Answer
$\dfrac{8\sqrt 2 \pi}{3}$
Work Step by Step
The spherical coordinates system can be expressed as follows:
$x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$;
and $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
Since, $x^2+y^2+z^2 =4 \implies (\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2 + (\rho \cos \phi)^2 =4 $ and $\rho=2$
Also, the jacobian for spherical coordinates can be expressed as: $\rho^2 \sin \rho$, therefore, $\iiint{E} \sqrt {x^2+y^2+z^2} dV =\iiint_{E} \rho \rho^2 \sin \rho d\rho d\theta d\phi$
$E= \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{2} \rho^2 \sin \phi d\rho d\theta d\phi$
or, $=\int_0^{2 \pi} \int_{\pi/4}^{\pi/2} [\dfrac{\rho^3}{3}]_0^2 d\theta $
or, $=\dfrac{8}{3} \int_0^{2 \pi} [-\cos\phi]_{\pi/4}^{\pi/2} d \theta]$
or, $=\dfrac{8}{3} \int_0^{2 \pi} [\dfrac{4\sqrt 2}{3} d \theta$
or, $=\dfrac{8\sqrt 2 \pi}{3}$
