Answer
$\dfrac{\pi}{4} \approx 0.2244$
Work Step by Step
Apply spherical coordinates: $x =\rho \sin \phi \cos \theta$ and $y =\rho \sin \phi \sin \theta$ and $\rho^2 =x^2+y^2+z^2$
Consider $I=\int_{-1}^{ 1}\int_{-\sqrt {1-x^2}} ^{-\sqrt {1-x^2}}\int_{0} ^{\sqrt {1-x^2-y^2} } z^3(x^2+y^2+z^2)^{1/2} \ dz \ dy \ dx$
Now, $I=\int_{0}^{\pi/2}\int_{0} ^{2 \pi} \int_0^1 \rho^6 \cos^3 \phi \sin \phi d\rho d\theta d\phi =\int_{0}^{\pi/2} \cos^3 \phi \sin \phi d\phi \int_{0} ^{2 \pi} d\theta \int_0^1 \rho^6 d\phi =\dfrac{1}{4}[\cos^4 \phi ]_0^{\pi/2} [\theta]_0^{2 \pi} \dfrac{1}{7}\times \rho^6 d\rho =\dfrac{\pi}{4} \approx 0.2244$
