Problem 10 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The area of the region enclosed by the polar curve \(r=1-\sin\theta\) above the polar axis is \( \frac{7\pi}{12} - \frac{\sqrt{3}}{4}\) square units.

Step 1: Graph and Identify Limits of Integration

Plot the polar equation \(r=1-\sin\theta\) and distinguish the sector that lies above the polar axis. The polar graph is a circle with radius 1/2 and center at (1/2,0) in polar coordinates. Observe that the circle intersects the polar axis at \(\theta=7\pi/6\) and \(\theta=11\pi/6\). These are the limits of integration.

Step 2: Set up the Integral

The formula for the area \(A\) in polar coordinates is \( A = 1/2\int_{a}^{b} r(\theta)^2 d\theta \). For the given function, \(r=1-\sin\theta\), the integral becomes \( A = 1/2\int_{7\pi/6}^{11\pi/6} [1-\sin(\theta)]^2 d\theta \).

Step 3: Evaluate the Integral

Simplify the integrand and integrate: \[\int [1-\sin(\theta)]^2 d\theta = \int (1 - 2sin\theta + \sin^2\theta) d\theta \]. Firstly, evaluate \(\int 1 d\theta\), Secondly, evaluate \(\int -2\sin\theta d\theta\), Then, evaluate \(\int\sin^2\theta d\theta\) using power-reduction formula. Add these results and then evaluate the complete integral from \(\theta=7\pi/6\) to \(\theta=11\pi/6\).

Step 4: Simplify and Finish

After evaluating definite integral, simplify the final expression to get the area of the region.

Key Concepts

Polar Equation Graphing

Understanding how to graph polar equations is key to solving various problems in polar coordinate systems. The polar equation in the exercise, r = 1 - \(sin \theta\), generates a curve known as a limaçon. To graph it, we plot points for different values of \(\theta\) and connect them to visualize the shape. For this equation, there is symmetry about the polar axis, and this symmetry can simplify graphing. Once plotted, we see that the graph intersects the polar axis at specific points, which plays a vital role in identifying the limits of integration for calculating the area.

Limits of Integration

The limits of integration define the interval over which we integrate to calculate area. To establish these for polar equations, we observe the points of intersection on the polar axis or other boundaries of interest. From the solution provided, the equation r = 1 - \(sin \theta\) intersects the polar axis at \(\theta = 7\pi/6\) and \(\theta = 11\pi/6\). These points of intersection indicate the bounds within which we will integrate to find the area of the polar region above the polar axis. Recognizing these limits is crucial because they confine the section of the graph we are concerned with.

Definite Integral Evaluation

The process of definite integral evaluation is about calculating the area under a curve within specific limits. When handling polar coordinates, we use the formula A = 1/2 \(\int_{a}^{b} r(\theta)^2 d\theta\). In our exercise, this translates into integrating [1 - \(sin(\theta)\)]^2 from \(\theta = 7\pi/6\) to \(\theta = 11\pi/6\). Breaking down the integrand and evaluating each part sequentially simplifies the process. The sum of these evaluations gives the area bounded by the polar curve. Students should take care to solve each part of the integral methodically and combine them to obtain the correct area.

Power-Reduction Formula

The power-reduction formula is an important tool in integration, especially when dealing with trigonometric functions. It is used to convert higher powers of sine and cosine into expressions involving the first power, which can be integrated using basic rules. In the context of the given problem, to integrate \(sin^2\theta\), we use the power-reduction formula, \(sin^2\theta = (1 - cos(2\theta))/2\). By applying this formula, we simplify the integral into a form that can be evaluated easily. It is essential for students to familiarize themselves with trigonometric identities like the power-reduction formula, as they are vital to simplifying and solving integrals in calculus.