Answer
$\dfrac{9\pi}{2}$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq 3$ and $0 \leq \theta \leq \dfrac{\pi}{2}$
$\iint_{D} f(x,y) \ dA=\int_0^{\frac{\pi}{2}} \int_0^{3} (r) (r) \ dr \ d\theta$
Now, we have:
$\int_0^{\frac{\pi}{2}} \int_0^{3} (r) (r) \ dr \ d\theta=\int_0^{\pi/2} [\dfrac{r^3}{3}]_0^{3} d\theta$
or. $=9 \times [\theta]_0^{\pi/2}$
or. $=9[\dfrac{\pi}{2}-0]$
or, $=\dfrac{9\pi}{2}$
