Answer
$$\dfrac{\pi}{6}$$
Work Step by Step
$$Volume; V=\int_{0}^{2 \pi} \int_0^{1} \int_{r^2}{r} r dz d \theta$$
Consider $x= r \cos \theta ; y=r \sin \theta $
Now, $$Volume=\int_{0}^{2 \pi} d \theta \times \int_0^{1} r(r-r^2) dr \\=2 \pi (\dfrac{1}{3}-\dfrac{1}{4}) \\ =\dfrac{2\pi}{3}-\dfrac{2 \pi}{4}=\\=\dfrac{\pi}{6}$$
