Problem 338 - Calculus Volume 3

Question

Find the area of the region. $$ r=1-\sin \theta \text { in the first quadrant } $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the area of the region defined by the polar equation \( r = 1 - \sin \theta \) within the first quadrant. This implies \( 0 \leq \theta \leq \frac{\pi}{2} \).

Step 2: Set up the Integral

The area \( A \) in polar coordinates is given by the integral \( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). Here, \( r = 1 - \sin \theta \), \( \alpha = 0 \), and \( \beta = \frac{\pi}{2} \). Therefore, the integral becomes:\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \sin \theta)^2 \, d\theta \]

Step 3: Expand the Integrand

Expand \((1 - \sin \theta)^2\) to simplify the integration:\( (1 - \sin \theta)^2 = 1 - 2 \sin \theta + \sin^2 \theta \).

Step 4: Integrate Term by Term

The integral becomes:\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( 1 - 2 \sin \theta + \sin^2 \theta \right) \, d\theta \]This can be separated into three integrals:\[ A = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - 2 \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta + \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \right) \]

Step 5: Compute Each Integral

Compute the integrals individually:1. \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left. \theta \right|_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \)2. \( \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = \left. -\cos \theta \right|_{0}^{\frac{\pi}{2}} = -0 - (-1) = 1 \)3. For \( \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \), use the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \): \[ \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos 2\theta \, d\theta \right) \] Calculating separately: - \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2} \) - \( \int_{0}^{\frac{\pi}{2}} \cos 2\theta \, d\theta = \left. \frac{1}{2} \sin 2\theta \right|_{0}^{\frac{\pi}{2}} = 0 \) Giving: \[ \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \]

Step 6: Combine Results

Substitute these results back into the overall equation for \( A \):\[ A = \frac{1}{2} \left( \frac{\pi}{2} - 2 \cdot 1 + \frac{\pi}{4} \right) = \frac{1}{2} \left( \frac{\pi}{2} + \frac{\pi}{4} - 2 \right) \]Combine terms:\[ A = \frac{1}{2} \left( \frac{2\pi}{4} + \frac{\pi}{4} - 2 \right) = \frac{1}{2} \left( \frac{3\pi}{4} - 2 \right) \]Final result:\[ A = \frac{3\pi}{8} - 1 \]

Step 7: Final Step: State the Result

The area of the region defined by \( r = 1 - \sin \theta \) in the first quadrant is \( \frac{3\pi}{8} - 1 \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with finding the quantities when the rate of change is known. One of its primary applications is to calculate areas under curves, volumes, and other related concepts through integration.
This method involves adding up an infinite number of small pieces to determine a whole. For example, calculating area or volume requires summing infinitesimally small units.
In the context of polar coordinates, we use a specific formula to find areas within curves expressed in terms of radius \( r \) and angle \( \theta \). The formula we use is:
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]
In our particular problem, the equation\( r = 1 - \sin \theta \)defined the curve in polar coordinates. By applying this formula, we compute the area enclosed by the curve in the specified region. Each component of our integrand contributes to calculating the infinitesimal changes in area as we transition through angle \( \theta \).

• Conceptually, we break down the curve into tiny segments through integration
• This process accumulates data to provide the entire area enclosed

Understanding this concept is crucial for a wide array of applications in physics, engineering, and other sciences.

Area Calculation

Calculating area involves integrating functions that define boundaries of regions. In the realm of polar coordinates, this becomes an interesting twist as curves can easily loop or spiral, requiring different considerations than Cartesian coordinates.
We calculate the area by setting up the integral that describes the region's boundary. Here, it's done using:
\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left(1 - \sin \theta \right)^2 \, d\theta \]
This formula accounts for both the radial distance (\( r \)) and angular rotation (\( \theta \)) to sum the infinitesimal area slices.
In the example, the original curve,\( r = 1 - \sin \theta \), was squared to manage the calculation through:
\[ \left(1 - \sin \theta \right)^2 = 1 - 2\sin \theta + \sin^2 \theta \]
We simplified this into separate integrals manageable individually.

• The process involved recognizing the need to solve each part of the expanded expression.
• Each piece represented a unique contribution to the total enclosed area.

By evaluating these integrals, we realized our final result, efficiently and accurately.

Trigonometric Identities

Trigonometric identities are essential tools that simplify expressions involving trigonometric functions. They play a critical role in solving integrals, especially those that appear in polar coordinates.
One crucial identity used is \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \). This enables us to rewrite expressions to facilitate integration. By substituting \( \sin^2 \theta \) with this identity, the process of integrating becomes easier because the expressions become somewhat linear in form.
Another useful identity in this context involves \( \cos 2\theta \), which appears during processes like double angle integration in trigonometric terms.
These identities help in converting complex trigonometric integrals into manageable parts. They allow transformation from products or powers of trigonometric functions to sums, which are usually more straightforward to integrate.

• Using such identities reduces complexity and streamlines calculations.
• They are particularly helpful in problems involving polar coordinates where expressions may otherwise become unwieldy.

Mastering these identities is invaluable for efficiently solving a wide range of calculus problems.