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Q1E

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Essential Calculus: Early Transcendentals
Found in: Page 384
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

(a)To determine the difficulty to use slicing to find the volume, \(V\) of solid \(S\).

(b)To sketch the typical approximating shell.

(c)To find the circumference, height and volume using the method of shell.

The circumference, height and volume of shell as follows, \(C = 2\pi x\;,\;\;H = x{(x - 1)^2}\), and \(V = \frac{\pi }{{15}}\).

See the step by step solution

Step by Step Solution

Given

The function is \(y = x{(x - 1)^2}\)

The region lies between 0 to 1.

The Concept ofvolume by the use of the method of shell

The volume by the use of the method of shell is \(V = \int_a^b 2 \pi x(f(x))dx\)

Evaluate the volume

Show the equation as below:

\(y = x{(x - 1)^2}\) ……… (1)

Plot a graph for the equation \(y = x{(x - 1)^2}\)by the use of the calculation as follows:

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

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

\(\begin{aligned}{}dy = 0{(0 - 1)^2}\\ = 0\end{aligned}\)

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

Calculate \(y\) value by the use ofEquation (1).

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

\(\begin{aligned}{}dy = 1{(1 - 1)^2}\\ = 0\end{aligned}\)

Hence, the co-ordinate of (x, y) is \((1,0)\).

Draw the region as shown in Figure

Draw the shell

Refer Figure 2

The radius of shell is \(x\).

Calculate the circumference of shell:

\(C = 2\pi r\) ………. (2)

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

\(C = 2\pi x\)

Calculate the height of shell:

\(H = y\) ..……. (3)

Substitute \(x{(x - 1)^2}\) for \(y\) in Equation (3).

\(H = x{(x - 1)^2}\)

Calculate the volume by the use ofthe method ofshell:

\(V = \int_a^b 2 \pi x(f(x))dx\) …….. (4)

Substitute0 for a, 1 for \(b\), and \(x{(x - 1)^2}\) for \((f(x))\) in Equation (4)\(V = \int_0^1 2 \pi x\left( {x{{(x - 1)}^2}} \right)dx\)

\(V = 2\pi \int_0^1 {{x^2}} \left( {{x^2} - 2x + 1} \right)dx\) …….. (5)

Integrate Equation (5).

\(\begin{aligned}{}V = 2\pi \left( {\frac{{{x^{4 + 1}}}}{{4 + 1}} - \frac{{2{x^{3 + 1}}}}{{3 + 1}} + \frac{{{x^{2 + 1}}}}{{2 + 1}}} \right)_0^1\\ = 2\pi \left( {\frac{{{x^5}}}{5} - \frac{{{x^4}}}{2} + \frac{{{x^3}}}{3}} \right)_0^1\\ = \frac{\pi }{{15}}\end{aligned}\)

Therefore, the circumference, height and volume of shell as follows, \(C = 2\pi x,\;\;\;H = x{(x - 1)^2}\), and \(V = \frac{\pi }{{15}}\).

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