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