/*! 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 6 Find the volume of the solid obt... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. $$y=\frac{1}{4} x^{2}, y=5-x^{2} ; \quad \text { about the } x-axis$$

Short Answer

Expert verified
The volume is approximately \( V = \frac{448\pi}{15} \).

Step by step solution

01

Understand the Problem

We need to find the volume of the solid formed by rotating the area between the curves \( y = \frac{1}{4} x^2 \) and \( y = 5 - x^2 \) around the x-axis. This involves using the washer method for rotating a region bounded by two curves.
02

Find the Points of Intersection

To determine the bounds of integration for the volume, we first find where the curves intersect by equating \( \frac{1}{4}x^2 = 5 - x^2 \). Solving this, we rearrange to get \( \frac{5}{4}x^2 = 5 \), leading to \( x^2 = 4 \). Thus, \( x = \pm 2 \). These are the points of intersection x-values.
03

Set Up the Integral for Volume

Using the washer method, the volume of the solid is calculated by the integral from \(-2\) to \(2\) of \( \pi \left[ (5-x^2)^2 - \left(\frac{1}{4}x^2\right)^2 \right] \, dx \). The outer radius is \( 5-x^2 \) and the inner radius is \( \frac{1}{4}x^2 \).
04

Simplify the Expression Inside the Integral

Expand the expression: \( (5-x^2)^2 = 25 - 10x^2 + x^4 \) and \( \left(\frac{1}{4}x^2\right)^2 = \frac{1}{16}x^4 \). Substitute these into the integral formula to get \( \pi \int_{-2}^{2} (25 - 10x^2 + x^4 - \frac{1}{16}x^4) \, dx \). Simplify further to \( \pi \int_{-2}^{2} \left( 25 - 10x^2 + \frac{15}{16}x^4 \right) \, dx \).
05

Integrate the Expression

Integrate term by term: \( \int 25 \, dx = 25x \), \( \int -10x^2 \, dx = -\frac{10}{3}x^3 \), and \( \int \frac{15}{16}x^4 \, dx = \frac{15}{80}x^5 \). Thus, the integral becomes \[ \pi \left[ 25x - \frac{10}{3}x^3 + \frac{15}{80}x^5 \right]_{-2}^{2} \].
06

Evaluate the Integral

Evaluate from \(-2\) to \(2\). Substitute \(2\) into the antiderivative: \( 25(2) - \frac{10}{3}(2^3) + \frac{15}{80}(2^5) \). Repeat for \(-2\): \( 25(-2) - \frac{10}{3}(-2)^3 + \frac{15}{80}(-2)^5 \). Calculate these values and subtract the second from the first to find the total volume.
07

Calculate the Result

Perform the arithmetic to find the value: \( V = \pi (50 - \frac{80}{3} + \frac{480}{80}) - [-50 + \frac{80}{3} - \frac{480}{80}] \). Calculate each term and simplify the result to get the final volume.

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 91Ó°ÊÓ!

Key Concepts

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

Washer Method
The washer method is a powerful technique in calculus used to find the volume of a solid of revolution. This method is particularly useful when the area being rotated around an axis is between two curves. Here's a simple way to visualize the process and understand how it works.

When you rotate a region around the x-axis, each slice perpendicular to the axis looks like a washer. A washer is a disk with a hole in the center. The outer edge of this washer corresponds to one of the curves, while the inner edge corresponds to the other curve. The formula to find the volume of these washers integrates the difference of the squares of the radii of the outer and inner curves into the calculation.

The integral for this volume, given two functions, takes the form:
  • Outer radius: the larger function (further from the axis of rotation),
  • Inner radius: the smaller function (closer to the axis of rotation).
The formula is: \[V = \pi \int_{a}^{b} \left( R_{outer}(x)^2 - R_{inner}(x)^2 \right) \, dx\]Here, \( R_{outer}(x) \) is the outer radius and \( R_{inner}(x) \) is the inner radius. This method neatly extends the disc method by accounting for the hollow section in the middle of each washer.
Integral Calculus
Integral calculus is an essential branch of mathematics that deals with the concept of integration. Integration helps us find quantities like area, volume, and displacement, among others. In the context of the washer method, integration is used to sum up many infinitesimally thin washers to find the total volume of a rotated solid.

