Answer
$\dfrac{\pi}{14} \approx 0.2244$
Work Step by Step
Use spherical coordinates: $x =\rho \sin \phi \cos \theta; y =\rho \sin \phi \sin \theta$ and $\rho^2 =x^2+y^2+z^2$
Let $V=\int_{-1}^{ 1}\int_{-\sqrt {1-x^2}} ^{-\sqrt {1-x^2}}\int_{0} ^{\sqrt {1-x^2-y^2} } z^3(x^2+y^2+z^2)^{1/2} dz dy dx \\
=\int_{0}^{\pi/2} \cos^3 \phi \sin \phi d\phi \int_{0} ^{2 \pi} d\theta \int_0^1 \rho^6 d\phi \\=\dfrac{1}{4}[\cos^4 \phi ]_0^{\pi/2} [\theta]_0^{2 \pi} [(\dfrac{1}{7}) \rho^6 d\rho] \\=\dfrac{\pi}{4} \\ \approx 0.2244$
