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Q8E
Expert-verifiedTo 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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