The problem involves a double integral over a region $D$ with the integrand:

$\iint_D \frac{x + y}{x^2 + y^2} \, dx \, dy$

The region $D$ is defined as:

$D = \{(x, y) \in \mathbb{R}^2 \, | \, y > 0 \text{ and } x^2 + y^2 \leq y - x\}.$

Step-by-step Analysis:

1. Integral Expression: The integrand is $\frac{x+y}{x^2+y^2}$, a rational function of $x$ and $y$.

2. Region $D$:

   • The condition $y > 0$ implies we are working in the upper half-plane.
   • The condition $x^2 + y^2 \leq y - x$ is a quadratic inequality. To analyze this:

     • Rewrite $y - x$ as a shifted circle: $x^2 + y^2 - y + x \leq 0.$

     • Complete the square for $x$ and $y$ terms:

       • For $y^2 - y$, rewrite as $(y - \frac{1}{2})^2 - \frac{1}{4}$,
       • For $x^2 + x$, rewrite as $(x + \frac{1}{2})^2 - \frac{1}{4}$.

     • The inequality becomes: $\left(x + \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 \leq \frac{1}{2}.$ This is a circle centered at $(-1/2, 1/2)$ with radius $\sqrt{1/2}$.

     • Combined with $y > 0$, the region $D$ is the upper half of this circle.

3. Simplify Using Polar Coordinates:

   • Switch to polar coordinates: $x = r \cos \theta, \quad y = r \sin \theta, \quad x^2 + y^2 = r^2.$
     • The integrand becomes: $\frac{x + y}{x^2 + y^2} \, dx \, dy \to \frac{r \cos \theta + r \sin \theta}{r^2} \cdot r \, dr \, d\theta = \frac{\cos \theta + \sin \theta}{r} \, dr \, d\theta.$
     • The region $D$ is now bounded by $0 \leq r \leq \sqrt{1/2}$ and $\theta$ corresponding to the upper semicircle (analyzing the geometry gives $-\pi/4 \leq \theta \leq 3\pi/4$).

4. Final Polar Integral: Substitute everything into the double integral: $\int_{-\pi/4}^{3\pi/4} \int_{0}^{\sqrt{1/2}} \frac{\cos \theta + \sin \theta}{r} \, r \, dr \, d\theta.$ Simplify: $\int_{-\pi/4}^{3\pi/4} \int_{0}^{\sqrt{1/2}} (\cos \theta + \sin \theta) \, dr \, d\theta.$

   • Compute $\int_{0}^{\sqrt{1/2}} 1 \, dr = \sqrt{1/2}$,
   • The remaining integral is: $\sqrt{1/2} \int_{-\pi/4}^{3\pi/4} (\cos \theta + \sin \theta) \, d\theta.$
   • Separate into two terms: $\sqrt{1/2} \left[\int_{-\pi/4}^{3\pi/4} \cos \theta \, d\theta + \int_{-\pi/4}^{3\pi/4} \sin \theta \, d\theta\right].$ The integrals of $\cos \theta$ and $\sin \theta$ can be computed directly.

Would you like me to compute the full solution explicitly? Let me know!

Follow-up Questions:

1. How does completing the square help define regions in integration problems?
2. Why is switching to polar coordinates useful for circular regions?
3. What are the bounds for $\theta$ in this problem, and why?
4. How do we interpret the geometry of the region $D$?
5. Could symmetry simplify this integral further?

Tip:

Always sketch the region $D$ for better visualization—it simplifies setting up the integral bounds.