Answer
$\dfrac{8\pi}{3}+\dfrac{128}{15}$
Work Step by Step
The given integral is $I=\iiint_E(x+y+z)dV$
We know that the Conversion of rectangular to cylindrical coordinate system gives $r^2=x^2+y^2;tan \theta=\dfrac{y}{x} ; z=z$
and $x=r \cos \theta; y=r \sin \theta, z=z$
This implies that
$\iiint_E(x+y+z)dV=\int_0^{\pi/2} \int_{0}^{2}\int_{0}^{4-r^2} (x+y+z) r dr dz d\theta$
$=\int_0^{\pi/2} \int_{0}^{2}\int_{0}^{4-r^2} (r \cos \theta+r \sin \theta+z) r dr dz d\theta$
This implies that
$=\int_0^{\pi/2} \int_{0}^{2}\int_{0}^{4-r^2}[r^2( \cos \theta+ \sin \theta)+rz] dz dr d\theta$
$=\int_0^{(\pi/2)} \int_{0}^{2}[r^2( \cos \theta+ \sin \theta)(z)+\dfrac{1}{2}(rz^2)]_{0}^{4-r^2} drd\theta$
$=\int_0^{(\pi/2)}[\dfrac{64}{15}(\cos \theta+ \sin \theta)+\dfrac{16}{3}] d\theta$
Thus, we have $\iiint_E(x+y+z)dV=\dfrac{8\pi}{3}+\dfrac{128}{15}$
