Answer
$\dfrac{15 \pi}{2}$
Work Step by Step
We write the region in the polar co-ordinates as: $1 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$
$\iint_{D} f(x,y) \ dA=\int_0^{2\pi} \int_0^{\sqrt 2} (r) (r) \ dr \ d\theta$
Now, we have:
$\int_0^{2\pi} \int_1^{2} (r^2) (r) \ dr \ d\theta=\int_0^{2\pi} [\dfrac{r^4}{4}]_1^{2} d\theta$
or, $=\int_0^{2\pi} [\dfrac{2^4}{4}-\dfrac{1}{4}] d\theta$
or. $=[\dfrac{15}{4} \theta]_0^{2\pi}$
or. $=\dfrac{15 \pi}{2}$
