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