Answer
$$\dfrac{4\pi(2\sqrt{2}-1)}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume=8\int^{\pi/2}_0 \int^\sqrt{2}_1 \int^r_0 dz r dr \space d\theta \\=(8) \int^{\pi/2}_0 \int^{\sqrt{2}}_1 r^2 \space dr \space d\theta\\=(8) \times (\dfrac{2\sqrt{2}-1}{3}) \times \int^{\pi/2}_0 (1) d\theta \\=\dfrac{4\pi(2\sqrt{2}-1)}{3}$$
