Convert the following integral to polar coordinates and evaluate. $$ \int_{0}^{2 \sqrt{2}} \int_{v}^{\sqrt{16-y^{2}}} x y d x d y $$

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To convert the given integral to polar coordinates, we need to follow these steps:

1. **Identify the region of integration**: The limits of integration suggest that $$ y $$ ranges from $$ 0 $$ to $$ 2\sqrt{2} $$ and for each fixed $$ y $$, $$ x $$ ranges from $$ v $$ to $$ \sqrt{16 - y^2} $$.

2. **Convert to polar coordinates**: In polar coordinates, we have:
   - $$ x = r \cos(\theta) $$
   - $$ y = r \sin(\theta) $$
   - The Jacobian of the transformation is $$ r $$.

3. **Determine the new limits**: 
   - The upper limit for $$ y $$ suggests that $$ r \sin(\theta) $$ goes from $$ 0 $$ to $$ 2\sqrt{2} $$.
   - The upper limit for $$ x $$ suggests that $$ r \cos(\theta) $$ goes from $$ v $$ to $$ \sqrt{16 - (r \sin(\theta))^2} $$.

4. **Set up the integral**: The integral in polar coordinates becomes:
   $$
   \int \int r^2 \cos(\theta) \sin(\theta) \, r \, dr \, d\theta
   $$

Now, let's convert the limits and evaluate the integral.

First, we need to find the limits for $$ r $$ and $$ \theta $$. The region defined by $$ y $$ and $$ x $$ can be analyzed to find the appropriate bounds.

Let's proceed with the conversion and evaluation of the integral.

The integral in polar coordinates will be:
$$
\int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r^2 \cos(\theta) \sin(\theta) \, dr \, d\theta
$$

Next, we will determine the limits and evaluate the integral.

Let's calculate the integral.
It seems there was an issue with the calculation. Let's break down the steps more clearly and evaluate the integral step by step.

1. **Convert the integral to polar coordinates**:
   The original integral is:
   $$
   \int_{0}^{2 \sqrt{2}} \int_{v}^{\sqrt{16-y^{2}}} x y \, dx \, dy
   $$

In polar coordinates:
   - $$ x = r \cos(\theta) $$
   - $$ y = r \sin(\theta) $$
   - The Jacobian is $$ r $$.

Thus, the integrand $$ x y $$ becomes:
   $$
   x y = (r \cos(\theta))(r \sin(\theta)) = r^2 \cos(\theta) \sin(\theta)
   $$

2. **Determine the limits**:
   - The limits for $$ y $$ are from $$ 0 $$ to $$ 2\sqrt{2} $$.
   - The limits for $$ x $$ are from $$ v $$ to $$ \sqrt{16 - y^2} $$.

The equation $$ y = 2\sqrt{2} $$ corresponds to $$ r \sin(\theta) = 2\sqrt{2} $$, giving $$ r = \frac{2\sqrt{2}}{\sin(\theta)} $$.

The equation $$ x = \sqrt{16 - y^2} $$ corresponds to the circle $$ x^2 + y^2 = 16 $$, which in polar coordinates is $$ r^2 = 16 $$ or $$ r = 4 $$.

3. **Set up the integral**:
   The integral in polar coordinates becomes:
   $$
   \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r^2 \cos(\theta) \sin(\theta) \, r \, dr \, d\theta
   $$

The limits for $$ r $$ will be from $$ 0 $$ to $$ 4 $$ and for $$ \theta $$ from $$ 0 $$ to $$ \frac{\pi}{4} $$.

4. **Evaluate the integral**:
   The integral can be expressed as:
   $$
   \int_{0}^{\frac{\pi}{4}} \int_{0}^{4} r^3 \cos(\theta) \sin(\theta) \, dr \, d\theta
   $$

Now, let's evaluate this integral.

First, we will compute the inner integral with respect to $$ r $$:
$$
\int_{0}^{4} r^3 \, dr
$$

Then, we will compute the outer integral with respect to $$ \theta $$:
$$
\int_{0}^{\frac{\pi}{4}} \cos(\theta) \sin(\theta) \, d\theta
$$

Let's perform these calculations.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{\frac{\pi }{4}} \cos\left(\theta \right)\sin\left(\theta \right) d\theta$$
- step1: Evaluate the integral:
  $$\int \cos\left(\theta \right)\sin\left(\theta \right) d\theta$$
- step2: Evaluate the integral:
  $$-\frac{\cos\left(2\theta \right)}{4}$$
- step3: Simplify the expression:
  $$-\frac{1}{4}\cos\left(2\theta \right)$$
- step4: Return the limits:
  $$\left(-\frac{1}{4}\cos\left(2\theta \right)\right)\bigg |_{0}^{\frac{\pi }{4}}$$
- step5: Calculate the value:
  $$\frac{1}{4}$$
Calculate or simplify the expression $$ \int_{0}^{4} r^3 \, dr $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{4} r^{3} dr$$
- step1: Evaluate the integral:
  $$\int r^{3} dr$$
- step2: Evaluate the integral:
  $$\frac{r^{3+1}}{3+1}$$
- step3: Add the numbers:
  $$\frac{r^{4}}{3+1}$$
- step4: Add the numbers:
  $$\frac{r^{4}}{4}$$
- step5: Return the limits:
  $$\left(\frac{r^{4}}{4}\right)\bigg |_{0}^{4}$$
- step6: Calculate the value:
  $$64$$
The calculations yield the following results:

1. The inner integral:
   $$
   \int_{0}^{4} r^3 \, dr = 64
   $$

2. The outer integral:
   $$
   \int_{0}^{\frac{\pi}{4}} \cos(\theta) \sin(\theta) \, d\theta = \frac{1}{4}
   $$

Now, we can combine these results to evaluate the entire integral:
$$
\int_{0}^{\frac{\pi}{4}} \int_{0}^{4} r^3 \cos(\theta) \sin(\theta) \, dr \, d\theta = 64 \cdot \frac{1}{4} = 16
$$

Thus, the value of the integral is $$ 16 $$.
The value of the integral is 16.

Convert the following integral to polar coordinates and evaluate.

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