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