Answer
$\displaystyle{V=2\pi\ln\left(2+\sqrt3\right)-\frac{\sqrt3\pi}{8}-\frac{\pi^2}{6}}\\$
Work Step by Step
$\displaystyle{A(x)=\pi\left(1+\sec x\right)^2-\pi\left(1+\cos x\right)^2}\\
\displaystyle{A(x)=\pi\left(2\sec x+\sec^2x-2\cos x-\frac{1}{2}\cos 2x-\frac{1}{2}\right)}$
$\displaystyle{V=\int_0^{\frac{\pi}{3}}A(x)\ dx}\\
\displaystyle{V=\int_0^{\frac{\pi}{3}}\pi\left(2\sec x+\sec^2x-2\cos x-\frac{1}{2}\cos 2x-\frac{1}{2}\right)\ dx}\\
\displaystyle{V=\pi\int_0^{\frac{\pi}{3}}2\sec x+\sec^2x-2\cos x-\frac{1}{2}\cos 2x-\frac{1}{2}\ dx}\\
\displaystyle{V=\pi\left[2\ln|\sec x+\tan x|+\tan x-2\sin x-\frac{1}{4}\sin2x-\frac{1}{2}x\right]_0^{\frac{\pi}{3}}}\\
\displaystyle{V=\pi\left(2\ln\left(2+\sqrt3\right)+\sqrt3-\sqrt3-\frac{\sqrt3}{8}-\frac{\pi}{6}\right)}\\
\displaystyle{V=2\pi\ln\left(2+\sqrt3\right)-\frac{\sqrt3\pi}{8}-\frac{\pi^2}{6}}\\$
