Answer
$V = 12\pi {\text{ uni}}{{\text{t}}^3}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {4x + 2} ,{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 2 \cr
& {\text{Using the disk method}} \cr
& V = \int_a^b {\pi {{\left[ {f\left( x \right)} \right]}^2}dx} \cr
& V = \int_0^2 {\pi {{\left[ {\sqrt {4x + 2} } \right]}^2}dx} \cr
& V = \pi \int_0^2 {\left( {4x + 2} \right)dx} \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {2{x^2} + 2x} \right]_0^2 \cr
& V = \pi \left[ {2{{\left( 2 \right)}^2} + 2\left( 2 \right)} \right] - \pi \left[ {2{{\left( 0 \right)}^2} + 2\left( 0 \right)} \right] \cr
& V = \pi \left( {12} \right) - \pi \left( 0 \right) \cr
& V = 12\pi {\text{ uni}}{{\text{t}}^3} \cr} $$
