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Magazine Chapter 15: Problem 17
Find the area of the region cut from the first quadrant by the curve \(r=2(2-\sin 2 \theta)^{1 / 2} .\)
Short Answer
Expert verified
The area of the region is \( 2\pi - 2 \).
Step by step solution
01
Identify the region of interest
This curve is given in polar coordinates as \( r = 2(2 - \sin 2\theta)^{1/2} \) and we need to find the area of the region it encloses in the first quadrant. Since the curve repeats for \( 0 \leq \theta \leq \pi/2 \), we only consider this range of \( \theta \).
02
Set up the integral for the area
The area \( A \) of a region in polar coordinates is given by: \[ A = \frac{1}{2} \int_{a}^{b} r^2 \ d\theta \] where the range of \( \theta \) is from \( a \) to \( b \). For our problem, \( r = 2(2 - \sin 2\theta)^{1/2} \) and \( \theta \) ranges from 0 to \( \pi/2 \).
03
Square the polar equation
Square the given polar equation to substitute in the integral: \( r^2 = [2(2 - \sin 2\theta)^{1/2}]^2 = 4(2 - \sin 2\theta) \).
04
Substitute and simplify the integral
Replace \( r^2 \) in the area formula with the expression we found: \[ A = \frac{1}{2} \int_{0}^{\pi/2} 4(2 - \sin 2\theta) \ d\theta \]. Simplify this to \[ A = 2 \int_{0}^{\pi/2} (2 - \sin 2\theta) \ d\theta \].
05
Evaluate the integral
Now solve the integral: \[ A = 2 \left( \int_{0}^{\pi/2} 2\ d\theta - \int_{0}^{\pi/2} \sin 2\theta\ d\theta \right) \].First part: \( \int_{0}^{\pi/2} 2\ d\theta = 2\theta \bigg|_{0}^{\pi/2} = 2(\pi/2 - 0) = \pi \).Second part: For \( \int_{0}^{\pi/2} \sin 2\theta\ d\theta \), use substitution \( u = 2\theta \), \( du = 2\,d\theta \), so \( d\theta = du/2 \). Change limits: when \( \theta = 0 \), \( u = 0 \), when \( \theta = \pi/2 \), \( u = \pi \).Thus, \( \int_{0}^{\pi} \frac{1}{2}\sin u\ du = \frac{1}{2} \int_{0}^{\pi} \sin u\ du = \frac{1}{2} [-\cos u]\bigg|_{0}^{\pi} = \frac{1}{2} [1 - (-1)] = 1 \).
06
Compute the final area
Combine the results from evaluating the integrals: \[ A = 2(\pi - 1) = 2\pi - 2 \]. Thus, the area of the region in the first quadrant enclosed by the given polar curve is \( 2\pi - 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area of Polar Regions
When dealing with curves described in polar coordinates, calculating the area of regions they enclose can seem daunting. However, with a good understanding of the formula and careful setup, it becomes manageable. The general formula to find the area of a region bounded by a polar curve is: \[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta \] Here, \( r \) represents the polar function and \( \theta \) is the angle measuring from a given line (often the positive x-axis). This formula helps us compute the area enclosed by the curve between two angles, \( a \) and \( b \).
- Determine the bounds: In our example, the region is within the first quadrant, meaning \( \theta\) runs from \(0 \) to \( \pi/2 \).
- Calculate \( r^2 \): Squaring the polar function gives us an expression we can integrate.
Definite Integral
Definite integrals play a crucial role in calculating areas under curves, especially those described by unique equations like polar coordinates. The definite integral evaluates the accumulation of quantities, such as area, between two bounds. This is key to area calculations:- **Integrate with limits**: The limits \( a \) and \( b \) define your interval, like \( 0 \) to \( \pi/2 \) in this polar problem.- **Apply to polar regions**: For an accurate area computation, we integrate the squared polar function \( r^2 \). The definite integral over a specified interval gives us the net area.For our problem, once \( r^2 \) is integrated between the limits, the definite integral neatly delivers the area, combining different parts of the function effectively. Thus, mastering definite integrals is indispensable for understanding area problems involving polar coordinates.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals, especially those that involve trigonometric functions. In our scenario involving this area problem: Using trigonometric functions often simplifies the integration process, particularly when dealing with functions like \( \sin \) or \( \cos \). Here's how it helped in the problem:
- For example, take the integral \( \int_{0}^{\pi/2} \sin 2\theta \, d\theta \). Normally, directly integrating \( \sin 2\theta \) can be a bit tricky, so substitution helps.
- **Set substitution**: Here, using \( u = 2\theta \) simplifies the integral. As a result, \( du = 2 \, d\theta\) helps in smoothly adjusting the integration limits and transforms the integration process into a standard recognizable form.
