Question

To determine the volume of the resulting solid by method of cylindrical shell.

Short answer

The volume of the solid is \(25.13\).

Step-by-step solution

Given information

We have been given the equation \(y = - {x^2} + 6x - 8\), \(y = 0\) about the y-axis.

The concept of method of washer

The shell method calculates the volume of the full solid of revolution by sum of the volumes of these thin cylindrical shells as the thickness \(\Delta x\backslash \) Delta \(x\Delta x\) goes to 000 in the limit: \(V = \int d V = \int a b2\pi xydx = \int a b2\pi xf(x)dx\).

Calculate the co-ordinates

Show the equation as below:

\(y = - {x^2} + 6x - 8\) …….(1)

Plot a graph for the equation \(y = - {x^2} + 6x - 8\) by the use of the calculation as follows:

Calculate \(y\) value by the use of Equation (1)

Substitute 0 for \(x\) in Equation (1).

\(\begin{aligned}{}y = - {0^2} + 6(0) - 8\\y = 0\end{aligned}\)

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

Calculate \(y\) value by the use of Equation (1)

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

\(\begin{aligned}{}y = - {1^2} + 6(1)8\\ = - 1 + 6 - 8\\ = - 3\end{aligned}\)

Hence, the co-ordinate of \((x,y)\) is \((1, - 3)\).

Calculate y value by the use of Equation (1)

Substitute 2 for \(x\) in Equation (1).

\(\begin{aligned}{}y = - {2^2} + 6(2) - 8\\ = - 4 + 12 - 8\\ = 0\end{aligned}\)

The co-ordinate of(x, y) is \((2,0)\).

Calculate y value by the use of Equation (1)

Substitute 3 for \(x\) in Equation (1).

\(\begin{aligned}{}y = - {3^2} + 6(3) - 8\\ = - 9 + 18 - 8\\ = 1\end{aligned}\)

The co-ordinate of (x, y) is \((3,1)\).

Calculate y value by the use of Equation (1)

Substitute 4 for \(x\) in Equation (1).

\(\begin{aligned}{}y = - {4^2} + 6(4) - 8\\ = - 16 + 24 - 8\\ = 0\end{aligned}\)

The co-ordinate of(x, y) is \((4,0)\).

Draw the graph

Draw the shell as Figure 1.

Calculate the volume

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

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

Substitute 2 for a, 4 for \(b\), and \(\left( { - {x^2} + 6x - 8} \right)\) for \((f(x))\) in Equation (2).

\(\begin{aligned}{}\int_a^b 2 \pi x(f(x))dx = \int_2^4 2 \pi x\left( { - {x^2} + 6x - 8} \right)dx\\ = 2\pi \int_2^4 {\left( { - {x^3} + 6{x^2} - 8x} \right)} dx\end{aligned}\) ……. (3)

Integrate Equation (3).

\(\begin{aligned}{}2\pi \int_2^4 {\left( { - {x^3} + 6{x^2} - 8x} \right)} dx = 2\pi \left( { - \left( {\frac{{{x^{3 + 1}}}}{{3 + 1}}} \right) + 6\left( {\frac{{{x^{2 + 1}}}}{{2 + 1}}} \right) - 8\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right)_2^4\\ = 2\pi \left( { - \left( {\frac{{{x^4}}}{4}} \right) + 6\left( {\frac{{{x^3}}}{3}} \right) - 8\left( {\frac{{{x^2}}}{2}} \right)} \right)_2^4\\ = \left\{ {2\pi \left( {\begin{aligned}{{}{}}{\left( { - \frac{{{4^4}}}{4} + \frac{{6 \times {4^3}}}{3} - \frac{{8 \times {4^2}}}{2}} \right)}\\{ - \left( { - \frac{{{2^4}}}{4} + \frac{{6 \times {2^3}}}{3} - \frac{{8 \times {2^2}}}{2}} \right)}\end{aligned}} \right)} \right\}\\ = 2\pi ( - 64 + 128 - 64 + 4 - 16 + 16)\end{aligned}\)

Thus, \(2\pi \int_2^4 {\left( { - {x^3} + 6{x^2} - 8x} \right)} dx = 25.13\).

The volume of the solid is \(25.13\).