Answer
$$V = \pi $$
Work Step by Step
$$\eqalign{
& y = \sqrt {{{\cos }^{ - 1}}x} ,{\text{ }}x = 0{\text{ and }}y = 0 \cr
& y = \sqrt {{{\cos }^{ - 1}}x} \to {\cos ^{ - 1}}x = {y^2},{\text{ }}x = \cos \left( {{y^2}} \right) \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& \cos \left( {{y^2}} \right) > 0{\text{ on the interval }}\left[ {0,\sqrt {\pi /2} } \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^{\sqrt {\pi /2} } {2\pi y\left[ {\cos \left( {{y^2}} \right)} \right]} dy \cr
& V = \pi \int_0^{\sqrt {\pi /2} } {2y\cos \left( {{y^2}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\sin \left( {{y^2}} \right)} \right]_0^{\sqrt {\pi /2} } \cr
& V = \pi \left[ {\sin {{\left( {\sqrt {\frac{\pi }{2}} } \right)}^2} - \sin {{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( 0 \right)} \right] \cr
& V = - \pi \left[ {1 - 0} \right] \cr
& V = \pi \cr} $$
