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Q12E

Expert-verifiedFound in: Page 379

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

**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**

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

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 \(x = y,x = {y^2}\)

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 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 \(y = 1\).

Volume of the region

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

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

Graph of the solid is

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