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Q16E

Expert-verifiedFound in: Page 379

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\(\frac{{16\pi }}{{15}}.\)

**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**

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

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 \(x = 1 - {y^2}\).

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 \(y = 1\).

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 \(\frac{{16\pi }}{{15}}\).

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