Question

To find: The Volume of the solid which is obtained on rotating the region bounded by the given curves about the specified line.

Short answer

The volume of the solid is \(\frac{{13}}{3}\pi \).

Step-by-step solution

Given information

The function is \(x = {y^2} + 1,x = 2\).

The concept of volume by cylindrical shell method

Definition of Volume by cylindrical shell method:

Consider that the solid\(S\)is obtained by rotation of the region under the curve\(y = f(x)\)about the\(y\)-axis from\(a\)to\(b\)is\(V = \int_a^b 2 \pi xf(x)dx\).

Calculate the height

The bounded curves of a region are assumed to be \(x = {y^2} + 1,x = 2\).

The specific line around which the region is rotated is\(y = - 2\)which is parallel to\(x\)-axis.

Obtain the value of\(y\)as follows:

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

Here, the shell radius is \( - 2 + y\) and the shell height is \({y^2} + 1 - 2 = {y^2} - 1\).

Calculate the volume

The region lies between \(y = 0\), and \(y = 1\), so by definition of cylindrical shell, the volume becomes

\(\begin{aligned}{}V = \int_0^1 2 \pi ( - 2 + y)\left( {{y^2} - 1} \right)dy\\ = 2\pi \int_0^1 {\left( { - 2{y^2} + 2 + {y^3} - y} \right)} dy\\ = 2\pi \left( { - \frac{{2{y^3}}}{3} + 2y + \frac{y}{4} - \frac{{{y^2}}}{2}} \right)_0^1\\ = 2\pi \left( {\frac{2}{3} - 2 + \frac{1}{4} - \frac{1}{2}} \right)\end{aligned}\)

On further simplifications, the followed value is obtained.

\(\begin{aligned}{}V = 2\pi \left( {\frac{{ - 8 + 24 + 3 - 6}}{{12}}} \right)\\ = 2\pi \left( {\frac{{13}}{{12}}} \right)\\ = \frac{{13\pi }}{6}\end{aligned}\)

Here, the shell radius is\( - 2 - y\)and the shell height is\({y^2} + 1 - 2 = {y^2} - 1\).

The region lies between \(y = - 1\), and \(y = 0\), so by definition of cylindrical shell, the volume becomes

\(\begin{aligned}{}V = \int_{ - 1}^0 2 \pi ( - 2 - y)\left( {{y^2} - 1} \right)dy\\ = 2\pi \int_{ - 1}^0 {\left( { - 2{y^2} + 2 - {y^3} + y} \right)} dy\\ = 2\pi \left( { - \frac{{2{y^3}}}{3} + 2y - \frac{y}{4}4 + \frac{{{y^2}}}{2}} \right)_{ - 1}^0\\ = 2\pi \left( { - \frac{2}{3} + 2 + \frac{1}{4} - \frac{1}{2}} \right)\end{aligned}\)

On further simplifications,

\(\begin{aligned}{}V = 2\pi \left( {\frac{{ - 8 + 24 + 3 - 6}}{{12}}} \right)\\ = 2\pi \left( {\frac{{13}}{{12}}} \right)\\ = \frac{{13\pi }}{6}\end{aligned}\)

Thus, the total volume becomes \(\frac{{13}}{6}\pi + \frac{{13}}{6}\pi = \frac{{13}}{3}\pi \).

Therefore, the volume of the solid is \(\frac{{13}}{3}\pi \).