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Find the area of one leaf of the "four-petaled rose" r = 6 sin 20 shown in the following figure: Answer: A x=16₂ 0 = 1 X With ro = 6

Find the area of one leaf of the "four-petaled rose" r = 6 sin 20 shown in the following figure: Answer: A x=16₂ 0 = 1 X With ro = 6

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.5: Polar Coordinates
Problem 45E
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### Problem Statement: Find the area of one leaf of the "four-petaled rose" \( r = 6 \sin 2\theta \) shown in the following figure: ### Diagram Description: - The diagram shows a four-petaled rose plotted on a polar coordinate system with the \( x \)-axis and \( y \)-axis clearly marked. - The rose consists of four symmetrical petals centered around the origin. - The polar curve \( r = 6 \sin 2\theta \) is represented. - One leaf of the rose is highlighted. - The point \( A \) where \( r = r_0 \), \( \theta = \frac{\pi}{4} \) is indicated on the diagram. - The parameter \( r_0 \) is given as 6. ### Instructions for Solving: To find the area of one leaf of the rose, integrate the area in polar coordinates for one leaf of the equation \( r = 6 \sin 2\theta \). ### Mathematical Steps: The area \( A \) of one leaf of the polar curve can be found using the integral: \[ A = \int_{\alpha}^{\beta} \frac{1}{2} (r(\theta))^2 \, d\theta \] For the curve \( r = 6 \sin 2\theta \): - Determine the limits of integration (\( \alpha \) and \( \beta \)) for one leaf. - These limits will be from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) since one petal completes over this range. Therefore, \[ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (6 \sin 2\theta)^2 \, d\theta \] ### Input for Answer: Write down the solution to the integral to find the area of one petal of the four-petaled rose in the given answer box. ```plaintext Answer: ``` By these steps, the area of one leaf can be calculated and the final answer can be entered in the provided text box for the educational purposes of solving this mathematical problem.
Transcribed Image Text:### Problem Statement: Find the area of one leaf of the "four-petaled rose" \( r = 6 \sin 2\theta \) shown in the following figure: ### Diagram Description: - The diagram shows a four-petaled rose plotted on a polar coordinate system with the \( x \)-axis and \( y \)-axis clearly marked. - The rose consists of four symmetrical petals centered around the origin. - The polar curve \( r = 6 \sin 2\theta \) is represented. - One leaf of the rose is highlighted. - The point \( A \) where \( r = r_0 \), \( \theta = \frac{\pi}{4} \) is indicated on the diagram. - The parameter \( r_0 \) is given as 6. ### Instructions for Solving: To find the area of one leaf of the rose, integrate the area in polar coordinates for one leaf of the equation \( r = 6 \sin 2\theta \). ### Mathematical Steps: The area \( A \) of one leaf of the polar curve can be found using the integral: \[ A = \int_{\alpha}^{\beta} \frac{1}{2} (r(\theta))^2 \, d\theta \] For the curve \( r = 6 \sin 2\theta \): - Determine the limits of integration (\( \alpha \) and \( \beta \)) for one leaf. - These limits will be from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) since one petal completes over this range. Therefore, \[ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (6 \sin 2\theta)^2 \, d\theta \] ### Input for Answer: Write down the solution to the integral to find the area of one petal of the four-petaled rose in the given answer box. ```plaintext Answer: ``` By these steps, the area of one leaf can be calculated and the final answer can be entered in the provided text box for the educational purposes of solving this mathematical problem.
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