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Magazine Chapter 5: Problem 38
For the following exercises, use Green's theorem to find the area. Use Green's theorem to find the area of one loop of a four-leaf rose \(r=3 \sin 2 \theta .\) (Hint: \(\left.x d y-y d x=\mathbf{r}^{2} d \theta\right)\).
Short Answer
Expert verified
The area of one loop of the four-leaf rose is \(\frac{9\pi}{8}\).
Step by step solution
01
Understand Green's Theorem for Area
Green's Theorem can be used to relate a line integral around a simple closed curve to a double integral over the plane region bounded by the curve. In the context of finding the area, we use the special form that states: \( A = \frac{1}{2} \oint_C (x \, dy - y \, dx) \). For polar coordinates, this can be expressed using the hint provided as \( \frac{1}{2} \int_C r^2 \, d\theta \).
02
Identify the Parameters of the Curve
For the given four-leaf rose curve \( r = 3 \sin 2\theta \), we need to determine the range of \( \theta \) that describes one loop of the rose. Each loop occurs when the sine function completes a full cycle from 0 to \( \pi \). Thus, for one loop of the rose, \( \theta \) ranges from 0 to \( \frac{\pi}{2} \).
03
Set Up the Integral for the Area
Using the adapted Green's Theorem for polar coordinates, plug \( r = 3 \sin 2\theta \) into the area integral: \( A = \frac{1}{2} \int_0^{\frac{\pi}{2}} r^2 \, d\theta \). Substitute \( r^2 = (3 \sin 2\theta)^2 = 9 \sin^2 2\theta \). The integral becomes \( A = \frac{1}{2} \int_0^{\frac{\pi}{2}} 9 \sin^2 2\theta \, d\theta \).
04
Simplify and Evaluate the Integral
Simplify the integral: \( A = \frac{9}{2} \int_0^{\frac{\pi}{2}} \sin^2 2\theta \, d\theta \). Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) for substitution. The integral becomes \( A = \frac{9}{2} \int_0^{\frac{\pi}{2}} \frac{1 - \cos 4\theta}{2} \, d\theta \). Simplify further: \( A = \frac{9}{4} \left[ \int_0^{\frac{\pi}{2}} 1 \, d\theta - \int_0^{\frac{\pi}{2}} \cos 4\theta \, d\theta \right] \).
05
Calculate Each Part of the Integral
Evaluate \( \int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2} \). For the second part, \( \int_0^{\frac{\pi}{2}} \cos 4\theta \, d\theta \), use the antiderivative: \( \frac{1}{4} \sin 4\theta \). Evaluate from 0 to \( \frac{\pi}{2} \): \( \left[ \frac{1}{4} \sin 4\theta \right]_0^{\frac{\pi}{2}} = 0 \).
06
Combine Results
By combining the parts, the integral is \( A = \frac{9}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{9\pi}{8} \). This is the area of one loop of the four-leaf rose \( r = 3 \sin 2\theta \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
polar coordinates
Polar coordinates are a way of describing a point in the plane using a distance and an angle. This system is widely used where situations have some form of symmetry about a central point. In polar coordinates, each point on the plane is determined by the radial distance from the origin, known as \( r \), and the angle, \( \theta \), measured from the positive x-axis.
- Radius \( r \): The distance from the origin to the point.
- Angle \( \theta \): The angle formed with the positive x-axis.
four-leaf rose
A four-leaf rose is a type of mathematical curve whose shape resembles a flower with four petals. In polar coordinates, this rose can be defined with equations like \( r = a \sin n\theta \) or \( r = a \cos n\theta \), where the number of petals depends on the value of \( n \).
- If \( n \) is even, the rose has \( 2n \) petals.
- If \( n \) is odd, the rose has \( n \) petals.
double integral
A double integral involves integrating a function over a two-dimensional region. It's analogous to finding the volume under a surface in three-dimensional space. In our context, the double integral is used to find the area of the region enclosed by the curve using Green's Theorem.Consider the expression: \[ \int_C (x \cdot dy - y \cdot dx) = \int_{R} 1 \, dA \] where \( C \) is a closed curve and \( R \) is the region it encloses.
- This can also be transformed for polar coordinates into \( \frac{1}{2} \int_0^{2\pi} r^2 \ d\theta \).
- The integral calculates the area by considering little "slices" of the region through \( \theta \) from \( 0 \) to \( \frac{\pi}{2} \) for one petal.
area calculation
To calculate the area of one loop of the four-leaf rose, we apply the adapted form of Green's Theorem for polar coordinates. This theorem simplifies the area calculation by transforming it into a single integral.The process begins by setting up the integral: \[ A = \frac{1}{2} \int_0^{\frac{\pi}{2}} (3 \sin 2\theta)^2 \, d\theta \]Using some trigonometric identities, the integral simplifies further. Apply \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to handle the \( \sin^2 2\theta \) terms:
- Simplify to: \( \frac{9}{2} \int_0^{\frac{\pi}{2}} \sin^2 2\theta \, d\theta \)
- Convert further to \( \frac{9}{4} \left( \int_0^{\frac{\pi}{2}} 1 \, d\theta - \int_0^{\frac{\pi}{2}} \cos 4\theta \, d\theta \right) \)
