Math Problem Statement

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

requals=sine 7 thetasin7θ.

Question content area bottom

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.

Upper A Subscript leafAleafequals=Integral from 0 to StartFraction pi Over 7 EndFraction Integral from 0 to sine left parenthesis 7 theta right parenthesis r dr font size decreased by 5 d theta∫00StartFraction pi Over 7 EndFractionπ7∫00sine left parenthesis 7 theta right parenthesissin(7θ)r dr dθ

​(Type exact​ answers, using

piπ

as​ needed.)

Part 2

Find the area of the region.

Upper A Subscript roseAroseequals=StartFraction 2 Over 21 EndFraction221

units squaredunits2

​(Type an exact​ answer, using

piπ

as​ needed.)

Solution

Part 1: Setting up the Double Integral for the Leaf Closest to the Positive xx-axis

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

The formula for the area in polar coordinates is:

A=αβ0r(θ)rdrdθA = \int_{\alpha}^{\beta} \int_0^{r(\theta)} r \, dr \, d\theta

For the leaf closest to the positive xx-axis, we are looking at the range from θ=0\theta = 0 to θ=π7\theta = \frac{\pi}{7}, which corresponds to one petal of the rose.

So, the integral becomes:

Aleaf=0π70sin(7θ)rdrdθA_{\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 θ\theta): The limits of integration for θ\theta are from 00 to π7\frac{\pi}{7}.
  2. Inner integral (with respect to rr): The radius rr goes from 0 to sin(7θ)\sin(7\theta), which is the equation of the rose.
  3. Integrand: The area element in polar coordinates is rdrdθr \, dr \, d\theta.

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

Aleaf=0π70sin(7θ)rdrdθA_{\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(7θ)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 rr:

0sin(7θ)rdr=[r22]0sin(7θ)=sin2(7θ)2\int_0^{\sin(7\theta)} r \, dr = \left[ \frac{r^2}{2} \right]_0^{\sin(7\theta)} = \frac{\sin^2(7\theta)}{2}

  1. Now, the outer integral becomes:

Aleaf=0π7sin2(7θ)2dθA_{\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):

Arose=7×Aleaf=7×0π7sin2(7θ)2dθA_{\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:

Arose=221units2A_{\text{rose}} = \frac{2}{21} \, \text{units}^2

Thus, the area of the entire rose is 221\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 sin2(θ)\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(kθ)r = \sin(k\theta)?
  5. How would the area calculation change if the equation were r=cos(7θ)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!

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Double Integration
Area Calculation

Formulas

Area in polar coordinates: A = ∫∫ r dr dθ
Equation of the rose curve: r = sin(7θ)

Theorems

Symmetry in Polar Curves
Integration in Polar Coordinates

Suitable Grade Level

Grades 11-12, College