Question

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

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

Short answer

The volume of the curve is \({\rm{V = }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}\).

Step-by-step solution

Curve.

The region of the curve

\(\begin{aligned}{}{{\rm{y}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{ = 1 \ldots \ldots \ldots \ldots \ldots \ldots (1)}}\\{\rm{y = 2}}\end{aligned}\)

No need to consider the lower part of the hyperbola because of no intersection only

\({\rm{y = 2}}\).

Volume of the curve line.

From the graph \({\rm{r = x}}\), in equation \({\rm{(1)}}\)get \({\rm{x = \pm }}\sqrt {{{\rm{y}}^{\rm{2}}}{\rm{ - 1}}} \).

So, the radius is \({\rm{r = }}\sqrt {{{\rm{y}}^{\rm{2}}}{\rm{ - 1}}} \)

Therefore, the area \({\rm{A(y) = \pi }}{{\rm{r}}^{\rm{2}}}{\rm{ = \pi }}\left( {{{\rm{y}}^{\rm{2}}}{\rm{ - 1}}} \right)\).

Then the Volume is,

\(\begin{aligned}{}{\rm{V = }}\int_{\rm{1}}^{\rm{2}} {\rm{A}} {\rm{(y)dy}}\\{\rm{ = }}\int_{\rm{1}}^{\rm{2}} {\rm{\pi }} \left( {{{\rm{y}}^{\rm{2}}}{\rm{ - 1}}} \right){\rm{dy}}\\{\rm{ = \pi }}\int_{\rm{1}}^{\rm{2}} {\left( {{{\rm{y}}^{\rm{2}}}{\rm{ - 1}}} \right)} {\rm{dy}}\\\left. {{\rm{ = \pi }}\left( {\frac{{\rm{1}}}{{\rm{3}}}{{\rm{y}}^{\rm{3}}}{\rm{ - y}}} \right)} \right)_{\rm{1}}^{\rm{2}}\\{\rm{ = \pi }}\left( {\frac{{\rm{1}}}{{\rm{3}}}{{\rm{2}}^{\rm{3}}}{\rm{ - 2 - }}\left( {\frac{{\rm{1}}}{{\rm{3}}}{{\rm{1}}^{\rm{3}}}{\rm{ - 1}}} \right)} \right)\\{\rm{ = \pi }}\left( {\frac{{\rm{8}}}{{\rm{3}}}{\rm{ - 2 - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ + 1}}} \right)\\{\rm{ = }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}\end{aligned}\)

Therefore, the volume of the curve is \({\rm{V = }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}\).