Answer
$V = \left( {\sqrt 2 - 1} \right)\pi $
Work Step by Step
$$\eqalign{
& x = 0,{\text{ }}y = \cos {x^2},{\text{ }}y = \sin {x^2},\;{\text{ }}x = \frac{{\sqrt \pi }}{2} \cr
& {\text{Using the shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& \cos {x^2} \geqslant \sin {x^2}{\text{ on the interval }}\left[ {0,\frac{{\sqrt \pi }}{2}} \right]{\text{ }} \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^{\sqrt \pi /2} {2\pi x\left( {\cos {x^2} - \sin {x^2}} \right)} dx \cr
& V = \pi \int_0^{\sqrt \pi /2} {\left( {2x\cos {x^2} - 2x\sin {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\sin {x^2} + \cos {x^2}} \right]_0^{\sqrt \pi /2} \cr
& V = \pi \left[ {\sin {{\left( {\frac{{\sqrt \pi }}{2}} \right)}^2} + \cos {{\left( {\frac{{\sqrt \pi }}{2}} \right)}^2}} \right] - \pi \left[ {\sin {{\left( 0 \right)}^2} + \cos {{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left[ {\sin \left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4}} \right)} \right] - \pi \left[ 1 \right] \cr
& V = \pi \left( {\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}} \right) - \pi \cr
& V = \pi \left( {\sqrt 2 } \right) - \pi \cr
& V = \left( {\sqrt 2 - 1} \right)\pi \cr} $$
