Answer
$$V = 2\pi \left( {\frac{\pi }{3} - \frac{1}{{\sqrt 3 }}} \right)$$
Work Step by Step
$$\eqalign{
& x = \frac{4}{{y + {y^3}}},{\text{ }}x = \frac{1}{{\sqrt 3 }},{\text{ and }}y = 1 \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& \frac{4}{{y + {y^3}}} > \frac{1}{{\sqrt 3 }}{\text{ on the interval }}\left[ {1,\sqrt 3 } \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_1^{\sqrt 3 } {2\pi y\left( {\frac{4}{{y + {y^3}}} - \frac{1}{{\sqrt 3 }}} \right)} dy \cr
& V = 2\pi \int_1^{\sqrt 3 } {\left( {\frac{{4y}}{{y + {y^3}}} - \frac{y}{{\sqrt 3 }}} \right)} dy \cr
& V = 2\pi \int_1^{\sqrt 3 } {\left( {\frac{4}{{1 + {y^2}}} - \frac{y}{{\sqrt 3 }}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {4{{\tan }^{ - 1}}y - \frac{{{y^2}}}{{2\sqrt 3 }}} \right]_1^{\sqrt 3 } \cr
& V = 2\pi \left[ {4{{\tan }^{ - 1}}\left( {\sqrt 3 } \right) - \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{{2\sqrt 3 }}} \right] - 2\pi \left[ {4{{\tan }^{ - 1}}\left( 1 \right) - \frac{{{{\left( 1 \right)}^2}}}{{2\sqrt 3 }}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left[ {4\left( {\frac{\pi }{3}} \right) - \frac{{\sqrt 3 }}{2}} \right] - 2\pi \left[ {4\left( {\frac{\pi }{4}} \right) - \frac{1}{{2\sqrt 3 }}} \right] \cr
& V = \frac{{8{\pi ^2}}}{3} - \pi \sqrt 3 - 2{\pi ^2} + \frac{\pi }{{\sqrt 3 }} \cr
& V = \frac{{2{\pi ^2}}}{3} - \frac{{2\pi }}{{\sqrt 3 }} \cr
& V = 2\pi \left( {\frac{\pi }{3} - \frac{1}{{\sqrt 3 }}} \right) \cr} $$
