Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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.

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Essential Calculus: Early Transcendentals

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Short Answer

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

Step by Step Solution

Given data

Concept used of Barrel

Solve to find the volume

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Given data

Concept used of Barrel

Solve to find the volume

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