Evaluate the integral

D
r dr d𝜃,
where D is the region bounded by the part of the four-leaved rose
r = sin(2𝜃)
situated in the first quadrant.

To evaluate the integral

\[
\int \int_D r \, dr \, d\theta,
\]

where \( D \) is the region bounded by the curve \( r = \sin(2\theta) \) in the first quadrant, we will go through the following steps:

### Step 1: Understanding the Region of Integration \( D \)
The equation \( r = \sin(2\theta) \) describes a four-leaved rose. Each "leaf" corresponds to a different range of \( \theta \). In the first quadrant, we are concerned with the part of the rose where:
1. \( r \geq 0 \),
2. \( \theta \in \left[0, \frac{\pi}{2}\right] \),
3. The function \( r = \sin(2\theta) \) describes the radial distance for each angle \( \theta \) in this range.

### Step 2: Setting Up the Integral
In polar coordinates, the area element \( dA = r \, dr \, d\theta \). Hence, the integral can be written as:

\[
\int_0^{\pi/2} \int_0^{\sin(2\theta)} r \, dr \, d\theta.
\]

### Step 3: Evaluating the Integral
1. **Inner Integral (in \( r \)):**
   \[
   \int_0^{\pi/2} \int_0^{\sin(2\theta)} r \, dr \, d\theta.
   \]
   Integrate with respect to \( r \):
   \[
   \int_0^{\pi/2} \left[ \frac{r^2}{2} \right]_0^{\sin(2\theta)} \, d\theta = \int_0^{\pi/2} \frac{\sin^2(2\theta)}{2} \, d\theta.
   \]

2. **Outer Integral (in \( \theta \)):**
   Now we need to evaluate
   \[
   \frac{1}{2} \int_0^{\pi/2} \sin^2(2\theta) \, d\theta.
   \]
   Using the identity \( \sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2} \), we get:
   \[
   \frac{1}{2} \int_0^{\pi/2} \frac{1 - \cos(4\theta)}{2} \, d\theta = \frac{1}{4} \int_0^{\pi/2} (1 - \cos(4\theta)) \, d\theta.
   \]
   Splitting the integral:
   \[
   \frac{1}{4} \left( \int_0^{\pi/2} 1 \, d\theta - \int_0^{\pi/2} \cos(4\theta) \, d\theta \right).
   \]
   
   Evaluating each part:
   - For \( \int_0^{\pi/2} 1 \, d\theta = \frac{\pi}{2} \).
   - For \( \int_0^{\pi/2} \cos(4\theta) \, d\theta \), since the integral of cosine over a full period is zero, it also evaluates to 0.

   Therefore:
   \[
   \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}.
   \]

### Final Answer
\[
\int \int_D r \, dr \, d\theta = \frac{\pi}{8}.
\]

Would you like a more detailed explanation on any step, or have any questions?

---

1. How would the limits change if we included the entire four-leaved rose?
2. How does the integral change if the region \( D \) is in a different quadrant?
3. What role does the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) play in solving such integrals?
4. How do polar coordinates simplify integration for certain regions?
5. What is the geometric interpretation of \( \int \int_D r \, dr \, d\theta \) in this context?

**Tip:** For integrals involving symmetrical shapes, use symmetry to simplify computations by considering only one section and multiplying if necessary.

Evaluate the integral D r dr dθ, where D is the region bounded by the part of the four-leaved rose r = sin(2θ) situated in the first quadrant.

This problem involves evaluating a double integral over a region bounded by the polar curve r = sin(2θ) in the first quadrant. The solution explores the use of polar coordinates, trigonometric identities, and integration limits to determine the area of a rose curve. Suitable for students in Calculus II.

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