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Q17E

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Found in: Page 385

### Essential Calculus: Early Transcendentals

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

# The volume of the resulting solid by the region bounded by given curve by the use of the method of washer

The volume of the solid by the region bounded by given curve is $$\underline {21.75}$$.

See the step by step solution

## Given information

The equation is $${y^2} - {x^2} = 1,y = 2$$; about the $$x$$-axis.

## The concept of method of washer

Show the theorem 7 (Integrals of symmetric Functions):

Apply the condition as follows if $$f$$ is continuous on $$( - a,a)$$.

a) If $$f$$ is even function $$(f( - x) = f(x))$$, then $$\int_{ - a}^a f (x)dx = 2\int_0^a f (x)dx$$.

b) If $$f$$ is odd function $$(f( - x) = f(x))$$, then $$\int_{ - a}^a f (x)dx = 0$$.

## Calculate the co-ordinates

Show the equation as below:

$${y^2} - {x^2} = 1$$ …….(1)

Plot a graph for the equation $${y^2} - {x^2} = 1$$ 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}{}{y^2} - {(0)^2} = 1\\{y^2} = 1\\y = 1\end{aligned}

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

Calculate y value by the use of Equation (1)

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

\begin{aligned}{}{y^2} - {(1)^2} = 1\\{y^2} = 1 + 1\\y = \sqrt 2 = \pm 1.41\end{aligned}

Hence, the co-ordinate of $$(x,{\rm{ }}y)$$ is $$(1, \pm 1.41)$$.

Calculate $$y$$ value by the use of Equation (1)

Substitute 2 for $$x$$ in Equation (1).

\begin{aligned}{}{y^2} - {(2)^2} = 1\\{y^2} = 1 + 4\\y = \sqrt 5 = \pm 2.24\end{aligned}

The co-ordinate of $$(x,{\rm{ }}y)$$ is $$(2, \pm 2.24)$$.

Calculate $$x$$ value by the use of Equation (1)

Substitute 2 for $$y$$ in Equation (1).

\begin{aligned}{}{(2)^2} - {x^2} = 1\\{x^2} = - 1 + 4\\x = \pm \sqrt 3 \end{aligned}

The co-ordinate of $$(x,{\rm{ }}y)$$ is $$( \pm \sqrt 3 ,2)$$.

## Draw the graph

Draw the washer as shown in Figure 1:

## Calculate the volume

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

$$V = \int_a^b \pi \left\{ {|f(x){|^2} - |g(x){|^2}} \right\}dx$$ …….(2)

Rearrange the Equation (1).

$$y = \sqrt {{x^2} + 1}$$

Substitute $$- \sqrt 3$$ for $$a,\sqrt 3$$ for b, 2 for $$(f(x))$$, and $$\sqrt {{x^2} + 1}$$ for in Equation (2).

\begin{aligned}{{}{}}{V = \int_{ - \sqrt 3 }^{\sqrt 3 } \pi \left( {{{(2)}^2} - \left( {\sqrt {{x^2} + 1} } \right)} \right.}\\{ = \pi \int_{ - \sqrt 3 }^{\sqrt 3 } 4 - \left( {{x^2} + 1} \right)dx}\\{ = \pi \int_{ - \sqrt 3 }^{\sqrt 3 } {\left( { - {x^2} + 3} \right)} dx}\end{aligned}

Consider $$f(x) = - {x^2} + 3$$ …….(4)

Substitute $$( - x)$$ for $$x$$ in Equation (4).

\begin{aligned}{}f( - x) = - {( - x)^2} + 3\\ = - {x^2} + 3\\ = f(x)\end{aligned}

Hence the function is even function.

Apply the theorem 7 in Equation (3).

$$V = 2\pi \int_0^{\sqrt 3 } {\left( { - {x^2} + 3} \right)} dx$$ …….(5)

Integrate Equation (5).

\begin{aligned}{}V = 2\pi \left( { - \left( {\frac{{{x^{2 + 1}}}}{{2 + 1}}} \right) + 3x} \right)_0^{\sqrt 3 }\\ = 2\pi \left( { - \frac{{{x^3}}}{3} + 3x} \right)_0^{\sqrt 3 }\\ = 2\pi \left( {\left( { - \frac{{{{(\sqrt 3 )}^3}}}{3} + 3 \times \sqrt 3 } \right) - 0} \right)\\ = 2\pi \left( { - \frac{{3\sqrt 3 }}{3} + 3\sqrt 3 } \right)\\ = 2\pi ( - \sqrt 3 + 3\sqrt 3 )\\ = 2\pi (2\sqrt 3 )\\ = 21.75\end{aligned}