Answer
$$\dfrac{486}{5}$$
Work Step by Step
Set $x= r \cos \theta ; y=r \sin \theta $
Consider $I=\int_{-\pi/2}^{ \pi/2} \int_0^{3} (r^3 \cos^3 \theta+r^3 \cos \theta \sin^2 \theta) r dr d \theta=\int_{-\pi/2}^{ \pi/2} \int_0^{3} r^4 \ \cos \theta dr \ d \theta$
and, $I=\int_{-\pi/2}^{ \pi/2} \cos \theta dr d \theta \int_0^{3} r^4 dr= \dfrac{1}{5} \times [\sin \theta ]_{-\pi/2}^{ \pi/2}[r^5]_0^3$
$\implies=2 \times \dfrac{243}{5}$
$\implies =\dfrac{486}{5}$
