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.