1. **Problem statement:** Calculate the double integral $$\iint \frac{xy}{\sqrt{x^2 + y^2}} dA$$ over the region defined by $$x^2 + y^2 = 1$$ (the unit disk). 2. **Formula and approach:** Since the region is a circle, it is convenient to use polar coordinates where $$x = r\cos\theta$$, $$y = r\sin\theta$$, and $$dA = r dr d\theta$$. 3. **Rewrite the integrand in polar coordinates:** $$\frac{xy}{\sqrt{x^2 + y^2}} = \frac{(r\cos\theta)(r\sin\theta)}{r} = r \cos\theta \sin\theta$$ 4. **Set up the integral in polar coordinates:** $$\int_0^{2\pi} \int_0^1 r \cos\theta \sin\theta \cdot r dr d\theta = \int_0^{2\pi} \int_0^1 r^2 \cos\theta \sin\theta dr d\theta$$ 5. **Separate the integrals:** $$\left( \int_0^{2\pi} \cos\theta \sin\theta d\theta \right) \left( \int_0^1 r^2 dr \right)$$ 6. **Evaluate the radial integral:** $$\int_0^1 r^2 dr = \left[ \frac{r^3}{3} \right]_0^1 = \frac{1}{3}$$ 7. **Evaluate the angular integral:** Use the identity $$\sin(2\theta) = 2 \sin\theta \cos\theta$$, so $$\int_0^{2\pi} \cos\theta \sin\theta d\theta = \frac{1}{2} \int_0^{2\pi} \sin(2\theta) d\theta = \frac{1}{2} \left[ -\frac{\cos(2\theta)}{2} \right]_0^{2\pi} = \frac{1}{2} \cdot 0 = 0$$ 8. **Multiply results:** $$0 \times \frac{1}{3} = 0$$ **Final answer:** $$\boxed{0}$$