Answer
$\dfrac{8\pi}{3}+\dfrac{128}{15}$
Work Step by Step
$I=\int_0^{(\pi/2)} \int_{0}^{2}\int_{0}^{4-r^2} (r \cos \theta+r \sin \theta+z) \times (r dr dz d\theta)$
$=\int_0^{(\pi/2)} \int_{0}^{2}\int_{0}^{4-r^2}[r^2( \cos \theta+ \sin \theta+rz) dz dr d\theta$
$=\int_0^{(\pi/2)} \int_{0}^{2}[r^2( \cos \theta+ \sin \theta)\times (z)+\dfrac{z^2 r}{2}]_{0}^{(4-r^2)} dr d\theta$
$=\int_0^{(\pi/2)} (\dfrac{64}{15}) \times (\cos \theta+ \sin \theta)+(\dfrac{16}{3}) d\theta$
$=\dfrac{8\pi}{3}+\dfrac{128}{15}$
