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The given integral is:

\[
\int_0^2 \int_0^{\sqrt{4 - y^2}} (x^2 + y^2) \, dA
\]

This represents a double integral over a region in the first quadrant bounded by a circle of radius 2 centered at the origin. Let's break down the steps to solve it:

### Step 1: Identify the region
The integral is in Cartesian coordinates, where the inner integral (with respect to \( x \)) ranges from \( 0 \) to \( \sqrt{4 - y^2} \). This represents a semicircle of radius 2 in the first quadrant.

### Step 2: Change to Polar Coordinates
Since the region is circular, it's easier to switch to polar coordinates. The relationships are:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( dA = r \, dr \, d\theta \)

The expression \( x^2 + y^2 = r^2 \) in polar coordinates.

### Step 3: Set up the integral in polar coordinates
- \( r \) ranges from \( 0 \) to \( 2 \) (the radius of the circle).
- \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \) (first quadrant).

Thus, the integral becomes:

\[
\int_0^{\frac{\pi}{2}} \int_0^2 r^2 \cdot r \, dr \, d\theta = \int_0^{\frac{\pi}{2}} \int_0^2 r^3 \, dr \, d\theta
\]

### Step 4: Evaluate the inner integral
\[
\int_0^2 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} = 4
\]

### Step 5: Evaluate the outer integral
\[
\int_0^{\frac{\pi}{2}} 4 \, d\theta = 4\theta \Bigg|_0^{\frac{\pi}{2}} = 4 \times \frac{\pi}{2} = 2\pi
\]

### Final Answer
The value of the integral is:

\[
2\pi
\]

Would you like further details or have any questions?

### Related Questions:
1. What is the geometric interpretation of this integral in Cartesian coordinates?
2. How would the integral change if the region were in the entire circle rather than just the first quadrant?
3. How do polar coordinates simplify integration over circular regions?
4. Could this integral be computed using symmetry properties instead of changing coordinates?
5. How would this integral look if expressed in cylindrical or spherical coordinates?

### Tip:
When integrating over circular or spherical regions, switching to polar, cylindrical, or spherical coordinates often simplifies the problem significantly.

Evaluate the double integral over the region in the first quadrant bounded by a circle of radius 2 centered at the origin, where the integrand is (x^2 + y^2).

Explore the calculation of a double integral over a circular region using polar coordinates. This guide provides a step-by-step solution for the integral ∫∫ R (x^2 + y^2) dA where R is the first quadrant of a circle of radius 2. Learn about the transformation to polar coordinates and the application of the change of variables theorem.

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