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