Answer
$ V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx $
Work Step by Step
{Step 1 of 3}
Consider the functions
$ y=cos^{2}x, \vert x \vert \leq \frac{ \pi }{2},y=\frac{1}{4} $ is shown below
{Step 2 of 3}
The shaded region p when rotated about $ x=\frac{ \pi }{2} $ will have a radius $ \left( \frac{ \pi }{2}-x \right) $ and its height is $ \left( cos^{2}x-\frac{1}{4} \right) $,like this the solid will form and will integrate it from $ x=-\frac{ \pi }{3} $ to $ x=\frac{ \pi }{3} $
{Step 3 of 3}
Since the solid lies between $ x=-\frac{ \pi }{3} $ and $ x=\frac{ \pi }{3} $
, its integral for the volume is
$ V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx $
