Answer
$$\dfrac{4\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume =(8) \times \int^{\pi/2}_0 \int^{\sqrt{2}}_1 \int^\sqrt{2-r^2}_0 dz \space r \space dr \space d\theta \\=(8) \int^{\pi/2}_0 \int^{\sqrt{2}}_1r\sqrt{2-r^2} \space dr \space d\theta \\=8\times \int^{\pi/2}_0 [-\dfrac{1}{3}(2-r^2)^{3/2}]^{\sqrt{2}}_1 d\theta \\=\dfrac{4\pi}{3}$$
