Answer
$\dfrac{6\sqrt{2} \pi+40\sqrt{2}-64}{9}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Volume=4\int^{\pi/4}_0 \int^{\sqrt{2 \cos 2\theta}} r\sqrt{2-r^2} \space dr \space d\theta $
or, $=-\dfrac{4}{3} \times \int^{\pi/4}_0 [(2-2 \cos2\theta)^{3/2}-2^{3/2}] \space d\theta $
or, $=\dfrac{2\pi\sqrt{2}}{3} -\dfrac{32}{3} \times \int^{\pi/4}_0 (1-\cos^2\theta) \space \sin\theta \space d\theta $
or, $=\dfrac{2\pi\sqrt{2}}{3}-\dfrac{32}{3} [\dfrac{\cos^3\theta}{4}-\cos\theta]^{\pi/4}_0$
or, $=\dfrac{6\sqrt{2} \pi+40\sqrt{2}-64}{9}$