We are tasked with evaluating the integral $\int \int_R ye^x \, dA$, where $R$ is the region in the first quadrant enclosed by the circle $x^2 + y^2 = 36$.

Step 1: Converting to Polar Coordinates

The given region is a quarter of a circle centered at the origin with a radius of 6 (since $\sqrt{36} = 6$).

In polar coordinates:

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

The equation of the circle $x^2 + y^2 = 36$ in polar coordinates becomes: $r^2 = 36 \quad \text{or} \quad r = 6$

The region $R$ in the first quadrant is where $\theta$ ranges from $0$ to $\frac{\pi}{2}$ and $r$ ranges from $0$ to $6$.

Step 2: Rewriting the Integral

We substitute the expressions for $x$ and $y$ in polar coordinates into the integral: $\int_0^{\frac{\pi}{2}} \int_0^6 \left( r\sin\theta \cdot e^{r\cos\theta} \right) r \, dr \, d\theta$ This simplifies to: $\int_0^{\frac{\pi}{2}} \int_0^6 r^2 \sin\theta e^{r\cos\theta} \, dr \, d\theta$

Step 3: Solving the Integral

Inner Integral (over $r$):

We first focus on the inner integral: $\int_0^6 r^2 e^{r\cos\theta} \, dr$ This is a standard integral of the form $r^n e^{kr}$, and we solve it using integration by parts or a formula. The result is: $\frac{e^{r\cos\theta}}{(\cos\theta)^3} \left( (2r\cos\theta - r^2\cos^2\theta - 6)\right)$