Answer
$$V = \frac{{4{\pi ^2}}}{3} - \sqrt 3 \pi $$
Work Step by Step
$$\eqalign{
& y = \sec x{\text{ and }}y = 2 \cr
& {\text{Using the washer method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& 2 > \sec x{\text{ on the interval }}\left[ {0,\frac{\pi }{3}} \right],{\text{ }} \cr
& f\left( x \right) = 2{\text{ and }}g\left( x \right) = \sec x \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \pi \int_0^{\pi /3} {\left[ {{{\left( 2 \right)}^2} - {{\left( {\sec x} \right)}^2}} \right]} dx \cr
& V = \pi \int_0^{\pi /3} {\left( {4 - {{\sec }^2}x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {4x - \tan x} \right]_0^{\pi /3} \cr
& V = \pi \left[ {4\left( {\frac{\pi }{3}} \right) - \tan \left( {\frac{\pi }{3}} \right)} \right] - \pi \left[ {4\left( 0 \right) - \tan \left( 0 \right)} \right] \cr
& V = \pi \left( {\frac{{4\pi }}{3} - \sqrt 3 } \right) - \pi \left[ 0 \right] \cr
& V = \frac{{4{\pi ^2}}}{3} - \sqrt 3 \pi \cr} $$
