Answer
$ 2 \pi \int _{0}^{\frac{ \pi }{3}}x \left( tanx-x \right) dx $
Work Step by Step
{Step 1 of 2}
Consider the region enclosed by the graphs of $ y=tanx $, $ y=x $
and $ x=\frac{ \pi }{3}. $
{Step 2 of 2}
We use the method of cylindrical shells.A typical shell is shown above.The radius of the shell is r(x)=x and the height of the shell is $ h \left( x \right) =tanx-x $
. The volume V of the resulting solid of revolution is given by
$ V=2 \pi \int _{0}^{\frac{ \pi }{3}}r \left( x \right) h \left( x \right) dx $
$ =2 \pi \int _{0}^{\frac{ \pi }{3}}x \left( tanx-x \right) dx $
