Answer
$\dfrac{\pi (3 \sqrt 3-\pi)}{3}$
Work Step by Step
The area of each cross-section triangle is given as: $A=\pi r^2 =\pi \tan^2 x $
We integrate the integral to calculate the volume as follows:
$V= \pi \int_{0}^{\pi/3}(\tan^2 x) dx$
or, $=\pi \int_{0}^{\pi/3}(\sec^2 x-1) dx$
or, $=\dfrac{\pi (3 \sqrt 3-\pi)}{3}$
