/*! 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 4 In Exercises 3-6, find the volum... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 3-6, find the volume of the solid analytically. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) The cross sections perpendicular to the \(x\) -axis are circular disks whose diameters run from the parabola \(y=x^{2}\) to the parabola $y=2-x^{2} .

Short Answer

Expert verified
The volume of the solid can be found by integrating the area function \(A(x) = \pi (1 - x^2)^2\) with respect to \(x\) from \(-1\) to \(1\).

Step by step solution

01

Find the radius of the circular cross sections

The radius \(r\) of each cross section is half of the distance between the parabolas at the given \(x\). This can be calculated by subtracting \(y = x^2\) from \(y = 2 - x^2\) and dividing by 2, hence, \(r(x) = (2 - x^2 - x^2) / 2 = 1 - x^2\).
02

Write down the formula for the circular cross section area

The area \(A\) of a circle with radius \(r\) is \(\pi r^2\). Therefore, the area of our cross sections is \(A(x) = \pi (1 - x^2)^2\).
03

Integrate to find the volume

The volume \(V\) of the solid will be the integral of \(A(x)\) from \(x = -1\) to \(x = 1\). By the fundamental theorem of calculus, \(V = \int_{-1}^{1} \pi (1 - x^2)^2 dx\).

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.

Integrating Cross Sectional Areas
When calculating the volume of a solid, one of the key techniques used in calculus is integrating cross sectional areas. Imagine slicing the solid perpendicular to a given axis at many points, which results in numerous cross sections, each with its own area. These cross-sectional areas can vary along the axis.

To find the volume, one essentially adds up the areas of these cross sections over the length of the solid, and this is where integration plays a vital role. It allows us to sum up an infinite number of infinitesimally thin areas to find the total volume. In the exemplary exercise, each cross-sectional area is a circle, and integrating these areas between the specified boundaries gives the volume of the solid. Understanding how to find these areas and how to set up the corresponding integral is pivotal in solving problems involving volumes of solids of revolution.
Applications of Integration
Integration finds innumerable applications in the world of calculus, beyond just finding areas under curves. One of the most visually fascinating applications is in determining the volume of solids. This principle is not limited to simple geometric shapes but extends to solid figures that take on more complex forms.

The use of integration is especially helpful when dealing with irregular shapes, where traditional volume formulas are not applicable. By integrating cross-sectional areas, we can compute volumes for any solid whose cross sections are shapes that we can describe mathematically, such as squares, rectangles, triangles, or, as in our exercise, circles. Moreover, the underlying concept of integration extends to a multitude of real-world applications, such as in physics for computing work done by a force, in engineering for material strength analysis, and in economics for modeling cumulative growth over time.
Solids of Revolution
Solids of revolution are three-dimensional shapes created by revolving a two-dimensional curve about an axis. This method is significant in calculus when you consider any area bounded by a function revolved around a line (the x-axis or y-axis), creating intricate 3D shapes. These shapes can be visualized as pottery shapes formed on a potter’s wheel, rotating a profile curve around its axis.

In our exercise, the solids of revolution are produced by revolving the space between two parabolas around the x-axis. The curves of the parabolas determine the changing radius of the circular cross sections, and thus, the resulting solid's geometry. Understanding the way this rotation produces a solid and the relationship between the generating curve and the final volume is crucial in visualizing and solving these types of problems.
Volume by Disks Method
The volume by disks method is a particular integration technique used to calculate the volume of solids of revolution. This technique involves visualizing the solid as a stack of circular disks (or rings, in the case of hollow solids) with varying radii, where each disk's volume is a thin cylinder.

To apply this method, one must first derive the radius of a general disk as a function of position along the axis of revolution. In our example, this means expressing the radius of each disk as a function of x. Once the radius function, denoted as r(x), is determined, the area of each disk can be formulated as \( \pi r(x)^2 \). Finally, integrating this area over the entire range of x gives the total volume. By breaking down a complex solid into familiar shapes and summing them up methodically, the disks method transforms the problem into a manageable calculus exercise.

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 35-38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis. $$y=x, \quad y=-x / 2, \quad x=2$$

You may use a graphing calculator to solve the following problems. True or False A force is applied to compress a spring several inches. Assume the spring obeys Hooke's Law. Twice as much work is required to compress the spring the second inch than is required to compress the spring the first inch. Justify your answer.

$$ \begin{array}{l}{\text { Tunnel Construction Your engineering firm is bidding for }} \\ {\text { the contract to construct the tunnel shown on the next page. The }} \\ {\text { tunnel is } 300 \mathrm{ft} \text { long and } 50 \mathrm{ft} \text { wide at the base. The cross }} \\ {\text { section is shaped like one arch of the curve } y=25 \cos (\pi x / 50)}\end{array} $$ $$ \begin{array}{l}{\text { Upon completion, the tunnel's inside surface (excluding the }} \\ {\text { roadway) will be treated with a waterproof sealer that costs }} \\ {\$ 1.75 \text { per square foot to apply. How much will it cost to apply }} \\ {\text { the sealer? } }\end{array} $$

$$ \begin{array}{l}{\text { The Length of an Astroid The graph of the equation }} \\\ {x^{2 / 3}+y^{2 / 3}=1 \text { is one of the family of curves called astroids }} \\ {\text { (not "asteroids") because of their starlike appearance (see figure). }}\end{array} $$ $$ \begin{array}{l}{\text { Find the length of this particular astroid by finding the length of }} \\ {\text { half the first quadrant portion, } y=\left(1-x^{2 / 3}\right)^{3 / 2}, \sqrt{2} / 4 \leq x \leq 1} \\ {\text { and multiplying by } 8.}\end{array} $$

In Exercises 55-62, find the area of the surface generated by revolving the curve about the indicated axis. $$x=\sqrt{y}, \quad 0 \leq y \leq 2 ; \quad y$$

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.