Answer
\[V=\frac{4\pi }{\sqrt[4]{3}}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{\left( {{x}^{2}}-1 \right)}^{-1/4}},\text{ }\left( 1,2 \right] \\
& \text{we can calculate the volume using the Shell Method the }y\text{-Axis } \\
& V=\int_{a}^{b}{2\pi xf\left( x \right)dx,\text{ }} \\
& V=\int_{1}^{2}{2\pi x{{\left( {{x}^{2}}-1 \right)}^{-1/4}}dx} \\
& V=\int_{1}^{2}{\frac{2\pi x}{{{\left( {{x}^{2}}-1 \right)}^{1/4}}}dx} \\
& \text{The integrand is not defined for }x=1,\text{ then} \\
& V=\underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\int_{b}^{2}{\frac{2\pi x}{{{\left( {{x}^{2}}-1 \right)}^{1/4}}}dx} \\
& V=\pi \underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\int_{b}^{2}{\frac{2x}{{{\left( {{x}^{2}}-1 \right)}^{1/4}}}dx} \\
& \text{Integrating} \\
& V=\pi \underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\left[ \frac{{{\left( {{x}^{2}}-1 \right)}^{3/4}}}{3/4} \right]_{b}^{2} \\
& V=\frac{4\pi }{3}\underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\left[ {{\left( {{x}^{2}}-1 \right)}^{3/4}} \right]_{b}^{2} \\
& V=\frac{4\pi }{3}\underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\left[ {{\left( {{2}^{2}}-1 \right)}^{3/4}}-{{\left( {{2}^{2}}-1 \right)}^{3/4}} \right] \\
& V=\frac{4\pi }{3}\underset{b\to {{1}^{+}}}{\mathop{\lim }}\,\left[ {{3}^{3/4}}-{{\left( {{b}^{2}}-1 \right)}^{3/4}} \right] \\
& \text{Evaluate the limit} \\
& V=\frac{4\pi }{3}\left[ {{3}^{3/4}}-{{\left( {{1}^{2}}-1 \right)}^{3/4}} \right] \\
& V=\frac{4\pi }{3}\left( {{3}^{3/4}} \right) \\
& V=\frac{4\pi }{{{3}^{1/4}}} \\
& V=\frac{4\pi }{\sqrt[4]{3}} \\
\end{align}\]
