Answer
$\dfrac{64 \pi}{9}$
Work Step by Step
We have: $I=\int_{-2}^{2}\int_{0} ^{\sqrt {4-y^2}}\int_{-\sqrt {4-x^2-y^2}} ^{\sqrt {4-x^2-y^2}} y^2 \sqrt {x^2+y^2+z^2} dz dy dx$
Apply spherical coordinates: $x =\rho \sin \phi \cos \theta; y =\rho \sin \phi \sin \theta$ and $\rho^2 =x^2+y^2+z^2$
Now, $I=\int_{0}^{\pi} \sin^3 \phi \ d\phi \times \int_{-\pi/2} ^{\pi/2 }\sin^2 \theta d \theta \times \int_0^2 \rho^6 d\rho =\dfrac{4}{3} \times [\dfrac{-\sin 2 \theta -2 \theta}{4}|_{-\pi/2} ^{\pi/2 }\times [ \dfrac{ \rho^6 }{6}|_0^2 $
and, $I=(\dfrac{4}{3} )( \dfrac{\pi}{2})( \dfrac{32}{3})=\dfrac{64 \pi}{9}$
