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