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Expert-verified Found in: Page 380 ### Essential Calculus: Early Transcendentals

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.

See the 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. ### Want to see more solutions like these? 