In our exercise, setting up the integral correctly is key to finding the exact volume. The integration limits, also known as the bounds, are defined by where the curves intersect. We find these by solving the equation where the two functions are equal.

Then, we integrate the expression from the inner radius to the outer radius squared difference between the curves as shown below:
  • The integral bounds: based on points of intersection \(-2\) to \(+2\).
  • The expression inside the integral: difference of squares of the functions representing washer radii.
This step-by-step integration process uses the basic rules of calculus to solve for the collective volume of these washers over the defined interval. It's this summation via integration that provides the total volume of the solid.
Intersection of Curves
Understanding where two curves meet is not only a fundamental concept in calculus but also crucial in solving problems involving solids of revolution. This intersection tells us where to start and stop our integration, as it represents the limits of the region being rotated.

In our example, the curves intersect where their equations are equal: \( \frac{1}{4}x^2 = 5 - x^2 \). Solving this equation yields critical points, which are values of \( x \) where the two curves cross. These points, \( x = -2 \) and \( x = 2 \), are where the region of interest begins and ends.

Finding these intersection points involves solving a quadratic equation derived from setting the curve equations equal. It's crucial as these points dictate the bounds of our integral. Without them, determining the exact volume of the solid obtained by rotating the bounded region about the axis wouldn't be possible.

In summary, understanding and calculating the intersection of curves allow us to correctly apply the washer method and integral calculus to find the volume of a rotating solid. It's an essential step that ensures accuracy in defining the limits of the integral.

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

A sphere with radius 1 \(\mathrm{m}\) has temperature \(15^{\circ} \mathrm{C}\) . It lies inside a concentric sphere with radius 2 \(\mathrm{m}\) and temperature \(25^{\circ} \mathrm{C}\) . The temperature \(T(r)\) at a distance \(r\) from the common center of the spheres satisfies the differential equation $$\frac{d^{2} T}{d r^{2}}+\frac{2}{r} \frac{d T}{d r}=0$$ If we let \(S=d T / d r,\) then \(S\) satisfies a first-order differential equation. Solve it to find an expression for the temperature \(T(r)\) between the spheres.

Find an equation of the curve that passes through the point \((0,1)\) and whose slope at \((x, y)\) is \(x y .\)

Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function \(P(t)\) , the performance of someone learning a skill as a function of the training time \(t .\) The derivative \(d P / d t\) represents the rate at which performance improves. (a) If \(M\) is the maximum level of performance of which the learner is capable, explain why the differential equation $$\frac{d P}{d t}=k(M-P) \quad k$$ \(k\) a positive constant is a reasonable model for learning. (b) Solve the differential equation in part (a) to find an expression for \(P(t) .\) What is the limit of this expression?

Suppose that 2 \(\mathrm{J}\) of work is needed to stretch a spring from its natural length of 30 \(\mathrm{cm}\) to a length of 42 \(\mathrm{cm} .\) (a) How much work is needed to stretch the spring from 35 \(\mathrm{cm}\) to 40 \(\mathrm{cm}?\) (b) How far beyond its natural length will a force of 30 \(\mathrm{N}\) keep the spring stretched?

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is $$F=\frac{m g R^{2}}{(x+R)^{2}}$$ where \(x=x(t)\) is the object's distance above the surface at time \(t, R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and \(s o\) $$m \frac{d v}{d t}=-\frac{m g R^{2}}{(x+R)^{2}}$$ (a) Suppose a rocket is fired vertically upward with an initial velocity \(v_{0 .}\) Let \(h\) be the maximum height above the surface reached by the object. Show that $$v_{0}=\sqrt{\frac{2 g R h}{R+h}}$$ $$ [\text { Hint: By the Chain Rule, } m(d v / d t)=m v(d v / d x) .]$$ (b) Calculate \(v_{e}=\lim _{h \rightarrow \infty} v_{0 .}\) This limit is called the escape velocity for the earth. (c) Use \(R=3960 \mathrm{mi}\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) to calculate \(v_{e}\) in feet per second and in miles per second.

See all solutions

Recommended explanations on Math Textbooks

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.