Answer
\[V = 8\pi \sqrt {15} - 2\pi \ln \,\left( {4 + \sqrt {15} } \right)\]
Work Step by Step
\[\begin{gathered}
Curves\,\,y = \,{\left( {1 - {x^2}} \right)^{ - \frac{1}{2}}}\,\,and\,\,y = 4\,\,intersects\,\,\,at \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \pm \sqrt {\frac{{15}}{{16}}} \hfill \\
\hfill \\
by\,\,disk\,\,\,method\,\,required\,\,volume\,\,is \hfill \\
\hfill \\
V = \pi \,\,\left[ {\int_{ - \sqrt {\frac{{15}}{{16}}} }^{\sqrt {\frac{{15}}{{16}}} } {\,\left( {{4^2} - \frac{1}{{\,{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}} \right)dx} } \right] \hfill \\
\hfill \\
simplify\,\,\,{4^2} - \frac{1}{{\,{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}} \hfill \\
\hfill \\
V = \pi \,\,\left[ {\int_{ - \sqrt {\frac{{15}}{{16}}} }^{\sqrt {\frac{{15}}{{16}}} } {\,\left( {16 - \frac{1}{{1 - {x^2}}}} \right)dx} } \right] \hfill \\
\hfill \\
in\,tegrate\,\,and\,\,evaluate \hfill \\
\hfill \\
V = \pi \,\,\left[ {16x - \frac{1}{2}\left| {\frac{{1 + x}}{{1 - x}}} \right|} \right]_{ - \sqrt {\frac{{15}}{{16}}} }^{\sqrt {\frac{{15}}{{16}}} } \hfill \\
\hfill \\
V = \pi \,\,\left[ {8\sqrt {15} } \right] - \pi \ln \left| {\frac{{4 + \sqrt {15} }}{{4 - \sqrt {15} }}} \right| \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
V = 8\pi \sqrt {15} - 2\pi \ln \,\left( {4 + \sqrt {15} } \right) \hfill \\
\hfill \\
\end{gathered} \]
