We need to evaluate the double integral $\iint_S \frac{2}{1+x^2} dA$, where $S$ is the triangular region with vertices at $(0,0)$, $(2,2)$, and $(0,2)$.

1. Problem Description

We need to evaluate the double integral $\iint_S \frac{2}{1+x^2} dA$, where $S$ is the triangular region with vertices at $(0,0)$, $(2,2)$, and $(0,2)$.

2. Solution Steps

First, we need to find the equations of the lines that form the boundaries of the triangular region.

The vertices are $(0,0)$, $(2,2)$, and $(0,2)$.

The line connecting $(0,0)$ and $(2,2)$ is $y=x$.

The line connecting $(0,2)$ and $(2,2)$ is $y=2$.

The line connecting $(0,0)$ and $(0,2)$ is $x=0$.

We can describe the region $S$ as follows: $0 \le x \le 2$ and $x \le y \le 2$.

Now we can set up the double integral:

$\iint_S \frac{2}{1+x^2} dA = \int_0^2 \int_x^2 \frac{2}{1+x^2} dy dx$

First, integrate with respect to $y$:

$\int_x^2 \frac{2}{1+x^2} dy = \frac{2}{1+x^2} \int_x^2 dy = \frac{2}{1+x^2} [y]_x^2 = \frac{2}{1+x^2} (2-x)$

Now, integrate with respect to $x$:

$\int_0^2 \frac{2(2-x)}{1+x^2} dx = 4 \int_0^2 \frac{1}{1+x^2} dx - 2 \int_0^2 \frac{x}{1+x^2} dx$

The first integral is:

$\int_0^2 \frac{1}{1+x^2} dx = [\arctan(x)]_0^2 = \arctan(2) - \arctan(0) = \arctan(2)$

The second integral is:

$\int_0^2 \frac{x}{1+x^2} dx$. Let $u = 1+x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

When $x=0$, $u=1$. When $x=2$, $u=5$.

$\int_0^2 \frac{x}{1+x^2} dx = \int_1^5 \frac{1}{2u} du = \frac{1}{2} [\ln(u)]_1^5 = \frac{1}{2} (\ln(5) - \ln(1)) = \frac{1}{2} \ln(5)$

Putting it all together:

$4 \arctan(2) - 2(\frac{1}{2} \ln(5)) = 4 \arctan(2) - \ln(5)$

3. Final Answer

$4\arctan(2) - \ln(5)$