Answer
$$V = \left[ {8\ln \left( {2 + \sqrt 3 } \right) - \sqrt 3 } \right]\pi $$
Work Step by Step
$$\eqalign{
& y = \sec x,{\text{ }}y = 0,{\text{ }}0 \leqslant x \leqslant \frac{\pi }{3},{\text{ revolving about }}y = 4 \cr
& {\text{Let }}R\left( x \right) = 4 - 0{\text{ and }}r\left( x \right) = 4 - \sec x \cr
& {\text{Using the Washer Method}} \cr
& V = \pi \int_a^b {\left( {{{\left[ {R\left( x \right)} \right]}^2} - {{\left[ {r\left( x \right)} \right]}^2}} \right)} dx,{\text{ then}} \cr
& V = \pi \int_0^{\pi /3} {\left( {{{\left[ 4 \right]}^2} - {{\left[ {4 - \sec x} \right]}^2}} \right)} dx \cr
& V = \pi \int_0^{\pi /3} {\left( {16 - 16 + 8\sec x - {{\sec }^2}x} \right)} dx \cr
& V = \pi \int_0^{\pi /3} {\left( {8\sec x - {{\sec }^2}x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {8\ln \left| {\sec x + \tan x} \right| - \tan x} \right]_0^{\pi /3} \cr
& V = \pi \left[ {8\ln \left| {\sec \left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right)} \right| - \tan \left( {\frac{\pi }{3}} \right)} \right] - \pi \left[ {8\ln \left| 1 \right|} \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left[ {8\ln \left| {2 + \sqrt 3 } \right| - \sqrt 3 } \right] - \left[ {8\ln \left| 1 \right|} \right] \cr
& V = \left[ {8\ln \left( {2 + \sqrt 3 } \right) - \sqrt 3 } \right]\pi \cr
& V \approx 27.657 \cr} $$
