To evaluate the integral

$\iint_{R} y e^x \, dA$

where $R$ is the region in the first quadrant enclosed by the circle $x^2 + y^2 = 4$, we can simplify it by changing to polar coordinates.

Step 1: Set up the integral in polar coordinates

In polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $dA = r \, dr \, d\theta$

The region $R$ is the first quadrant part of the circle $x^2 + y^2 = 4$, so:

• $0 \leq r \leq 2$
• $0 \leq \theta \leq \frac{\pi}{2}$

Thus, the integral becomes:

$\iint_{R} y e^x \, dA = \int_{0}^{\pi/2} \int_{0}^{2} (r \sin \theta) e^{r \cos \theta} \cdot r \, dr \, d\theta$

or

$= \int_{0}^{\pi/2} \int_{0}^{2} r^2 \sin \theta \, e^{r \cos \theta} \, dr \, d\theta$

Step 2: Separate the integrals

Since $\sin \theta$ does not depend on $r$, we can separate the integral:

$= \int_{0}^{\pi/2} \sin \theta \, d\theta \int_{0}^{2} r^2 e^{r \cos \theta} \, dr$

Step 3: Evaluate the $r$-integral

To solve $\int_{0}^{2} r^2 e^{r \cos \theta} \, dr$, we use integration by parts or recognize that it involves differentiation with respect to $\cos \theta$. However, solving this in closed form requires further steps with special functions.

For simplicity, we conclude here that