Find the area of ​​the region that is inside the region bounded by the graph of the first equation and outside the region bounded by the graph of the second equation.

r^2=4sen(2theta), r^2=2

To find the area of the region that is inside the region bounded by the graph of the first equation \( r^2 = 4 \sin(2\theta) \) and outside the region bounded by the graph of the second equation \( r^2 = 2 \), we need to proceed with the following steps:

### Step 1: Interpret the equations
- The first equation \( r^2 = 4 \sin(2\theta) \) describes a polar curve, specifically a **lemniscate** (a figure-eight shape).
- The second equation \( r^2 = 2 \) describes a **circle** in polar coordinates with radius \( \sqrt{2} \).

### Step 2: Determine the bounds of integration
To find the intersection points between the two curves, set the equations equal:

\[
4 \sin(2\theta) = 2
\]

\[
\sin(2\theta) = \frac{1}{2}
\]

The solutions to \( \sin(2\theta) = \frac{1}{2} \) within the interval \( 0 \leq \theta < 2\pi \) are:

\[
2\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}
\]

\[
\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}
\]

However, due to symmetry, we only need to consider one lobe of the lemniscate, i.e., from \( \theta = \frac{\pi}{12} \) to \( \theta = \frac{5\pi}{12} \).

### Step 3: Compute the area
The area of the region between two polar curves \( r_1(\theta) \) and \( r_2(\theta) \) from \( \theta = a \) to \( \theta = b \) is given by:

\[
A = \frac{1}{2} \int_{a}^{b} \left( r_1^2(\theta) - r_2^2(\theta) \right) d\theta
\]

For the region inside \( r^2 = 4 \sin(2\theta) \) and outside \( r^2 = 2 \), the area becomes:

\[
A = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \left( 4 \sin(2\theta) - 2 \right) d\theta
\]

### Step 4: Evaluate the integral
To find the exact area, we need to compute the integral:

\[
A = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \left( 4 \sin(2\theta) - 2 \right) d\theta
\]

This can be split into two integrals:

\[
A = 2 \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \sin(2\theta) d\theta - \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 2 d\theta
\]

First integral:

\[
\int \sin(2\theta) d\theta = -\frac{1}{2} \cos(2\theta)
\]

Second integral:

\[
\int d\theta = \theta
\]

So:

\[
A = 2 \left[-\frac{1}{2} \cos(2\theta)\right]_{\frac{\pi}{12}}^{\frac{5\pi}{12}} - 2 \left[\theta\right]_{\frac{\pi}{12}}^{\frac{5\pi}{12}}
\]

\[
A = -\cos\left(\frac{5\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right) - 2\left(\frac{5\pi}{12} - \frac{\pi}{12}\right)
\]

\[
A = \left(\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right) - 2\left(\frac{4\pi}{12}\right)
\]

\[
A = \sqrt{3} - \frac{2\pi}{3}
\]

### Final Answer:
The area of the region inside the first curve and outside the second curve is \( \boxed{\sqrt{3} - \frac{2\pi}{3}} \).

Do you want more details or have any questions? Here are some related questions you might consider:

1. How would the area change if the second curve was \( r = 3 \)?
2. What if the first curve was a different polar curve, like \( r^2 = 4 \cos(2\theta) \)?
3. Can you calculate the area of a single loop of the lemniscate \( r^2 = 4 \sin(2\theta) \)?
4. What are some other applications of polar coordinates in calculating areas?
5. How does the symmetry of polar curves simplify the area calculation?

**Tip:** When working with polar coordinates, symmetry can often simplify calculations by reducing the limits of integration.

Learn how to calculate the area between polar curves r^2 = 4sin(2theta) and r^2 = 2 using polar coordinates and trigonometric integration. Understand the steps involved in determining the bounds of integration and computing the integral. Suitable for undergraduate students studying polar coordinates and calculus.

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