Answer
$\dfrac{8\sqrt 2 \pi}{3}$
Work Step by Step
Apply the spherical coordinates system as:
$x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$;
So, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
Since,
$x^2+y^2+z^2 =4 \implies (\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2 + (\rho \cos \phi)^2 =4 \implies \rho=2$
The jacobian for spherical coordinates is $\rho^2 \sin \rho$; therefore,
$\iiint{E} \sqrt {x^2+y^2+z^2} dV =\iiint_{E} \rho \rho^2 \sin \rho d\rho d\theta d\phi$
Now, we will consider
$I= \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{2} \rho^2 \sin \phi d\rho d\theta d\phi$
$=\int_0^{2 \pi} \int_{\pi/4}^{\pi/2} [\dfrac{\rho^3}{3}]_0^2 d\theta \\=\dfrac{8}{3} \int_0^{2 \pi} [-\cos\phi]_{\pi/4}^{\pi/2} d \theta] \\ =\dfrac{8}{3} \int_0^{2 \pi} [\dfrac{4\sqrt 2}{3} d \theta$
Hence, $I=\dfrac{8\sqrt 2 \pi}{3}$
