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Expert-verified Found in: Page 379 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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

## 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  ### Want to see more solutions like these? 