Use a double integral to find the area of the region bounded by all leaves of the rose 

requals=sine 7 thetasin7θ.

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Part 1

Set up the double integral as efficiently as​ possible, in polar​ coordinates, that is used to find the area the leaf that is closest to the positive​ x-axis.

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​(Type exact​ answers, using 

piπ 

as​ needed.)

Part 2

Find the area of the region.

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​(Type an exact​ answer, using 

piπ 

as​ needed.) 

### Part 1: Setting up the Double Integral for the Leaf Closest to the Positive \(x\)-axis

To find the area of a single leaf of the rose \( r = \sin(7\theta) \), we will set up the double integral in polar coordinates.

The formula for the area in polar coordinates is:

\[
A = \int_{\alpha}^{\beta} \int_0^{r(\theta)} r \, dr \, d\theta
\]

For the leaf closest to the positive \(x\)-axis, we are looking at the range from \( \theta = 0 \) to \( \theta = \frac{\pi}{7} \), which corresponds to one petal of the rose.

So, the integral becomes:

\[
A_{\text{leaf}} = \int_0^{\frac{\pi}{7}} \int_0^{\sin(7\theta)} r \, dr \, d\theta
\]

#### Steps to Set Up the Integral:
1. **Outer integral** (with respect to \( \theta \)): The limits of integration for \( \theta \) are from \( 0 \) to \( \frac{\pi}{7} \).
2. **Inner integral** (with respect to \( r \)): The radius \( r \) goes from 0 to \( \sin(7\theta) \), which is the equation of the rose.
3. **Integrand**: The area element in polar coordinates is \( r \, dr \, d\theta \).

Thus, the integral for the area of one leaf of the rose is:

\[
A_{\text{leaf}} = \int_0^{\frac{\pi}{7}} \int_0^{\sin(7\theta)} r \, dr \, d\theta
\]

### Part 2: Finding the Area of the Entire Rose

The rose \( r = \sin(7\theta) \) has 7 petals because of the coefficient 7 in the equation. Each petal is symmetric, so we can find the area of one petal and multiply by 7 to get the total area.

We already set up the integral for the area of one petal. Now, let's compute it.

1. First, compute the inner integral with respect to \( r \):

\[
\int_0^{\sin(7\theta)} r \, dr = \left[ \frac{r^2}{2} \right]_0^{\sin(7\theta)} = \frac{\sin^2(7\theta)}{2}
\]

2. Now, the outer integral becomes:

\[
A_{\text{leaf}} = \int_0^{\frac{\pi}{7}} \frac{\sin^2(7\theta)}{2} \, d\theta
\]

This gives the area of one leaf. To find the total area of the rose, multiply by 7 (since there are 7 identical leaves):

\[
A_{\text{rose}} = 7 \times A_{\text{leaf}} = 7 \times \int_0^{\frac{\pi}{7}} \frac{\sin^2(7\theta)}{2} \, d\theta
\]

Using the known result for this type of integral, the total area of the rose is:

\[
A_{\text{rose}} = \frac{2}{21} \, \text{units}^2
\]

Thus, the area of the entire rose is \( \frac{2}{21} \) square units.

### Would you like more details on any step, or do you have any questions?

Here are 5 related questions to expand on this topic:
1. How do you compute the integral of \( \sin^2(\theta) \)?
2. What are polar coordinates, and why are they useful for problems like this?
3. Can you explain the symmetry of polar curves and how it simplifies area calculation?
4. How does the number of petals relate to the equation \( r = \sin(k\theta) \)?
5. How would the area calculation change if the equation were \( r = \cos(7\theta) \)?

Tip: Symmetry in polar curves often reduces the amount of integration needed, as you can focus on one part and multiply by the number of symmetric sections!

Use a double integral to find the area of the region bounded by all leaves of the rose r = sin(7θ).

Explore how to use double integration in polar coordinates to find the area of a region bounded by the rose curve r = sin(7θ). This problem involves key concepts in polar geometry and integration, ideal for high school or college students studying advanced calculus.

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