Answer
\[\frac{{2\pi }}{3}\,\,\ln \,\,2\]
Work Step by Step
\[\begin{gathered}
Apply\,\,\,the\,\,\,disk\,\,method\,\,for\,\,volume \hfill \\
\hfill \\
V = \pi \,\,\left[ {\int_1^2 {\,\left( {\frac{1}{{\,{{\left( {\sqrt {x\,\left( {3 - x} \right)} } \right)}^2}}}} \right)} dx} \right] \hfill \\
\hfill \\
simplify\,\,\frac{1}{{\,{{\left( {\sqrt {x\,\left( {3 - x} \right)} } \right)}^2}}} \hfill \\
\hfill \\
V = \pi \int_1^2 {\,\left( {\frac{1}{{\,\left( {x\,\left( {3 - x} \right)} \right)}}} \right)} dx \hfill \\
\hfill \\
decompose\,\,\frac{1}{{\,x\,\left( {3 - x} \right)}}{\text{ into partial fractions}} \hfill \\
\hfill \\
V = \frac{\pi }{3}\int_1^2 {\,\left( {\frac{1}{x} + \frac{1}{{3 - x}}} \right)} dx \hfill \\
\hfill \\
in\,tegrate\,\,and\,\,evaluate \hfill \\
\hfill \\
V = \frac{\pi }{3}\,\,\left[ {\ln \left| x \right| - \ln \left| {x - 3} \right|} \right]_1^2 \hfill \\
\hfill \\
V = \frac{\pi }{3}\,\,\left[ {\ln \left| 2 \right| + \ln \left| 2 \right|} \right] \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
\frac{{2\pi }}{3}\,\,\ln \,\,2 \hfill \\
\end{gathered} \]
