I derived

$\displaystyle \int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(2x^2+2y^2+2z^2)^{-1/2}\,dz\,dy\,dx$

$\displaystyle =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 (2\rho^2)^{-\frac{1}{2}}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$

$\displaystyle =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho^{-1}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$

$\displaystyle =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho\,\sin \phi\,d\rho\,d\theta\,d\phi$

$\displaystyle =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{2\sqrt{2}}\sin \phi\,d\theta\,d\phi$

$\displaystyle =\int_0^{\frac{\pi}{2}} \frac{\pi}{4\sqrt{2}}\sin \phi\,d\phi$

$\displaystyle =\left(-\frac{\pi}{4\sqrt{2}}\cos \phi\right)_0^{\frac{\pi}{2}}$

$\displaystyle =-\frac{\pi}{4\sqrt{2}}\cos \left(\frac{\pi}{2}\right)-\left(-\frac{\pi}{4\sqrt{2}}\cos 0\right)$

$\displaystyle =0-\left(-\frac{\pi}{4\sqrt{2}}\right)=\frac{\pi}{4\sqrt{2}} .$