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Q16E

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Found in: Page 379

### Essential Calculus: Early Transcendentals

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

# The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.$$x = {y^2},x = 1;$$ about $$x = 1.$$

The Volume of the solid is$$\frac{{16\pi }}{{15}}.$$

See the step by step solution

## Volume of solid lies between curves

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{aligned}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{aligned}

## Given information from question

The curves that bound a region are $$x = {y^2},x = 1$$.

The specific line around which the region is rotated is $$x = 1$$ which is parallel to $$y$$ axis.

If the curve $$x = {y^2}$$ is rotated about the $$x = 1$$, a solid is obtained.

If is sliced at a point $$y$$, a disk is obtained whose radius is given by $$x = 1 - {y^2}$$.

## The area of the disk

Area of the disk is

\begin{aligned}A(y) = \pi {\left( {1 - {y^2}} \right)^2}\\ = \pi \left( {1 - 2{y^2} + {y^4}} \right)\end{aligned}

The region lies between $$y = - 1$$, and $$y = 1$$.

## Apply definition of volume to find volume

Volume of region is

\begin{aligned}V &= \int_{ - 1}^1 \pi \left( {1 - 2{y^2} + {y^4}} \right)dy\\ &= \pi \left( {y - \frac{2}{3}{y^3} + \frac{{{y^5}}}{5}} \right)_{ - 1}^1\\ &= \pi \left( {(1) - \frac{2}{3}{{(1)}^3} + \frac{{{{(1)}^5}}}{5} - ( - 1) + \frac{2}{3}{{( - 1)}^3} - \frac{{{{( - 1)}^5}}}{5}} \right)\end{aligned}

\begin{aligned}V &= 2\pi \left( {1 - \frac{2}{3} + \frac{1}{5}} \right)\\ &= 2\pi \left( {\frac{{15 - 10 + 3}}{{15}}} \right)\\ &= \frac{{16}}{{15}}\pi \end{aligned}

Therefore, the volume of the solid is $$\frac{{16\pi }}{{15}}$$.