Answer
$$\dfrac{8\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume =(4) \times \int^{\pi/2}_0 \int^1_0 \int^{4-4r^2}_{r^4-1} \space dz \space r dr \space d\theta \\=(4) \times \int^{\pi/2}_0 \int^1_0 (5r-4r^3-r^5)\space dr \space d\theta \\=(4) \int^{\pi/2}_0 (\dfrac{5}{2}-1-\dfrac{1}{6}) \times d\theta \\=(4) \int^{\pi/2}_0 d\theta \\=\dfrac{8\pi}{3}$$
