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 determine the bounds of integration. The region $S$ is a triangle.

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

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

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

So, the region $S$ can be described as $0 \le x \le y$ and $0 \le y \le 2$.

Therefore, the integral can be written as

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

Now, we evaluate the inner integral:

$\int_0^y \frac{2}{1+x^2} dx = 2 \arctan(x) \Big|_0^y = 2(\arctan(y) - \arctan(0)) = 2 \arctan(y)$.

Next, we evaluate the outer integral:

$\int_0^2 2 \arctan(y) dy$.

To evaluate this integral, we use integration by parts. Let $u = \arctan(y)$ and $dv = dy$. Then $du = \frac{1}{1+y^2} dy$ and $v = y$.

$\int 2 \arctan(y) dy = 2(y \arctan(y) - \int \frac{y}{1+y^2} dy)$.

Let $w = 1+y^2$. Then $dw = 2y dy$, so $y dy = \frac{1}{2} dw$.

$\int \frac{y}{1+y^2} dy = \int \frac{1}{w} \frac{1}{2} dw = \frac{1}{2} \ln|w| = \frac{1}{2} \ln(1+y^2)$.

Thus, $\int 2 \arctan(y) dy = 2y \arctan(y) - \ln(1+y^2)$.

Now, we evaluate the definite integral:

$\int_0^2 2 \arctan(y) dy = \Big[ 2y \arctan(y) - \ln(1+y^2) \Big]_0^2 = (4 \arctan(2) - \ln(5)) - (0 - \ln(1)) = 4 \arctan(2) - \ln(5)$.

3. Final Answer

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