Answer
$\dfrac{4096 \pi}{21}$
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 $\phi^2 \sin \phi$
Therefore,
$\int_0^{2 \pi} \int_0^{\pi/2} \int_0^{4 \cos \phi} \rho^3 i \rho^2 \rho^2 \sin \phi d\rho d \phi d \theta=\int_0^{2 \pi} \int_0^{\pi/2} [ \rho^6/6]_0^{4 \cos \phi} \sin \phi d\rho d \phi d \ \theta$
Now, set $ \cos \phi =a$
Thus,
$E=\dfrac{4^6}{6} \times 2 \pi (\dfrac{-\cos^7 \pi}{7})]_0^{\pi/2} = \dfrac{4096 \pi}{21}$