The given double integral is:

$\iint_R \sqrt{x^2 + y^2} \, dx \, dy$ where $R$ is the region enclosed between the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.

Step 1: Set up the integral in polar coordinates

In this case, the region $R$ is a circular annular region, so we convert the integral to polar coordinates, where:

• $x = r\cos\theta$
• $y = r\sin\theta$
• The differential element $dx\,dy = r\,dr\,d\theta$
• The integrand $\sqrt{x^2 + y^2} = \sqrt{r^2} = r$

In polar coordinates, the integral becomes:

$\iint_R \sqrt{x^2 + y^2} \, dx \, dy = \iint_R r \, r \, dr \, d\theta$

Thus, the integral simplifies to:

$\iint_R r^2 \, dr \, d\theta$

Step 2: Determine the limits

• The region $R$ is bounded by the circles $x^2 + y^2 = 1$ (which has a radius of 1) and $x^2 + y^2 = 4$ (which has a radius of 2).
• In polar coordinates, the limits for $r$ will range from 1 to 2.
• The angle $\theta$ spans a full circle, so $\theta$ ranges from 0 to $2\pi$.

Thus, the double integral is:

$\int_0^{2\pi} \int_1^2 r^2 \, dr \, d\theta$

Step 3: Solve the inner integral

We first solve the inner integral with respect to $r$:

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

Step 4: Solve the outer integral

Now, we evaluate the outer integral:

$\int_0^{2\pi} \frac{7}{3} \, d\theta = \frac{7}{3} \cdot \theta \Bigg|_0^{2\pi} = \frac{7}{3} \cdot 2\pi = \frac{14\pi}{3}$

Final Answer:

The value of the double integral is:

$\boxed{\frac{14\pi}{3}}$

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Would you like more details or clarifications on any of the steps?

Here are 5 related questions:

1. How do you convert a double integral from Cartesian to polar coordinates?
2. What is the general formula for converting $dx \, dy$ to polar coordinates?
3. How do you handle integrals over circular regions in polar coordinates?
4. Can you evaluate a double integral over a region with a non-circular boundary?
5. How does the Jacobian affect the conversion to polar coordinates?

Tip: When working with polar coordinates, always remember that $r^2$ appears in the integrand when $x^2 + y^2$ is involved, simplifying the expression for circular regions.