(a) Differentiate in the interior: \[y\prime=-\sin x\sqrt{\sin2x}+\frac{\cos x\cos2x}{\sqrt{\sin2x}}.\] At a stationary point, \[-\sin a\sin2a+\cos a\cos2a=0\Rightarrow\cos3a=0.\] The sole interior solution is \(a=\pi/6\); \(a=\pi/2\) is an endpoint outside this derivative calculation. The endpoints have \(y=0\) and the curve is positive inside, so \[a=\frac\pi6.\]

(b) \[V=\pi\int_0^{\pi/2}y^2\,dx=\pi\int_0^{\pi/2}\cos^2x\sin2x\,dx=2\pi\int_0^{\pi/2}\cos^3x\sin x\,dx.\] Thus \[V=\left[-\frac\pi2\cos^4x\right]_0^{\pi/2}=\frac\pi2.\]