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Essential Calculus: Early Transcendentals
Found in: Page 379
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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\(\frac{{8\pi }}{{15}}.\)

See the step by step solution

Step by Step Solution

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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