Answer
$\dfrac{\pi (9 \pi -16)}{2}$
Work Step by Step
The area of each cross-section triangle is given as: $A=\pi r^2 =\pi (2-\sin x)^2 $
We integrate the integral to calculate the volume as follows:
$V= \pi \int_{0}^{\pi/3} (2-\sin x)^2 dx$
or, $=\pi \int_{0}^{\pi}(4-4 \sin x +\dfrac{1-\cos 2x}{2} ) dx$
or, $=\pi [4x+4 \cos x +\dfrac{x}{2}-\dfrac{\sin 2x}{4}]_{0}^{\pi}$
or, $=\dfrac{\pi (9 \pi -16)}{2}$
