Find ∫∫D √(x²+y²) dxdy if the area between the two circles with radius 1 and 2 situated in the first quarter and the line x=y and x=0

To solve the given double integral, we first need to define the region \( D \) over which we are integrating. The region \( D \) is situated in the first quadrant between two circles with radii 1 and 2, and it is bounded by the lines \( x = y \) and \( x = 0 \).

### Step 1: Describe the Region \( D \)

1. **Circles**:
   - The equation of a circle centered at the origin with radius \( r \) is \( x^2 + y^2 = r^2 \).
   - The given circles have radii 1 and 2, so their equations are:
     \[
     x^2 + y^2 = 1 \quad \text{and} \quad x^2 + y^2 = 4.
     \]

2. **First Quadrant**:
   - The region is confined to the first quadrant, so \( x \geq 0 \) and \( y \geq 0 \).

3. **Lines**:
   - The line \( x = y \) and the vertical line \( x = 0 \).

Given these constraints, the region \( D \) is a sector between the two circles, bounded between the line \( x = y \) (or equivalently, \( \theta = \frac{\pi}{4} \) in polar coordinates) and the \( y \)-axis (or \( \theta = \frac{\pi}{2} \)).

### Step 2: Convert to Polar Coordinates

To evaluate the integral, we switch to polar coordinates. Recall that:

\[
x = r \cos \theta, \quad y = r \sin \theta, \quad \text{and} \quad dx \, dy = r \, dr \, d\theta.
\]

Also, the expression under the square root simplifies to:

\[
\sqrt{x^2 + y^2} = \sqrt{r^2} = r.
\]

Thus, the integral becomes:

\[
\iint_D \sqrt{x^2 + y^2} \, dx \, dy = \iint_D r \, r \, dr \, d\theta = \iint_D r^2 \, dr \, d\theta.
\]

### Step 3: Determine the Limits of Integration

In polar coordinates, the region \( D \) is defined by:

- Radius \( r \) ranges from 1 to 2 (the radii of the two circles).
- Angle \( \theta \) ranges from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) (between the line \( x = y \) and the \( y \)-axis).

So, the integral can be rewritten as:

\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{1}^{2} r^2 \, dr \, d\theta.
\]

### Step 4: Evaluate the Integral

First, integrate with respect to \( r \):

\[
\int_{1}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.
\]

Next, integrate with respect to \( \theta \):

\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{7}{3} \, d\theta = \frac{7}{3} \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = \frac{7}{3} \cdot \frac{\pi}{4} = \frac{7\pi}{12}.
\]

### Final Answer

The value of the double integral is:

\[
\iint_D \sqrt{x^2 + y^2} \, dx \, dy = \frac{7\pi}{12}.
\]

Would you like further details on any step, or do you have any questions?

### Related Questions

1. How would the integral change if the region were in the second quadrant?
2. What would the integral be if the region were between the radii 1 and 3 instead?
3. How does the result change if the circles were centered at a different point?
4. What happens if the line \( x = y \) is replaced with \( y = mx \) where \( m \neq 1 \)?
5. Can this problem be solved using Cartesian coordinates directly?

### Tip
Converting to polar coordinates is useful when dealing with circular or sector-shaped regions, as it simplifies the bounds and expressions in integrals.

Solve the double integral ∫∫D √(x²+y²) dxdy over a sector between circles of radius 1 and 2 in the first quadrant. Learn how to convert to polar coordinates and calculate integrals in this detailed step-by-step solution. Suitable for university-level calculus students.

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