Problem 9 - Essential Calculus

Question

Find the area of the region. Interior of \(r=1-\sin \theta\)

Step-by-Step Solution

Answer

The area of the region defined by \(r=1-\sin\theta\) is 0.75\(\pi\).

Step 1: Identify the range of \theta and graph

First, the given equation is \(r=1-\sin\theta\). As \(\sin\theta\) ranges between -1 and 1, the equation \(1 - \sin\theta\) will range from 0 to 2. If we plot this, we will notice that the plot is a circle that starts at the pole, goes to the point (0,1), around to (0,-1), and back to the pole. Hence, the limits of \(\theta\) will be from 0 to \(\pi\)

Step 2: Set Up the Integral

Using the formula for the area \(A = 0.5 \int_{a}^{b} r^2 d \theta\) where the limits of the integral are the range values of \(\theta\), the area integral becomes \(A = 0.5 \int_{0}^{\pi} (1-\sin\theta)^2 d \theta\)

Step 3: Evaluate the Integral

Calculating \(A = 0.5 \int_{0}^{\pi} (1-\sin\theta)^2 d \theta\), which equals \(A = 0.5 \int_{0}^{\pi} (1-2\sin\theta + \sin^2\theta) d \theta\). Using power-reduction identity, the integral can be simplified to \(A = 0.5 \int_{0}^{\pi} (1-2\sin\theta + 0.5 -0.5 \cos 2\theta) d \theta\). Now, calculating the integral, we get \(A = 0.5[ \theta - 2\cos\theta + 0.5\theta - (1/4)\sin 2\theta]\) evaluated from 0 to \(\pi\)

Step 4: Simplify the Expression

The result of the integral gives us \(A = 0.5[\pi - 0 + 0.5\pi - 0] - 0.5[0 - 2 + 0 - 2]\). Simplifying this, we end up with 0.75\(\pi\)

Key Concepts

Area Calculation

When calculating the area of a region in polar coordinates, we use a specific formula that differs from the Cartesian system. This is because polar coordinates are based on the radial distance from the origin and the angle, rather than horizontal and vertical distances. The formula for calculating the area enclosed by a curve in polar coordinates is:

• \( A = \frac{1}{2} \int_a^b r^2 \, d\theta \)

This integral sums up the area of infinitesimally small sectors defined by the angle \( \theta \) and radius \( r \). For each small slice, the area is approx \( \frac{1}{2} r^2 \, \Delta \theta \), and by integrating, we find the total area for the range of angles from \( a \) to \( b \).
For this problem, specifically involving the equation \( r = 1 - \sin \theta \), we find that the limits of \( \theta \) are from 0 to \( \pi \). This is seen by examining the behavior of \( \sin \theta \), which provides a complete description of the shape as it sweeps through this range.
The integration provides a comprehensive area enclosed by the curve, efficiently calculated using the above formula.

Trigonometric Integrals

Integrals involving trigonometric functions can often appear complex, but they usually follow a methodical approach. In our exercise, we have to evaluate the integral of \( (1 - \sin \theta)^2 \). Trigonometric integrals like this require breaking down the expression into simpler components to integrate easily.
Initially, expand the square: \( (1 - \sin \theta)^2 = 1 - 2 \sin \theta + \sin^2 \theta \). Each term is then dealt with independently:

• The integral of \( 1 \) with respect to \( \theta \) is simply \( \theta \).
• The integral of \( \sin \theta \) relates to \( -\cos \theta \), noting the standard integration rule \( \int \sin \theta \, d\theta = -\cos \theta \). The coefficient complicates it slightly, requiring factoring during integration.
• For \( \sin^2 \theta \), the integral is simplified using identities, specifically, the power-reduction identity covered next.

Understanding these small pieces makes evaluating the integral step-by-step more efficient and less daunting.

Power-Reduction Identity

The power-reduction identity is a valuable trigonometric tool that simplifies integrals involving squared trigonometric functions. Such identities help express powers of trig functions as functions of double angles, making them easier to integrate.
In our exercise, we encounter \( \sin^2 \theta \), which doesn't integrate directly and needs simplification. We use the power-reduction identity:

• \[ \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta) \]

This identity rewrites \( \sin^2 \theta \) into a form that is straightforward to integrate since it consists of easier terms like \( 1 \) and \( \cos 2\theta \).
When rewritten, the integral \( \int \sin^2 \theta \, d\theta \) becomes \( \int \frac{1}{2}(1 - \cos 2\theta) \, d\theta \), which is simply \( \frac{1}{2} \left( \theta - \frac{1}{2} \sin 2\theta \right) \).
These transformations through identities are crucial for simplifying and solving the integral efficiently. Once understood and applied, they make such integrals far more approachable.