## Calculus: Early Transcendentals 8th Edition

$V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx$
{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 $V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx$