We need to evaluate the double integral $\iint_S y \, dA$, where $S$ is the region in the first quadrant that is inside the circle $x^2 + y^2 = 4$ and outside the circle $x^2 + y^2 = 1$. We are instructed to use polar coordinates.

1. Problem Description

We need to evaluate the double integral $\iint_S y \, dA$, where $S$ is the region in the first quadrant that is inside the circle $x^2 + y^2 = 4$ and outside the circle $x^2 + y^2 = 1$. We are instructed to use polar coordinates.

2. Solution Steps

First, we convert the equations to polar coordinates. We have $x = r\cos\theta$ and $y = r\sin\theta$.

The region $S$ is described by $1 \le x^2 + y^2 \le 4$, which translates to $1 \le r^2 \le 4$, or $1 \le r \le 2$. Since we are in the first quadrant, we have $0 \le \theta \le \pi/2$.

Also, $dA = r \, dr \, d\theta$, so $y \, dA = r\sin\theta \cdot r \, dr \, d\theta = r^2 \sin\theta \, dr \, d\theta$.

The integral becomes

$$\iint_S y \, dA = \int_0^{\pi/2} \int_1^2 r^2 \sin\theta \, dr \, d\theta = \int_0^{\pi/2} \sin\theta \left( \int_1^2 r^2 \, dr \right) d\theta$$

First, we evaluate the inner integral:

$$\int_1^2 r^2 \, dr = \left[ \frac{r^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$

Now, we evaluate the outer integral:

$$\int_0^{\pi/2} \sin\theta \left( \frac{7}{3} \right) d\theta = \frac{7}{3} \int_0^{\pi/2} \sin\theta \, d\theta = \frac{7}{3} \left[ -\cos\theta \right]_0^{\pi/2} = \frac{7}{3} \left( -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) \right) = \frac{7}{3} ( -0 - (-1) ) = \frac{7}{3} (1) = \frac{7}{3}$$

3. Final Answer

$\frac{7}{3}$