We are tasked with evaluating the following double integral:

$I = \iint_D \frac{1}{\sqrt{x^2 + y^2}} \, dA$ where the region $D$ is the area inside the circle $x^2 + y^2 = 4x$ and outside the circle $x^2 + y^2 = 2x$.

Step 1: Convert to Polar Coordinates

The equations of the circles are not in standard form, so let's first rewrite them in a more manageable form.

• The equation of the first circle: $x^2 + y^2 = 4x \quad \Rightarrow \quad x^2 - 4x + y^2 = 0 \quad \Rightarrow \quad (x - 2)^2 + y^2 = 4$ This represents a circle centered at $(2, 0)$ with a radius of 2.

• The equation of the second circle: $x^2 + y^2 = 2x \quad \Rightarrow \quad x^2 - 2x + y^2 = 0 \quad \Rightarrow \quad (x - 1)^2 + y^2 = 1$ This represents a circle centered at $(1, 0)$ with a radius of 1.

Step 2: Polar Coordinates Setup

In polar coordinates, we have:

• $x = r \cos \theta$
• $y = r \sin \theta$
• The differential area element is $dA = r \, dr \, d\theta$

The integral becomes: $I = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} \frac{1}{r} r \, dr \, d\theta = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} 1 \, dr \, d\theta$ where:

• The inner radius $r_1(\theta)$ is the radius of the smaller circle.
• The outer radius $r_2(\theta)$ is the radius of the larger circle.

Step 3: Finding the Radii

For the larger circle $(x-2)^2 + y^2 = 4$, we have the radius $r_2 = 2$. For the smaller circle $(x-1)^2 + y^2 = 1$, we have the radius $r_1 = 1$.

Step 4: Setting the Integral Limits

We are looking for the region between the two circles, so:

• The angular limits are $\theta_1 = 0$ to $\theta_2 = 2\pi$ because the circles are symmetric about the origin.
• The radial limits are $r_1 = 1$ and $r_2 = 2$.

Thus, the double integral becomes: $I = \int_0^{2\pi} \int_1^2 1 \, dr \, d\theta$

Step 5: Evaluating the Integral

Now, we can evaluate the double integral:

$I = \int_0^{2\pi} \left[ r \right]_1^2 \, d\theta = \int_0^{2\pi} (2 - 1) \, d\theta = \int_0^{2\pi} 1 \, d\theta$

Finally, integrating with respect to $\theta$: $I = \left[ \theta \right]_0^{2\pi} = 2\pi$

Final Answer:

The value of the double integral is $\boxed{2\pi}$.

Would you like any further clarification on this problem or the process involved? Here are some related questions for you:

1. How would the integral change if the integrand was $\frac{1}{(x^2 + y^2)^{3/2}}$?
2. What are the parametric equations for a circle in polar coordinates?
3. How do you derive the equation of a circle in polar coordinates?
4. How would you compute the area of the region inside a circle using polar coordinates?
5. What is the significance of the Jacobian in converting from Cartesian to polar coordinates?

Tip: For problems involving regions defined by circles, converting to polar coordinates often simplifies the problem because the integrals typically become easier to evaluate when the geometry is radial.