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?

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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.