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Magazine Chapter 8: Problem 27
Use Green's theorem to find the area of one loop of the four-leafed rose \(r=3 \sin 2 \theta .\) (HINT: \(x d y-y d x=r^{2} d \theta\)
Short Answer
Expert verified
The area of one loop is \(\frac{9\pi}{8}\).
Step by step solution
01
Understanding Green's Theorem for Area
Green's Theorem relates a line integral around a simple closed curve to a double integral over the plane region bounded by the curve. Specifically, for area calculations, Green's Theorem can be simplified to: \[ A = \frac{1}{2} \oint_C (x \, dy - y \, dx) \] where \(A\) is the area enclosed by curve \(C\). The hint given corroborates this formula as it equates \(x \, dy - y \, dx\) to \(r^2 \, d\theta\).
02
Polar Coordinates Area Representation
The parametric representation of a curve in polar coordinates involves the transformation: \[ x = r \cos \theta, \quad y = r \sin \theta \] The expression provided, \(x \, dy - y \, dx = r^2 \, d\theta\), suggests that we can compute the area enclosed by just integrating \(\frac{1}{2} r^2 \, d\theta\) over the appropriate limits of \(\theta\).
03
Analyzing the Function and Finding Limits
Given the polar equation \(r = 3 \sin 2\theta\), the function describes a four-leafed rose. To find the area of one loop, we consider the corresponding interval of \(\theta\) that traces out one 'leaf'. For \(r = 3 \sin 2\theta\), each loop occurs over \(\theta = 0\) to \(\theta = \frac{\pi}{2}\).
04
Setting up the Integral
Using the formula derived for area, we now set up the integral specifically for one petal. The area of one loop is thus: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (3 \sin 2\theta)^2 \, d\theta \] Simplifying, \[ = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 9 \sin^2 2\theta \, d\theta \]
05
Solving the Integral
To solve \[ \int \sin^2 2\theta \, d\theta \], we use the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\): \[ = \frac{9}{2} \int_{0}^{\frac{\pi}{2}} \left(\frac{1 - \cos 4\theta}{2}\right) \, d\theta \] \[ = \frac{9}{4} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos 4\theta \, d\theta \right) \]
06
Calculating the Result
Integrating separately, \[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}, \quad \int_{0}^{\frac{\pi}{2}} \cos 4\theta \, d\theta = 0 \] Thus, \[ A = \frac{9}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{9\pi}{8} \] So the area of one loop of the four-leafed rose is \(\frac{9\pi}{8}\).
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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 system where each point on a plane is defined by a distance from a reference point and an angle from a reference direction. More specifically,
To convert polar coordinates \( (r, \theta) \) to Cartesian coordinates \( (x, y) \):
- Reference point: Known as the pole, similar to the origin in Cartesian coordinates.
- Angle: Measured counterclockwise from the positive x-axis, known as \( \theta \).
- Radius: The distance from the pole to the point, represented by \( r \).
To convert polar coordinates \( (r, \theta) \) to Cartesian coordinates \( (x, y) \):
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Line Integral
A line integral is a type of integral where we sum a field over a curve. In our problem, it involves integrating a function along a path or contour, like the boundary of a region.
Using Green's Theorem, which connects line integrals to area calculations, we reduce our line integral to simpler components. Here, the line integral is expressed as:
Using Green's Theorem, which connects line integrals to area calculations, we reduce our line integral to simpler components. Here, the line integral is expressed as:
- \( \oint_C (x \, dy - y \, dx) \)
- \( \frac{1}{2} \oint_C r^2 \, d\theta \)
Double Integral
A double integral extends regular integration to calculate properties like area and volume over a given region. In essence, it's a way to "add up" values over a two-dimensional space.
Employing Green's Theorem, we use a double integral comprising two separate integrals. For polar regions, traditional Cartesian double integrals are complex. However,
Employing Green's Theorem, we use a double integral comprising two separate integrals. For polar regions, traditional Cartesian double integrals are complex. However,
- By relating them to line integrals, Green's Theorem offers a way to substitute for the area enclosed by the curve.
- It changes the problem from calculating a region's area directly to integrating straightforward terms over angular paths: \( \frac{1}{2} \int 9 \sin^2 2\theta \, d\theta \).
Area Calculation
Calculating the area within unusual regions, like loops of a parametric figure such as a four-leafed rose, is often complex using basic geometry.
Green's Theorem helps by transforming our area calculation task into easier-to-handle integrals. The steps are:
Green's Theorem helps by transforming our area calculation task into easier-to-handle integrals. The steps are:
- Recognize that \( r = 3 \sin 2\theta \) describes our "four-leafed rose" pattern, a key to knowing which areas to consider.
- Convert through suggested formula, focusing on \( \frac{1}{2} \int 9 \sin^2 2\theta \, d\theta \), covering each loop over \( \theta = 0 \) to \( \frac{\pi}{2} \).
- Dealing further with identities, simplifying and evaluating terms like \( 1 - \cos 4\theta \).
