We are tasked with evaluating the double integral:

$\iint_D \frac{x^2}{\sqrt{x^2 + y^2}} \, dx \, dy$

where $D$ is the region bounded by the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.

Step 1: Change to polar coordinates

In polar coordinates: $x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dx\,dy = r \, dr \, d\theta$

The integral becomes: $\iint_D \frac{x^2}{\sqrt{x^2 + y^2}} \, dx \, dy = \iint_D \frac{(r\cos\theta)^2}{\sqrt{r^2}} \cdot r \, dr \, d\theta$

Simplify: $\frac{(r\cos\theta)^2}{\sqrt{r^2}} = \frac{r^2\cos^2\theta}{r} = r\cos^2\theta$

Thus, the integral in polar coordinates is: $\int_0^{2\pi} \int_1^2 r \cos^2\theta \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_1^2 r^2 \cos^2\theta \, dr \, d\theta$

Step 2: Separate the integral

We separate the $r$- and $\theta$-dependent parts: $\int_0^{2\pi} \int_1^2 r^2 \cos^2\theta \, dr \, d\theta = \left( \int_0^{2\pi} \cos^2\theta \, d\theta \right) \cdot \left( \int_1^2 r^2 \, dr \right)$

Part 1: Compute $\int_0^{2\pi} \cos^2\theta \, d\theta$

Using the trigonometric identity: $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ $\int_0^{2\pi} \cos^2\theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int_0^{2\pi} 1 \, d\theta + \frac{1}{2} \int_0^{2\pi} \cos(2\theta) \, d\theta$

The first term is: $\frac{1}{2} \int_0^{2\pi} 1 \, d\theta = \frac{1}{2} \cdot 2\pi = \pi$

The second term is zero because $\cos(2\theta)$ is periodic over $[0, 2\pi]$: $\int_0^{2\pi} \cos(2\theta) \, d\theta = 0$

Thus: $\int_0^{2\pi} \cos^2\theta \, d\theta = \pi$

Part 2: Compute $\int_1^2 r^2 \, dr$

$\int_1^2 r^2 \, dr = \left[\frac{r^3}{3}\right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$

Step 3: Combine the results

$\iint_D \frac{x^2}{\sqrt{x^2 + y^2}} \, dx \, dy = \left(\pi\right) \cdot \left(\frac{7}{3}\right) = \frac{7\pi}{3}$

Final Answer:

$\boxed{\frac{7\pi}{3}}$

Would you like details on a specific step, or have any other questions?

Related Questions:

1. How does the Jacobian factor affect integrals in polar coordinates?
2. What is the role of symmetry in simplifying double integrals?
3. Can this problem be approached using cylindrical coordinates?
4. How would the integral change if the region were an ellipse instead of a circle?
5. What is the physical significance of $\frac{x^2}{\sqrt{x^2 + y^2}}$ in a practical context?

Tip:

Always use symmetry to reduce computation in integrals over circular or elliptical regions.