Answer
$\dfrac{27}{8}$
Work Step by Step
Consider $I=\iiint_E z dV$
We need to use polar co-ordinates.
$I= \int_{0}^{ \pi/2}\int_0^3 \int_0^{1/3} z dx r dr d\theta$
or, $=\int_{0}^{ \pi/2}\int_0^3 (z) [x]_0^{1/3} r dr d\theta$
or, $=\int_{0}^{ \pi/2}\int_0^3 r \sin \theta[(1/3) r \cos \theta]r dr d\theta$
or, $=\int_{0}^{ \pi/2}\int_0^3 (r^3/3) \sin \theta \cos \theta dr d\theta$
or, $=(1/12) \int_{0}^{ \pi/2}[r^4]_0^3 \sin \theta \cos \theta d\theta$
or, $=\dfrac{81}{12} \int_{0}^{ \pi/2} \sin \theta \cos \theta d\theta$
Plug $a=\sin \theta \implies d\theta=da/\cos \theta$
Thus, we have
$\dfrac{81}{12} \int_{0}^{ \pi/2} \sin \theta \cos \theta d\theta=\dfrac{81}{12} \int_{0}^{ \pi/2} a da$
or, $=\dfrac{27}{8}$
