1. **State the problem:** Find the average value of the function $$F(x,y) = \frac{x^2 + 2xy + y^2}{x^2 + y^2}$$ over the region in the first quadrant bounded by the coordinate axes and the line $$x + y = 2$$. 2. **Set up the region and integral:** The region is the triangle with vertices at $$(0,0), (2,0), (0,2)$$. 3. **Simplify the function:** Note that $$x^2 + 2xy + y^2 = (x + y)^2$$, so $$F(x,y) = \frac{(x + y)^2}{x^2 + y^2}.$$ 4. **Find the area of the region:** The triangle area is $$\text{Area} = \frac{1}{2} \times 2 \times 2 = 2.$$ 5. **Set up the average value formula:** $$\text{Average} = \frac{1}{\text{Area}} \iint_{R} F(x,y) \, dA = \frac{1}{2} \int_0^2 \int_0^{2 - x} \frac{(x + y)^2}{x^2 + y^2} \, dy \, dx.$$ 6. **Change to polar coordinates:** Let $$x = r \cos \theta$$, $$y = r \sin \theta$$, with $$r \geq 0$$, $$0 \leq \theta \leq \frac{\pi}{2}$$. The boundary $$x + y = 2$$ becomes $$r(\cos \theta + \sin \theta) = 2 \implies r = \frac{2}{\cos \theta + \sin \theta}.$$ 7. **Rewrite the function in polar form:** $$F(r, \theta) = \frac{(r \cos \theta + r \sin \theta)^2}{r^2} = \frac{r^2 (\cos \theta + \sin \theta)^2}{r^2} = (\cos \theta + \sin \theta)^2.$$ 8. **Set up the integral in polar coordinates:** $$\text{Average} = \frac{1}{2} \int_0^{\pi/2} \int_0^{\frac{2}{\cos \theta + \sin \theta}} (\cos \theta + \sin \theta)^2 r \, dr \, d\theta.$$ 9. **Integrate with respect to $$r$$:** $$\int_0^{\frac{2}{\cos \theta + \sin \theta}} r \, dr = \left[ \frac{r^2}{2} \right]_0^{\frac{2}{\cos \theta + \sin \theta}} = \frac{1}{2} \left( \frac{2}{\cos \theta + \sin \theta} \right)^2 = \frac{2}{(\cos \theta + \sin \theta)^2}.$$ 10. **Substitute back:** $$\text{Average} = \frac{1}{2} \int_0^{\pi/2} (\cos \theta + \sin \theta)^2 \times \frac{2}{(\cos \theta + \sin \theta)^2} \, d\theta = \frac{1}{2} \int_0^{\pi/2} 2 \, d\theta = \int_0^{\pi/2} 1 \, d\theta.$$ 11. **Integrate with respect to $$\theta$$:** $$\int_0^{\pi/2} 1 \, d\theta = \frac{\pi}{2}.$$ 12. **Final answer:** $$\boxed{\frac{\pi}{2}}.$$ This is the average value of $$F(x,y)$$ over the given region.