A curve $G$ has equation $y = \frac{1}{\cos x} + \cos x$. The region bounded by $G$, the line $x = \frac{\pi}{3}$, the $x$-axis and the $y$-axis is rotated through 4 right angles about the $x$-axis. Find the exact volume generated, giving your answer in terms of $\pi$.

1. Set up the definite integral for the volume of revolution.

$V = \pi \int_{0}^{\frac{\pi}{3}} y^2 \, dx$
$V = \pi \int_{0}^{\frac{\pi}{3}} \left( \frac{1}{\cos x} + \cos x \right)^2 \, dx$

2. Expand the squared expression and apply trigonometric identities.

$\left( \sec x + \cos x \right)^2 = \sec^2 x + 2\sec x \cos x + \cos^2 x$
$= \sec^2 x + 2 + \cos^2 x$
Using the double angle formula $\cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x$:
$V = \pi \int_{0}^{\frac{\pi}{3}} \left( \sec^2 x + \frac{5}{2} + \frac{1}{2}\cos 2x \right) \, dx$

3. Integrate the expression with respect to $x$.

$\int \left( \sec^2 x + \frac{5}{2} + \frac{1}{2}\cos 2x \right) \, dx = \tan x + \frac{5}{2}x + \frac{1}{4}\sin 2x$

4. Evaluate the definite integral using exact trigonometric values.

Evaluate from $x = 0$ to $x = \frac{\pi}{3}$:
At $x = \frac{\pi}{3}$:
$\tan\left(\frac{\pi}{3}\right) + \frac{5}{2}\left(\frac{\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) = \sqrt{3} + \frac{5\pi}{6} + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)$
$= \sqrt{3} + \frac{5\pi}{6} + \frac{\sqrt{3}}{8} = \frac{9\sqrt{3}}{8} + \frac{5\pi}{6}$
At $x = 0$:
$\tan(0) + \frac{5}{2}(0) + \frac{1}{4}\sin(0) = 0$
$V = \pi \left( \frac{9\sqrt{3}}{8} + \frac{5\pi}{6} \right) - 0 = \pi \left( \frac{9\sqrt{3}}{8} + \frac{5\pi}{6} \right)$