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Found in: Page 385

### Essential Calculus: Early Transcendentals

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

# To determinethe volume by the method of slicing and cylindrical shells.

The volume of the solid by slicing is 0.942

The volume of the solid by cylindrical shell is 0.942.

See the step by step solution

## Given

The equations are $$y = \sqrt x$$ and $$y = {x^2}$$.

## The Concept ofvolume by the use of the method of shell and slicing

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

The volume by the use of the method of slicing $$V = \pi {\int\limits_a^b {\left( {g\left( y \right)} \right)} ^2} - {\left( {f\left( y \right)} \right)^{2\;\;}}dy$$

## Determine the Co-ordinates

Show the equations as below:

$$y = \sqrt x$$ ……….. (1)

$$y = {x^2}$$ …………. (2)

Plot a graph for the equations $$y = \sqrt x$$ and $$y = {x^2}$$by the use of the calculation as follows:

Calculate yvalue by the use of equation $$\{ 1\}$$

Substitute O for $$x$$ in equation (1).

$$y{\rm{ }} = \sqrt 0 = 0$$

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

Substitute 1 for $$x$$ in equation (1).

$$y{\rm{ }} = \sqrt T = 1$$

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

Substitute O for $$x$$ in equation $$(2)$$.

$$y{\rm{ }} = {0^2} = 0$$

The co-ordinate of(x, y) is $$(0,5)$$.

Substitute 1 for $$x$$ in equation (2).$$y{\rm{ }} = {1^2} = 1$$

The co-ordinate of(x, y) is $$(1,1)$$.

Similarly calculate the coordinate values up to the bounded region in the graph

## Draw the regions

Draw the region as shown in Figure1.

Draw the slicing of figure

Draw the cylindrical shell

## Evaluate the volumeby the use of the methof of Slicing

Calculate the volume by the use of the method of slicing:

$$V = \pi {\int\limits_a^b {\left( {g\left( y \right)} \right)} ^2} - {\left( {f\left( y \right)} \right)^{2\;\;}}dy$$ ………….. (3)

Rearrange equation (1).

\begin{aligned}{}{y^2} = x\\x = {y^2}\end{aligned}

Rearrange equation (2).

\begin{aligned}{}\sqrt y = x\\x = \sqrt y \end{aligned}

Substitute 0 for a, 1 for $$b,\sqrt y$$ for$$(g(y))$$, and $${y^2}$$ for $$(f(y))$$ in equation (3).\begin{aligned}{}V = \pi \int_a^b {{{(\sqrt y )}^2}} - {\left( {{y^2}} \right)^2}\\ = \pi \int_a^b y - {y^4}dy\end{aligned}

Integrate Equation (4).

\begin{aligned}{}V = \pi \left( {\left( {\frac{{{y^{1 + 1}}}}{{1 + 1}}} \right) - \left( {\frac{{{y^{1 + 1}}}}{{4 + 1}}} \right)} \right)_0^1\\ = \pi \left( {\left( {\frac{{{y^2}}}{2}} \right) - \left( {\frac{{{y^3}}}{5}} \right)} \right)_0^1\\ = \pi \left( {\frac{{{1^2}}}{2} - \frac{{{1^3}}}{5}} \right)\\ = 0.942\end{aligned}

Hence, the volume of the solid by slicing is 0.942.

## Evaluate the volumeby the use of method of Cylindrical shell

Calculate the volume by the use of the method of cylindrical shell:

$$V = \int_a^b 2 \pi x(f(x))dx(5)$$

Consider $$f(x) = \sqrt x - y(6)$$

Substitute $${x^2}$$ for $$y$$ in Equation (4).

$$f(x) = \sqrt x - {x^2}$$

Substitute 0 for a, 1 for $$b$$, and $$\sqrt x - {x^2}$$ for $$(f(x))$$ in equation (5).

\begin{aligned}{}V = \int_0^1 2 \pi x\left( {\sqrt x - {x^2}} \right)dx\\ = 2\pi \int_0^1 {\left( {x{{(x)}^{\frac{1}{2}}} - {x^3}} \right)} dx(7)\\ = 2\pi \int_0^1 {\left( {{x^{\frac{3}{2}}} - {x^3}} \right)} dx\end{aligned}

Integrate equation (7).

\begin{aligned}{}{I_2} = 2\pi \left( {\left( {\frac{{{x^{\frac{3}{2} + 1}}}}{{\frac{3}{2} + 1}}} \right) - \left( {\frac{{{x^{3 + 1}}}}{{3 + 1}}} \right)} \right)_0^1\\ = 2\pi \left( {\left( {\frac{2}{5}{x^{\frac{3}{2}}}} \right) - \left( {\frac{{{x^4}}}{4}} \right)} \right)_0^1\\ = 2\pi \left( {\left( {\frac{2}{5} \times {{(1)}^{\frac{3}{2}}} - \frac{{{1^4}}}{4}} \right) - 0} \right)\quad \\ = 2\pi \left( {\frac{2}{5} - \frac{1}{4}} \right) = 0.942\end{aligned}

Hence, the volume of the solid by cylindrical shell is 0.942.