1. **State the problem:** We want to evaluate the double integral $$\iint_R \sqrt{x^2 + y^2} \, dA$$ where the region $$R$$ is the disk defined by $$x^2 + y^2 \leq 4$$. 2. **Identify the region and function:** The region $$R$$ is a circle centered at the origin with radius 2. The function inside the integral is $$f(x,y) = \sqrt{x^2 + y^2}$$, which represents the distance from the origin. 3. **Use polar coordinates:** Since the region is circular and the function depends on $$\sqrt{x^2 + y^2}$$, polar coordinates simplify the integral. Recall the transformations: $$x = r \cos \theta, \quad y = r \sin \theta$$ $$dA = r \, dr \, d\theta$$ 4. **Rewrite the integral in polar coordinates:** The radius $$r$$ ranges from 0 to 2, and the angle $$\theta$$ ranges from 0 to $$2\pi$$. The function becomes $$\sqrt{r^2} = r$$. Thus, the integral is: $$\int_0^{2\pi} \int_0^2 r \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2 \, dr \, d\theta$$ 5. **Evaluate the inner integral:** $$\int_0^2 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^2 = \frac{2^3}{3} - 0 = \frac{8}{3}$$ 6. **Evaluate the outer integral:** $$\int_0^{2\pi} \frac{8}{3} \, d\theta = \frac{8}{3} \times \left[ \theta \right]_0^{2\pi} = \frac{8}{3} \times 2\pi = \frac{16\pi}{3}$$ 7. **Final answer:** $$\boxed{\frac{16\pi}{3}}$$