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Found in: Page 384

### Essential Calculus: Early Transcendentals

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

# (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

## 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 $$V = \frac{\pi }{{15}}$$.