Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Calculate the volume of the solid obtained by rotating region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
\(y = x,y = \sqrt x ;\) about\(x = 2.\)

The volume of the solid is\(\frac{{8\pi }}{{15}}.\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Volume of solid lies between curves

Definition of volume:

Consider a solid that lies between the curves \(x = a\), and \(x = b\). If the cross section of \(S\) in the plane \({P_x}\), through \(x\) and perpendicular to the \(x\)-axis, is given by an integrable function \(A(x)\), then the volume of \(S\) is

\(\begin{array}{l}V = \mathop {\lim }\limits_{\max \Delta {x_i} \to 0} \sum\limits_{i = 1}^n A \left( {x_i^*} \right)\Delta {x_i}\\V = \int_a^b A (x)dx.\end{array}\)

Given information from question

It is given that the curves that bound a region are \(y = x,y = \sqrt x \).

The specific line around which the region is rotated is \(x = 2\) which is parallel to \(y\) axis.

The curves can be written as \(x = y,x = {y^2}\)

Calculate the values of \(y\)

The value of \(y\)is

\(\begin{array}{l}y = {y^2}\\{y^2} - y = 0\\y(y - 1) = 0\end{array}\)

Then, \(y = 0\) or \(y = 1\).

The curves become \(x = 2 - y,x = 2 - {y^2}\).

If the curves \(x = 2 - y\) and \(x = 2 - {y^2}\) are rotated about the \(x = 2\), a solid is obtained.

If is sliced at a point \(y\), a washer is obtained whose inner radius is given by \(x = 2 - {y^2}\) and the outer radius is given by \(x = 2 - y\).

Area of the washer

Area of the washer is

\(\begin{array}{l}A(y) = \pi {\left( {{x_2}} \right)^2} - \pi {\left( {{x_1}} \right)^2}\\ = \pi {(2 - y)^2} - \pi {\left( {2 - {y^2}} \right)^2}\\ = \pi \left( {4 - 4y + {y^2} - 4 + 4{y^2} - {y^4}} \right)\\ = \pi \left( {5{y^2} - 4y - {y^4}} \right)\end{array}\)

The region lies between \(y = 0\), and \(y = 1\).

Apply the definition of volume to find the volume

Volume of the region

\(\begin{array}{l}V = \int_1^0 \pi \left( {5{y^2} - 4y - {y^4}} \right)dx\\ = \pi \left( {\frac{{5{y^3}}}{3} - 2{y^2} - \frac{{{y^5}}}{5}} \right)_1^0\\ = - \pi \left( {\frac{{5{{(1)}^3}}}{3} - 2{{(1)}^2} - \frac{{{{(1)}^5}}}{5}} \right)\end{array}\)

\(\begin{array}{l}V = - \pi \left( {\frac{5}{3} - 2 - \frac{1}{5}} \right)\\ = - \pi \left( {\frac{{25 - 30 - 3}}{{15}}} \right)\\ = - \pi \left( {\frac{{ - 8}}{{15}}} \right)\\ = \frac{8}{{15}}\pi \end{array}\)

The graph of the solid which is rotated about the axis \(x = 2\)in order to obtain the solid.

Graph of the solid is

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Calculate the volume of the solid obtained by rotating region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
\(y = x,y = \sqrt x ;\) about\(x = 2.\)

Step by Step Solution

Volume of solid lies between curves

Given information from question

Calculate the values of \(y\)

Area of the washer

Apply the definition of volume to find the volume

The graph of the solid which is rotated about the axis \(x = 2\)in order to obtain the solid.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Calculate the volume of the solid obtained by rotating region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.\(y = x,y = \sqrt x ;\) about\(x = 2.\)

Step by Step Solution

Volume of solid lies between curves

Given information from question

Calculate the values of \(y\)

Area of the washer

Apply the definition of volume to find the volume

The graph of the solid which is rotated about the axis \(x = 2\)in order to obtain the solid.

Most popular questions for Math Textbooks

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Calculate the volume of the solid obtained by rotating region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
\(y = x,y = \sqrt x ;\) about\(x = 2.\)