-axis, we will use the disk method formula for volumes: $$V = \pi \int_{a}^{b} [r(x)]^2 dx$$ where V is the volume, a and b are the limits of integration, and r(x) is the radius of each disk. In our case, the radius of each disk is given by the x-coordinate, so \(r(x) = x\). The function \(f(x)\) represents the upper boundary of the region, so we have the interval (1,2] for the x-coordinates. The integral we need to compute is $$V = \pi \int_{1}^{2} [x]^2 [(x^2-1)^{-\frac{1}{4}}] dx$$.
02

Simplify the integral

Simplify the integrand by multiplying the constants and the functions: $$V = \pi \int_{1}^{2} x^2 (x^2-1)^{-\frac{1}{4}} dx$$
03

Perform integration

To integrate this function, we will use substitution. Let \(u = x^2 - 1\). Then, we have \(du = 2x dx\). So, we will need an extra \(\frac{1}{2}\) factor when we substitute. $$V = \frac{\pi}{2} \int_{0}^{1} u^{-\frac{1}{4}} du$$ Now, we can easily integrate: $$V = \frac{\pi}{2} \left[\frac{4}{3} u^{\frac{3}{4}} \right]_{0}^{1} $$
04

Evaluate the integral

Plugging in the limits, we find the volume of the solid of revolution: $$ V = \frac{\pi}{2} \left[\frac{4}{3}(1)^{\frac{3}{4}} - \frac{4}{3}(0)^{\frac{3}{4}} \right] = \frac{2\pi}{3}$$ So, the volume of the solid of revolution formed when the region bounded by \(f(x) = (x^2 - 1)^{-\frac{1}{4}}\) and the x-axis on the interval (1,2] is revolved around the y-axis is \(V = \frac{2\pi}{3}\).

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!

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

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x}{(x-1)\left(x^{2}+2 x+2\right)^{2}} d x$$

Evaluate \(\int \frac{d y}{y(\sqrt{a}-\sqrt{y})},\) for \(a > 0\). (Hint: Use the substitution \(u=\sqrt{y}\) followed by partial fractions.)

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x+1)^{-3 / 2}\) and the \(y\) -axis on the interval (-1,1] is revolved about the line \(x=-1.\)

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x^{3}+1}{x\left(x^{2}+x+1\right)^{2}} d x$$

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{x \sqrt{1+2 x}} ; 1+2 x=u^{2}$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

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.