Answer
$V = \frac{\pi }{{16}}\left( {1 + \frac{\pi }{2}} \right){\text{uni}}{{\text{t}}^3}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{4 + {x^2}}},{\text{ }}y = 0,{\text{ }}x = - 2,{\text{ }}x = 2 \cr
& {\text{Using the disk method }}V = \int_a^b {\pi {{\left[ {f\left( x \right)} \right]}^2}} dx,{\text{ then}} \cr
& V = \int_{ - 2}^2 {\pi {{\left[ {\frac{1}{{4 + {x^2}}}} \right]}^2}dx} \cr
& {\text{By symmetry}} \cr
& V = 2\pi \int_0^2 {\frac{1}{{{{\left( {4 + {x^2}} \right)}^2}}}dx} \cr
& {\text{Integrating by a CAS }}\left( {{\text{Wolfram Alpha Website}}} \right) \cr
& V = 2\pi \left( {\frac{1}{{16}}} \right)\left[ {\frac{{2x}}{{{x^2} + 4}} + {{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^2 \cr
& V = \frac{\pi }{8}\left[ {\frac{{2x}}{{{x^2} + 4}} + {{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^2 \cr
& {\text{Evaluating}} \cr
& V = \frac{\pi }{8}\left[ {\frac{{2\left( 2 \right)}}{{{{\left( 2 \right)}^2} + 4}} + {{\tan }^{ - 1}}\left( {\frac{2}{2}} \right)} \right] - \frac{\pi }{8}\left[ {\frac{{2\left( 0 \right)}}{{{{\left( 2 \right)}^2} + 4}} + {{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& V = \frac{\pi }{8}\left[ {\frac{1}{2} + \frac{\pi }{4}} \right] - \frac{\pi }{8}\left[ 0 \right] \cr
& V = \frac{\pi }{{16}}\left( {1 + \frac{\pi }{2}} \right){\text{uni}}{{\text{t}}^3} \cr
& V \approx 0.5047{\text{uni}}{{\text{t}}^3} \cr} $$
