To solve the given integral step by step, we are tasked with evaluating the double integral

$\iint_R |r - 1| \, dA$

where $R$ is the circular disk $r \leq 2$ in the $xy$-plane.

Step 1: Understand the Geometry

The region $R$ is a disk of radius 2 centered at the origin, and we are working with polar coordinates. Therefore:

• The limits for $r$ are $0$ to $2$,
• The limits for $\theta$ are $0$ to $2\pi$.

The integrand $|r - 1|$ introduces a piecewise behavior:

• When $0 \leq r \leq 1$, $|r - 1| = 1 - r$,
• When $1 \leq r \leq 2$, $|r - 1| = r - 1$.

Step 2: Set Up the Integral in Polar Coordinates

Using the area element $dA = r \, dr \, d\theta$, the integral is split as: $\iint_R |r - 1| \, dA = \int_0^{2\pi} \left[ \int_0^1 (1 - r) r \, dr + \int_1^2 (r - 1) r \, dr \right] d\theta$

Step 3: Solve the Inner Integrals

First Integral ($0 \leq r \leq 1$):

$\int_0^1 (1 - r) r \, dr = \int_0^1 (r - r^2) \, dr$ Split the terms and integrate: [ \int_0^1 r , dr = \left[\frac{r^2}{2}\right]_0^1 = \frac{1}{2}, \quad \int_0^1 r^2 , dr = \left[\frac{r^3}{3}\right]_0^1 = \frac{1}{3} ] So: $\int_0^1 (1 - r) r \, dr = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$

Second Integral ($1 \leq r \leq 2$):

$\int_1^2 (r - 1) r \, dr = \int_1^2 (r^2 - r) \, dr$ Split the terms and integrate: [ \int_1^2 r^2 , dr = \left[\frac{r^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}, \quad \int_1^2 r , dr = \left[\frac{r^2}{2}\right]_1^2 = 2 - \frac{1}{2} = \frac{3}{2}. ] So: $\int_1^2 (r - 1) r \, dr = \frac{7}{3} - \frac{3}{2} = \frac{14}{6} - \frac{9}{6} = \frac{5}{6}.$

Step 4: Combine the Results

Now combine both parts of the integral: $\int_0^{2\pi} \left[ \frac{1}{6} + \frac{5}{6} \right] d\theta = \int_0^{2\pi} \frac{6}{6} \, d\theta = \int_0^{2\pi} 1 \, d\theta.$

Step 5: Solve the Outer Integral

The outer integral is: $\int_0^{2\pi} 1 \, d\theta = \theta \Big|_0^{2\pi} = 2\pi.$

Final Answer:

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

Let me know if you want further clarifications or additional questions to practice on!