Answer
$\dfrac{8}{9} (3\pi-4)$
Work Step by Step
We evaluate the double integral as follows:
$ \iint{R} \sqrt {x^2+y^2} \ dA=\int^{0}_{\pi} \int_{2 \ sin\theta}^2{2} \ r \ r dr \ d\theta$
or, $=\int_{0}^{\pi} \int_{2 \ sin\theta}^2 (r^2 dr) \ d\theta $
or, $=\int_{0}^{\pi} [\dfrac{r^3}{3}]_{2 \sin\theta}^2 \space d\theta $
or, $=\int_{0}^{\pi} [\dfrac{8}{3}-\dfrac{8}{3} \sin^3 \theta] \ d\theta $
or, $=(\dfrac{8}{3})\int_{0}^{\pi} (1- \sin^3 \theta) \ d \theta $
or, $=(\dfrac{8}{3}) [\theta +\cos \theta -\dfrac{1 }{3} \times \cos^3 \theta]_0^{\pi}$
or, $=\dfrac{8}{3}\times (\pi-\dfrac{4}{3})$
or, $=\dfrac{8}{9} (3\pi-4)$
