Answer
\[\frac{{{\pi ^2}}}{2} - \pi {\tan ^{ - 1}}\,2\]
Work Step by Step
\[\begin{gathered}
the\,\,area\,\,is\,\,bounded\,\,by\,\,\,{\left( {{x^2} + 1} \right)^{\frac{1}{2}}} \hfill \\
and\,\,y \geqslant 0\,\,and\,\,x \geqslant 2\,\,is\,\,given\,by. \hfill \\
\hfill \\
\int_2^\infty {\frac{1}{{\sqrt {{x^2} + 1} }}dx} \hfill \\
\hfill \\
which\,\,diverges\,\,to\,\,\,infinity.\, \hfill \\
But\,the\,\,volume\,\,is\,\,given\,\,by. \hfill \\
\hfill \\
\int_2^\infty {\pi \,\,\,{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}\,dx} \hfill \\
\hfill \\
\int_2^\infty {\pi \,\,\,{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}\,dx} = \pi \int_2^\infty {\frac{1}{{{x^2} + 1}}} \,dx \hfill \\
\hfill \\
{\text{integrate}}\,\, \hfill \\
= \pi \,\,\left[ {{{\tan }^{ - 1}}\,\left( x \right)} \right]_2^\infty \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \pi \,\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\,2} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{{\pi ^2}}}{2} - \pi {\tan ^{ - 1}}\,2 \hfill \\
\hfill \\
\end{gathered} \]
