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Q17E

Expert-verifiedFound in: Page 385

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 \(\underline {21.75} \).

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

**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 **\(\int_{ - a}^a f (x)dx = 0\)**.**

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 washer as shown in Figure 1:

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).

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

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