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Magazine Chapter 6: Problem 102
For the following exercises, find the area of the region. $$ r=1-\sin \theta \text { in the first quadrant } $$
Short Answer
Expert verified
The area is \( \frac{3\pi}{8} - 1 \).
Step by step solution
01
Understand the Equation of the Curve
The given polar equation is \( r = 1 - \sin \theta \). This defines a lima莽on curve. Our task is to find the area of this region in the first quadrant, i.e., where both \( r \) and \( \theta \) are positive and \( \theta \) is between 0 and \( \frac{\pi}{2} \).
02
Set Up the Area Integral
The area \( A \) of a region in polar coordinates is given by the integral: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \] where \( r \) is the function of \( \theta \), and \( \alpha \) and \( \beta \) are the bounds of \( \theta \). For the first quadrant, we'll integrate from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \).
03
Substitute and Simplify the Integral
Substitute \( r = 1 - \sin \theta \) into the integral: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \sin \theta)^2 \, d\theta \] Expand the square: \( (1 - \sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta \), so the integral becomes:\[ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( 1 - 2\sin \theta + \sin^2 \theta \right) \, d\theta \].
04
Integrate Term by Term
Separate the integral:\[ A = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - 2\int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta + \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \right) \]Integrate each term:- \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left. \theta \right|_0^{\frac{\pi}{2}} = \frac{\pi}{2} \)- \( \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = -\cos \theta \Big|_0^{\frac{\pi}{2}} = -[0 - (-1)] = 1 \)- \( \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \theta - \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \).
05
Calculate the Final Area
Now combine the results:\[ A = \frac{1}{2} \left( \frac{\pi}{2} - 2(1) + \frac{\pi}{4} \right) = \frac{1}{2} \left( \frac{\pi}{2} - 2 + \frac{\pi}{4} \right) \]Simplify:\[ A = \frac{1}{2} \left( \frac{3\pi}{4} - 2 \right) = \frac{3\pi}{8} - 1 \].This is the area in the first quadrant for the given lima莽on curve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lima莽on Curve
A lima莽on curve is a special type of polar graph that is shaped like a snail or a lima莽on. The general form of a lima莽on is given by the equation \( r = a + b \, ext{sin} \theta \) or \( r = a + b \, ext{cos} \theta \), where \( a \) and \( b \) are constants. The appearance of the lima莽on depends on the relation between \( a \) and \( b \):
- If \( a = b \), the curve is a cardioid.
- If \( a > b \), the curve is a dimpled lima莽on (no loop).
- If \( a < b \), the lima莽on has an inner loop.
Area Calculation
When calculating the area enclosed by a curve in polar coordinates, the formula used is:
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]
Here, \( A \) represents the area, \( r \) is the polar equation of the curve, and \( \alpha \) to \( \beta \) are the angles defining the section of interest.
For example, in the case of the lima莽on curve \( r = 1 - \sin \theta \) from the exercise, we calculate the area in the first quadrant. Thus, \( \alpha \) and \( \beta \) are 0 and \( \frac{\pi}{2} \), respectively. Once the formula is set, it involves integrating the expression \( (1 - \sin \theta)^2 \), which expands to \( 1 - 2\sin \theta + \sin^2 \theta \).
Integrating each term separately helps to simplify the calculation. This specific kind of calculation is common, allowing us to find precise areas enclosed by curves defined in polar coordinates.
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]
Here, \( A \) represents the area, \( r \) is the polar equation of the curve, and \( \alpha \) to \( \beta \) are the angles defining the section of interest.
For example, in the case of the lima莽on curve \( r = 1 - \sin \theta \) from the exercise, we calculate the area in the first quadrant. Thus, \( \alpha \) and \( \beta \) are 0 and \( \frac{\pi}{2} \), respectively. Once the formula is set, it involves integrating the expression \( (1 - \sin \theta)^2 \), which expands to \( 1 - 2\sin \theta + \sin^2 \theta \).
Integrating each term separately helps to simplify the calculation. This specific kind of calculation is common, allowing us to find precise areas enclosed by curves defined in polar coordinates.
Integral Calculus
Integral calculus involves finding the accumulation of quantities, and in this context, it is used to calculate areas under curves. The process involves setting up an integral based on the function provided and then integrating to find the sum of infinitesimal areas.
When solving for the area using polar coordinates:
When solving for the area using polar coordinates:
- First, convert the function into a square form ready for integration, such as \( (1 - \sin \theta)^2 \).
- Next, expand the expression and integrate each component separately over the defined interval.
- The integral \( \int \, \sin^2 \theta \, d\theta \) requires a trigonometric identity \( \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta) \) for simplification.
