Mathematics

Problem

Consider the region that lies between

y=\sin x

and

y=\cos x

0\le x\le\dfrac{\pi}{4}

and between

y=\cos x

and the

x

-axis on

\dfrac{\pi}{4}\le x\le\dfrac{\pi}{2}

. When this region is revolved about the

x

-axis, what is the exact volume of the resulting solid? a)

\dfrac{\pi^2}{4}

\dfrac{\pi\sqrt{2}}{2}

\dfrac{\pi^2}{8}+\dfrac{\pi}{4}

\dfrac{\pi^2}{4}-\dfrac{\pi}{8}