To evaluate the double integral $\iint_{R} (x+y) \, dA$ where $R$ is the region inside $x^2 + y^2 = 4$ in the first quadrant, we can proceed asTo evaluate the double integral

$\iint_{R} (x+y) \, dA$

where $R$ is the region inside $x^2 + y^2 = 4$ in the first quadrant, we can proceed as follows:

1. Set up the region in polar coordinates: The given region is the quarter circle of radius 2 in the first quadrant. In polar coordinates, $x = r\cos\theta$ and ( y =To evaluate the double integral

$\iint_{R} (x+y) \, dA$

where $R$ is the region inside $x^2 + y^2 = 4$ in the first quadrant, we can proceed as follows:

1. Convert to polar coordinates: The region $R$ is a quarter circle of radius 2 in the first quadrant. In polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$, and the differential area element $dA = r \, dr \, d\theta$.

2. Set up the limits of integration: In the first quadrant, $\theta$ ranges from $0$ to $\frac{\pi}{2}$, and $r$ ranges from $0$ to $2$.

3. Transform the integrand: Substitute $x$ and $y$ with their polar coordinates expressions: $x + y = r \cos \theta + r \sin \theta = r (\cos \theta + \sin \theta)$

4. Set up the integral in polar coordinates: $\iint_{R} (x + y) \, dA = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r (\cos \theta + \sin \theta) \cdot r \, dr \, d\theta$ $= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^2 (\cos \theta + \sin \theta) \, dr \, d\theta$

5. Evaluate the inner integral: $\int_{0}^{2} r^2 (\cos \theta + \sin \theta) \, dr = (\cos \theta + \sin \theta) \int_{0}^{2} r^2 \, dr$ $= (\cos \theta + \sin \theta) \left[ \frac{r^3}{3} \right]_{0}^{2}$ $= (\cos \theta + \sin \theta) \left[ \frac{8}{3} \right]$ $= \frac{8}{3} (\cos \theta + \sin \theta)$

6. Evaluate the outer integral: $\int_{0}^{\frac{\pi}{2}} \frac{8}{3} (\cos \theta + \sin \theta) \, d\theta$ $= \frac{8}{3} \left[ \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta + \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta \right]$ $= \frac{8}{3} \left[ \sin \theta \Big|_{0}^{\frac{\pi}{2}} - \cos \theta \Big|_{0}^{\frac{\pi}{2}} \right]$ $= \frac{8}{3} \left[ \sin \left( \frac{\pi}{2} \right) - \sin (0) + \cos (0) - \cos \left( \frac{\pi}{2} \right) \right]$ $= \frac{8}{3} \left[ 1 - 0 + 1 - 0 \right]$ $= \frac{8}{3} \times 2 = \frac{16}{3}$

Therefore, the value of the double integral is

$\boxed{\frac{16}{3}}$

Would you like more details or have any questions about this solution?

Here are 8 additional questions to expand on this topic:

1. How do you convert rectangular coordinates to polar coordinates?
2. What is the geometric interpretation of the given region $R$?
3. How does the Jacobian determinant factor into converting to polar coordinates?
4. What are other common applications of double integrals in polar coordinates?
5. How would the integral change if the region was the full circle instead of the quarter circle?
6. Can this problem be solved using symmetry arguments? How?
7. What is the physical interpretation of the integral $\iint_{R} (x+y) \, dA$?
8. How does changing the limits of integration affect the outcome of the integral?

Tip: When converting integrals to polar coordinates, always remember to include the Jacobian determinant $r$ when transforming the area element $dA$.