Answer
$\text{The volume is}$
\begin{align}
V = \sqrt 3 -1
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
y = \sec x; x = \frac{\pi}{4} \ and \ x = \frac{\pi}{3}
\end{align}
$\text{To find the volume, we have to find the area of the cross section.}$
$\text{It is given that the cross section is squre, therefore:}$
\begin{align}
A(x) = y^2 = \sec^2 x
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sec^2 x \ dx = \left[\tan x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \sqrt 3 -1
\end{align}