1. **Problem:** Evaluate the integral $$\iint_D \frac{xy}{\sqrt{x^2 + y^2}}\,dxdy$$ over the region $$D = \{(x,y) : x^2 + y^2 \leq 1, xy \leq 0\}$$. 2. **Step 1: Understand the region and convert to polar coordinates.** - The region $$D$$ is the unit disk $$x^2 + y^2 \leq 1$$. - The condition $$xy \leq 0$$ means $$x$$ and $$y$$ have opposite signs or one is zero. - In polar coordinates: $$x = r\cos\theta$$, $$y = r\sin\theta$$, with $$r \in [0,1]$$. - The condition $$xy \leq 0$$ becomes $$r^2 \cos\theta \sin\theta \leq 0$$, or $$\cos\theta \sin\theta \leq 0$$. 3. **Step 2: Determine the angular bounds.** - $$\cos\theta \sin\theta \leq 0$$ means $$\sin(2\theta) \leq 0$$ since $$\sin(2\theta) = 2\sin\theta\cos\theta$$. - $$\sin(2\theta) \leq 0$$ on intervals where $$2\theta \in [\pi, 2\pi]$$ and $$[3\pi, 4\pi]$$ modulo $$2\pi$$. - For $$\theta \in [0, 2\pi]$$, this corresponds to $$\theta \in [\frac{\pi}{2}, \pi] \cup [\frac{3\pi}{2}, 2\pi]$$. 4. **Step 3: Express the integral in polar coordinates.** - The integrand: $$\frac{xy}{\sqrt{x^2 + y^2}} = \frac{(r\cos\theta)(r\sin\theta)}{r} = r\cos\theta\sin\theta$$. - The area element: $$dxdy = r dr d\theta$$. - Integral becomes: $$\int_{\theta=\frac{\pi}{2}}^{\pi} \int_0^1 r \cos\theta \sin\theta \cdot r dr d\theta + \int_{\theta=\frac{3\pi}{2}}^{2\pi} \int_0^1 r \cos\theta \sin\theta \cdot r dr d\theta$$ $$= \int_{\frac{\pi}{2}}^{\pi} \cos\theta \sin\theta \int_0^1 r^2 dr d\theta + \int_{\frac{3\pi}{2}}^{2\pi} \cos\theta \sin\theta \int_0^1 r^2 dr d\theta$$ 5. **Step 4: Evaluate the inner integral over $$r$$.** - $$\int_0^1 r^2 dr = \left[ \frac{r^3}{3} \right]_0^1 = \frac{1}{3}$$. 6. **Step 5: Integral reduces to:** $$\frac{1}{3} \left( \int_{\frac{\pi}{2}}^{\pi} \cos\theta \sin\theta d\theta + \int_{\frac{3\pi}{2}}^{2\pi} \cos\theta \sin\theta d\theta \right)$$ 7. **Step 6: Use substitution for angular integrals.** - Note: $$\cos\theta \sin\theta = \frac{1}{2} \sin(2\theta)$$. - So integral becomes: $$\frac{1}{3} \cdot \frac{1}{2} \left( \int_{\frac{\pi}{2}}^{\pi} \sin(2\theta) d\theta + \int_{\frac{3\pi}{2}}^{2\pi} \sin(2\theta) d\theta \right) = \frac{1}{6} \left( I_1 + I_2 \right)$$ 8. **Step 7: Evaluate $$I_1$$ and $$I_2$$:** - $$I_1 = \int_{\frac{\pi}{2}}^{\pi} \sin(2\theta) d\theta = \left[-\frac{\cos(2\theta)}{2}\right]_{\frac{\pi}{2}}^{\pi} = -\frac{\cos(2\pi) - \cos(\pi)}{2} = -\frac{1 - (-1)}{2} = -1$$ - $$I_2 = \int_{\frac{3\pi}{2}}^{2\pi} \sin(2\theta) d\theta = \left[-\frac{\cos(2\theta)}{2}\right]_{\frac{3\pi}{2}}^{2\pi} = -\frac{\cos(4\pi) - \cos(3\pi)}{2} = -\frac{1 - (-1)}{2} = -1$$ 9. **Step 8: Sum and final evaluation:** - $$I_1 + I_2 = -1 + (-1) = -2$$ - Integral value: $$\frac{1}{6} \times (-2) = -\frac{1}{3}$$ **Final answer:** $$\boxed{-\frac{1}{3}}$$