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Q45E

Expert-verifiedFound in: Page 380

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**To show the volume enclosed by the barrel \(V = \frac{1}{3}\pi h\left( {2{R^2} + {r^2} - \frac{2}{5}{d^2}} \right)\).**

The volume enclosed by the barrel \(V = \frac{1}{3}\pi h\left( {2{R^2} + {r^2} - \frac{2}{5}{d^2}} \right)\) is proved.

A barrel with height \(h\) and maximum radius \(R\).

The barrel is constructed by rotation about the \(x\)-axis the parabola \(y = R - c{x^2}, - \frac{h}{2} \le x \le \frac{h}{2}\).

Where, \(c\) is a positive constant.

The value of \(d = \frac{{c{h^2}}}{4}\).

**A barrel solid of revolution composed of parallel circular top and bottom with a common axis and a side formed by a smooth curve symmetrical about the midplane.**

Sketch the barrel with height \(h\) and radius \(R\) as shown in Figure 1.

Refer to Figure 1.

About the \(y\)-axis the barrel is symmetric, since for \(x > 0\) part, the volume of the barrel is twice the volume of that part of the barrel.

The barrel is constructed by rotation about the \(x\)-axis. Hence, the barrel is the volume of rotation.

Show the formula for the volume of rotation as shown below.

\(V = 2\int_a^b \pi {y^2}dx\) …..(1)

Here \(y\) is the Equation parabola.

Substitute \(\left( {R - c{x^2}} \right)\) for y, 0 for \(a\) and \(\frac{h}{2}\) for \(b\) in Equation (1).

\(\begin{aligned}{}V = 2\int_0^{\frac{h}{2}} \pi {\left( {R - c{x^2}} \right)^2}dx\\ = 2\pi \int_0^{\frac{h}{2}} {\left( {{R^2} + {c^2}{x^4} - 2Rc{x^2}} \right)} dx\\ = 2\pi \left( {{R^2}x + {c^2}\frac{{{x^5}}}{5} - 2Rc\frac{{{x^3}}}{3}} \right)_0^{\frac{h}{2}}\\ = 2\pi \left( {{R^2}\left( {\frac{h}{2}} \right) + {c^2}\left( {\frac{{{h^5}}}{{5 \times 32}}} \right) - 2Rc\left( {\frac{{{h^3}}}{{24}}} \right) - 0} \right)\end{aligned}\)

Which means, \(V = 2\pi \left( {\frac{1}{2}{R^2}h + \frac{1}{{160}}{c^2}{h^5} - \frac{1}{{12}}Rc{h^3}} \right)\) ….(2)

Divide and multiply both sides of the Equation (1) by 3 .

\(\begin{aligned}{}V = \frac{2}{3}\pi h\left( {\frac{3}{2}{R^2} + \frac{3}{{160}}{c^2}{h^4} - \frac{3}{{12}}Rc{h^2}} \right)\\ = \frac{1}{3}\pi h\left( {3{R^2} + \frac{3}{{80}}{c^2}{h^4} - \frac{1}{2}Rc{h^2}} \right)\\ = \frac{1}{3}\pi h\left( {2{R^2} + \left( {{R^2} + \frac{3}{{80}}{c^2}{h^4} - \frac{1}{2}Rc{h^2}} \right)} \right)\\ = \frac{1}{3}\pi h\left( {2{R^2} + \left( {{R^2} + \frac{1}{{16}}{c^2}{h^4} - \frac{1}{{40}}{c^2}{h^4} - \frac{1}{2}Rc{h^2}} \right)} \right)\end{aligned}\)

Simplify further,

\(\begin{aligned}{}V = \frac{1}{3}\pi h\left( {2{R^2} + {{\left( {R - \frac{1}{4}c{h^2}} \right)}^2} - \frac{1}{{40}}{c^2}{h^4}} \right)\\ = \frac{1}{3}\pi h\left( {2{R^2} + {{\left( {R - \frac{1}{4}c{h^2}} \right)}^2} - \frac{2}{5}{{\left( {\frac{{c{h^2}}}{4}} \right)}^2}} \right)\end{aligned}\) …..(3)

Substitute \(d\) for \(\frac{{c{h^2}}}{4}\) in Equation (3).

\(V = \frac{1}{3}\pi h\left( {2{R^2} + {{(R - d)}^2} - \frac{2}{5}{{(d)}^2}} \right)\) …..(4)

Substitute \(r\) for \(R - d\) in Equation (4).

\(V = \frac{1}{3}\pi h\left( {2{R^2} + {r^2} - \frac{2}{5}{{(d)}^2}} \right)\)

Therefore, the volume enclosed by the barrel \(V = \frac{1}{3}\pi h\left( {2{R^2} + {r^2} - \frac{2}{5}{d^2}} \right)\) is proved.

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