ye^x dA, where R is the region in the first quadrant enclosed by the circle x^2 + y^2 = 36

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:
\[
\frac{e^{r\cos\theta}}{(\cos\theta)^3} \left( (2r\cos\theta - r^2\cos^2\theta - 6)\right)
\]

This problem involves evaluating a double integral in polar coordinates for the function ye^x, over the region in the first quadrant enclosed by the circle x^2 + y^2 = 36. Detailed steps cover the conversion to polar coordinates and solving the integral. Suitable for college-level calculus students.

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