Evaluate the given integral by changing to polar coordinates.
$$
\iint_{\!R}\,ye^{x}\,dA ,
$$ 
where R is the region in the first quadrant enclosed by the circle x2 + y2 = 4.
The value of integral is 


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

Evaluate the given integral by changing to polar coordinates.
$$
\iint_{R} y e^x \, dA ,
$$
where R is the region in the first quadrant enclosed by the circle x^2 + y^2 = 4. The value of integral is

Learn how to evaluate the double integral $ \iint_{R} y e^x \, dA $ over a region in the first quadrant enclosed by a circle using polar coordinates. This problem involves multivariable calculus concepts, including polar coordinates, double integrals, and the transformation of areas. Ideal for undergraduate calculus students.

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