## Thomas' Calculus 13th Edition

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