The problem provided involves finding the integral over an annular region $R$, which lies between two circles defined by:

1. $x^2 + y^2 = 1$ (a circle with radius 1)
2. $x^2 + y^2 = 5$ (a circle with radius $\sqrt{5}$)

The integral to evaluate is:

$\iint_R (x^2 + y^2) \, dA$

Solution Approach:

We can simplify this problem by converting to polar coordinates since the region is circular (annular). In polar coordinates, the transformations are:

$x = r\cos\theta, \quad y = r\sin\theta, \quad \text{and} \quad x^2 + y^2 = r^2$

The area element $dA$ in polar coordinates becomes $r \, dr \, d\theta$.

Thus, the integral becomes:

$\iint_R (x^2 + y^2) \, dA = \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{5}} r^2 \cdot r \, dr \, d\theta$

This simplifies to:

$\int_0^{2\pi} \int_1^{\sqrt{5}} r^3 \, dr \, d\theta$

Step 1: Evaluate the radial part of the integral

$\int_1^{\sqrt{5}} r^3 \, dr = \left[ \frac{r^4}{4} \right]_1^{\sqrt{5}} = \frac{(\sqrt{5})^4}{4} - \frac{1^4}{4} = \frac{25}{4} - \frac{1}{4} = \frac{24}{4} = 6$

Step 2: Evaluate the angular part of the integral

$\int_0^{2\pi} d\theta = 2\pi$

Step 3: Combine the results

The total integral is:

$6 \times 2\pi = 12\pi$

Thus, the value of the integral is $\boxed{12\pi}$.

Would you like more details on any part of this solution or have further questions?

Related Questions:

1. What would the integral be if the annular region were between $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$?
2. How does the area element change in polar coordinates for other geometric shapes, such as ellipses?
3. How would this integral change if we included a weight function like $f(r,\theta) = r \sin(\theta)$?
4. What if the integral had been $\iint_R x \, dA$? How would the approach differ?
5. How can polar coordinates simplify the calculation of integrals over circular regions in general?

Tip:

When solving problems involving circular or radial symmetry, converting to polar coordinates often simplifies the calculation due to the natural geometry of the problem.