To solve the given problem, let us change the double integral to polar coordinates and evaluate it step by step. Here's the process:

Problem:

Evaluate
$\iint_R \frac{y^2}{x^2 + y^2} \, dA,$ where $R$ is the region between the circles $x^2 + y^2 = 16$ and $x^2 + y^2 = 100$.

Step 1: Change to Polar Coordinates

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

The integral becomes: $\iint_R \frac{y^2}{x^2 + y^2} \, dA = \iint_R \frac{r^2 \sin^2\theta}{r^2} r \, dr \, d\theta = \iint_R \sin^2\theta \, r \, dr \, d\theta.$

Step 2: Bounds for $r$ and $\theta$

• $r$ ranges from 4 (inner circle $r^2 = 16$) to 10 (outer circle $r^2 = 100$).
• $\theta$ ranges from $0$ to $2\pi$ (full circle).

Thus, the integral becomes: $\int_0^{2\pi} \int_4^{10} \sin^2\theta \, r \, dr \, d\theta.$

Step 3: Simplify and Separate Integrals

Since $\sin^2\theta$ is independent of $r$, we can separate the integral: $\int_0^{2\pi} \int_4^{10} \sin^2\theta \, r \, dr \, d\theta = \left( \int_0^{2\pi} \sin^2\theta \, d\theta \right) \left( \int_4^{10} r \, dr \right).$

Step 4: Evaluate Each Part

1. Evaluate $\int_0^{2\pi} \sin^2\theta \, d\theta$:

Use the identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$: $\int_0^{2\pi} \sin^2\theta \, d\theta = \int_0^{2\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int_0^{2\pi} 1 \, d\theta - \frac{1}{2} \int_0^{2\pi} \cos(2\theta) \, d\theta.$

• $\int_0^{2\pi} 1 \, d\theta = 2\pi$,
• $\int_0^{2\pi} \cos(2\theta) \, d\theta = 0$ (since $\cos(2\theta)$ is periodic over $[0, 2\pi]$).

Thus: $\int_0^{2\pi} \sin^2\theta \, d\theta = \frac{1}{2} (2\pi) = \pi.$

2. Evaluate $\int_4^{10} r \, dr$: $\int_4^{10} r \, dr = \left[ \frac{r^2}{2} \right]_4^{10} = \frac{10^2}{2} - \frac{4^2}{2} = \frac{100}{2} - \frac{16}{2} = 50 - 8 = 42.$

Step 5: Combine Results

$\iint_R \frac{y^2}{x^2 + y^2} \, dA = \left( \int_0^{2\pi} \sin^2\theta \, d\theta \right) \left( \int_4^{10} r \, dr \right) = \pi \cdot 42 = 42\pi.$

Final Answer:

$\boxed{42\pi}$

Would you like further clarification or additional examples?

Related Questions:

1. How do you convert a general Cartesian integral into polar coordinates?
2. What is the geometric interpretation of changing regions of integration to polar coordinates?
3. Can you explain why $\cos(2\theta)$ integrates to zero over a full period?
4. How do you find the limits for $r$ and $\theta$ in polar coordinates?
5. Why is $dA = r \, dr \, d\theta$ in polar coordinates?

Tip:

Always simplify trigonometric terms using identities like $\sin^2\theta + \cos^2\theta = 1$ or $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$ when integrating over full circular intervals!