Question

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

Short answer

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.

Step-by-step solution

Given data

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

Concept used of Barrel

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.

Solve to find the volume

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.