Answer
$$\pi(6\sqrt{2}-8)$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^1_0 \int^{(2-r^2)^{-1/2}}_r 3 \space dz \space r dr \space d\theta =3 \times \int^{2\pi}_0 \int^1_0 [r \times (2-r^2)^{-(1/2)}-r^2] \space dr \space d\theta \\=3 \times \int^{2\pi}_0[-(2-r^2)^{1/2}-\dfrac{1}{3} \times r^3]^1_0 \space d\theta \\=3 \times \int^{2\pi}_0 (\sqrt{2}-\dfrac{4}{3}) d\theta \\=\pi(6\sqrt{2}-8)$$
