Answer
$$0$$
Work Step by Step
The spherical coordinates system can be written as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta; z=\rho \cos \phi$;
and $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
Now, we will write the jacobian for spherical coordinates as $\phi^2 \sin \phi$.
Thus,
$\int_{-a}^{a} \int_{-\sqrt {a^2-y^2}} \int_{-\sqrt {a^2 -x^2-y^2}}^{\sqrt {a^2-x^2-y^2}} (x^2 z+y^2 z+z^3) \ dz dx dy =\int_0^{2 \pi} \int_0^{\pi} \int_0^{a} (\rho \cos \phi \rho^2 \rho^2 \sin \phi d\rho d \phi d \theta \\=\int_0^{2 \pi} d \theta \int_0^{\pi} \cos \phi \sin \phi d \phi \int_0^a \rho^5 d\rho$
Now, set $ \cos \theta =u$ and $ du=-\sin \theta $
Thus,
$E=2 \pi -\int_0^{\pi} u du [\rho^6/6]_0^u =2 \pi(\dfrac{\cos^2 \pi}{2} -\cos^2 0) \times \dfrac{a^6}{6}=0$
