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Q8E

Expert-verifiedFound in: Page 385

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**To determinethe volume by the method of slicing and cylindrical shells.**

The volume of the solid by slicing is 0.942

The volume of the solid by cylindrical shell is 0.942.

The equations are \(y = \sqrt x \) and \(y = {x^2}\).

**The volume by the use of the method of shell is **\(V = \int_a^b 2 \pi x(f(x))dx\)

**The volume by the use of the method of slicing **\(V = \pi {\int\limits_a^b {\left( {g\left( y \right)} \right)} ^2} - {\left( {f\left( y \right)} \right)^{2\;\;}}dy\)

Show the equations as below:

\(y = \sqrt x \) ……….. (1)

\(y = {x^2}\) …………. (2)

Plot a graph for the equations \(y = \sqrt x \) and \(y = {x^2}\)by the use of the calculation as follows:

Calculate yvalue by the use of equation \(\{ 1\} \)

Substitute O for \(x\) in equation (1).

\(y{\rm{ }} = \sqrt 0 = 0\)

Hence, the co-ordinate of(x, y) is \((0,0)\).

Substitute 1 for \(x\) in equation (1).

\(y{\rm{ }} = \sqrt T = 1\)

Hence, the co-ordinate of (x, y) is \((1,1)\).

Substitute O for \(x\) in equation \((2)\).

\(y{\rm{ }} = {0^2} = 0\)

The co-ordinate of(x, y) is \((0,5)\).

Substitute 1 for \(x\) in equation (2).\(y{\rm{ }} = {1^2} = 1\)

The co-ordinate of(x, y) is \((1,1)\).

Similarly calculate the coordinate values up to the bounded region in the graph

Draw the region as shown in Figure1.

Draw the slicing of figure

Draw the cylindrical shell

Calculate the volume by the use of the method of slicing:

\(V = \pi {\int\limits_a^b {\left( {g\left( y \right)} \right)} ^2} - {\left( {f\left( y \right)} \right)^{2\;\;}}dy\) ………….. (3)

Rearrange equation (1).

\(\begin{aligned}{}{y^2} = x\\x = {y^2}\end{aligned}\)

Rearrange equation (2).

\(\begin{aligned}{}\sqrt y = x\\x = \sqrt y \end{aligned}\)

Substitute 0 for a, 1 for \(b,\sqrt y \) for\((g(y))\), and \({y^2}\) for \((f(y))\) in equation (3).\(\begin{aligned}{}V = \pi \int_a^b {{{(\sqrt y )}^2}} - {\left( {{y^2}} \right)^2}\\ = \pi \int_a^b y - {y^4}dy\end{aligned}\)

Integrate Equation (4).

\(\begin{aligned}{}V = \pi \left( {\left( {\frac{{{y^{1 + 1}}}}{{1 + 1}}} \right) - \left( {\frac{{{y^{1 + 1}}}}{{4 + 1}}} \right)} \right)_0^1\\ = \pi \left( {\left( {\frac{{{y^2}}}{2}} \right) - \left( {\frac{{{y^3}}}{5}} \right)} \right)_0^1\\ = \pi \left( {\frac{{{1^2}}}{2} - \frac{{{1^3}}}{5}} \right)\\ = 0.942\end{aligned}\)

Hence, the volume of the solid by slicing is 0.942.

Calculate the volume by the use of the method of cylindrical shell:

\(V = \int_a^b 2 \pi x(f(x))dx(5)\)

Consider \(f(x) = \sqrt x - y(6)\)

Substitute \({x^2}\) for \(y\) in Equation (4).

\(f(x) = \sqrt x - {x^2}\)

Substitute 0 for a, 1 for \(b\), and \(\sqrt x - {x^2}\) for \((f(x))\) in equation (5).

\(\begin{aligned}{}V = \int_0^1 2 \pi x\left( {\sqrt x - {x^2}} \right)dx\\ = 2\pi \int_0^1 {\left( {x{{(x)}^{\frac{1}{2}}} - {x^3}} \right)} dx(7)\\ = 2\pi \int_0^1 {\left( {{x^{\frac{3}{2}}} - {x^3}} \right)} dx\end{aligned}\)

Integrate equation (7).

\(\begin{aligned}{}{I_2} = 2\pi \left( {\left( {\frac{{{x^{\frac{3}{2} + 1}}}}{{\frac{3}{2} + 1}}} \right) - \left( {\frac{{{x^{3 + 1}}}}{{3 + 1}}} \right)} \right)_0^1\\ = 2\pi \left( {\left( {\frac{2}{5}{x^{\frac{3}{2}}}} \right) - \left( {\frac{{{x^4}}}{4}} \right)} \right)_0^1\\ = 2\pi \left( {\left( {\frac{2}{5} \times {{(1)}^{\frac{3}{2}}} - \frac{{{1^4}}}{4}} \right) - 0} \right)\quad \\ = 2\pi \left( {\frac{2}{5} - \frac{1}{4}} \right) = 0.942\end{aligned}\)

Hence, the volume of the solid by cylindrical shell is 0.942.

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