Question

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

Short answer

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

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).

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