Answer
$\dfrac{15}{4} (2 -\sqrt 2) \pi$
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, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
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 $I=\int_0^{\pi/4} \int_{0}^{2\pi}\int_{1}^{2} \rho^3 \sin \phi d\rho d\theta d\phi$
$=\int_0^{\pi/4} \sin \phi d\phi \cdot \int_{0}^{2\pi} d\theta \cdot \int_{1}^{2} \rho^3 d\rho \\=[-\cos\phi]_0^{\pi/4} [ \theta]{0}^{2\pi} [\rho^4/4]_1^2 \\ =(-\cos (\pi/4)+\cos (0)) (2 \pi-0) (\dfrac{2^4}{4}-\dfrac{1}{4}] $
Hence, $I=(\dfrac{-\sqrt 2}{2}+1) \times 2 \pi \times (\dfrac{15}{4})=\dfrac{15}{4} (2 -\sqrt 2) \pi$
