Question

For the following exercises, find the area of the region. $$ r=1-\sin \theta \text { in the first quadrant } $$

Short answer

The area is \( \frac{3\pi}{8} - 1 \).

Step-by-step solution

Understand the Equation of the Curve

The given polar equation is \( r = 1 - \sin \theta \). This defines a limaçon 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 \( \frac{\pi}{2} \).

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 \( \theta = \frac{\pi}{2} \).

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:\[ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( 1 - 2\sin \theta + \sin^2 \theta \right) \, d\theta \].

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:- \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left. \theta \right|_0^{\frac{\pi}{2}} = \frac{\pi}{2} \)- \( \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = -\cos \theta \Big|_0^{\frac{\pi}{2}} = -[0 - (-1)] = 1 \)- \( \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \theta - \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \).

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 limaçon curve.

Key concepts

Limaçon Curve

A limaçon curve is a special type of polar graph that is shaped like a snail or a limaçon. The general form of a limaçon 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 limaçon depends on the relation between \( a \) and \( b \):

• If \( a = b \), the curve is a cardioid.
• If \( a > b \), the curve is a dimpled limaçon (no loop).
• If \( a < b \), the limaçon has an inner loop.

In our exercise, the equation is \( r = 1 - \text{sin} \theta \), which indicates it could be a dimpled limaçon or one with no loop, as \( a = 1 \) and \( b = 1 \). Understanding the shape of the curve helps when calculating areas, as it gives insight into the integration limits in polar coordinates.

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 limaçon 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:

• 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.

By solving these integrals, you find the total area of interest. Each part of the integration represents a contribution to the overall sum, showcasing how integral calculus allows for the precise calculation of areas defined by complex curves in polar coordinates.