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The problem shown in the image involves finding the area of a region defined by certain inequalities. The problem reads:

\[ R(x,y) = 2x + 2y = 2 \left(\sqrt{x^2 + y^2}\right) \text{ for } x, y \geq 0. \]
You are asked to find the area in polar coordinates.

Let's start by rewriting the equation and conditions in a more manageable form.

### Given:
- The equation is \( 2x + 2y = 2 \left(\sqrt{x^2 + y^2}\right) \).
- This equation simplifies to:
  \[
  x + y = \sqrt{x^2 + y^2}.
  \]

### Step 1: Simplify the Equation
We square both sides to eliminate the square root:
\[
(x + y)^2 = x^2 + y^2
\]
Expanding the left side:
\[
x^2 + 2xy + y^2 = x^2 + y^2
\]
Canceling out \(x^2 + y^2\) from both sides:
\[
2xy = 0
\]
This implies \(xy = 0\), meaning either \(x = 0\) or \(y = 0\).

### Step 2: Determine the Region in Polar Coordinates
The polar coordinates are given by:
\[
x = r \cos(\theta), \quad y = r \sin(\theta).
\]
Given \(xy = 0\), the region lies along the \(x\)-axis or the \(y\)-axis in the first quadrant (since \(x, y \geq 0\)).

This suggests that the region \(R(x, y)\) is bounded by the axes \(x = 0\) and \(y = 0\), which corresponds to the \(r\) axis from \(0\) to some maximum \(r\).

### Step 3: Area Calculation
The area of the region in polar coordinates can be calculated as:
\[
\text{Area} = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta
\]

Since \(r\) corresponds to the line \(x + y = r\) in the first quadrant, and considering \(x\) and \(y\) from 0 to 1 along the axes, the limits for \(\theta\) would be \(0\) to \(\frac{\pi}{2}\) and the radius \(r\) varies from 0 to 1.

### Final Area Calculation
\[
\text{Area} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left(\text{(Radius corresponding to \(\theta\))} \right)^2 d\theta
\]

Given the symmetry, the result simplifies, and upon solving, you find the exact area. However, based on this setup, without full integral evaluation, it can be understood that the problem describes a quarter of a circle of radius \(1\), so the area would be:

\[
\text{Area} = \frac{1}{4} \times \pi r^2 = \frac{\pi}{4}
\]

Would you like a more detailed breakdown of the steps, or have any specific questions? 

### Follow-Up Questions:
1. What would the area be if the region spanned the entire first quadrant?
2. How do the Cartesian and polar forms of the equation compare?
3. What would change if the inequality \( x + y \leq \sqrt{x^2 + y^2} \) were used?
4. How does symmetry play a role in simplifying this area calculation?
5. What other coordinate systems might simplify this kind of problem?

### Tip:
When converting problems involving symmetry or circular regions to polar coordinates, always consider how the boundaries translate and simplify the integrals by leveraging the symmetry of the problem.

Find the area of a region defined by the equation 2x + 2y = 2(sqrt(x^2 + y^2)) for x, y >= 0 in polar coordinates.

Learn how to calculate the area of a region defined by the equation 2x + 2y = 2(sqrt(x^2 + y^2)) for x, y >= 0 in polar coordinates. This solution involves using polar coordinate transformations, understanding integration, and leveraging geometric properties. Suitable for high school students in Grades 10-12.

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