Problem 30 - Essential Calculus

Question

Find the area of the region. Common interior of \(r=a \cos \theta\) and \(r=a \sin \theta\) where \(a>0\)

Step-by-Step Solution

Answer

The area of the region that is the common interior of \(r=a \cos \theta\) and \(r=a \sin \theta\) where \(a>0\) is given by \(a^2/8 * [-2\theta + \sin(2\theta)+2\theta]\) evaluated between \(0\) and \(\pi/4\), plus \(a^2/8 * [2\theta - \sin(2\theta)-2\theta]\) evaluated between \(\pi/4\) and \(\pi/2\).

Step 1: Understand the Graphs

Since both equations represent circles of radii \(a/2\) with centres at \(a/2\) along x and y axes, you should notice that these circles intersect at the origin and at points where \(\cos \theta = \sin \theta\), hence at \(\theta = \pi/4\) and \(5\pi/4\). However, since \(a > 0\) and \(\theta\)'s maximal value is \(\pi\), we limit our observation to quadrants I and IV.

Step 2: Setup the Integral

Please remember that when you calculate the area of a region in polar coordinates, you use the equation \(\frac{1}{2}\int_{\alpha}^{\beta} [f(\theta)]^2 d\theta\). Our integral will be \(\frac{1}{2}\int_{0}^{\pi/4} [(a \cos \theta)^2 - (a \sin \theta)^2] d\theta + \frac{1}{2}\int_{\pi/4}^{\pi/2} [(a \sin \theta)^2 - (a \cos \theta)^2] d\theta\). Here, the order of subtraction is important since \(a \cos \theta > a \sin \theta\) for \(\theta \in [0, \pi/4]\) and \(a \sin \theta > a \cos \theta\) for \(\theta \in [\pi/4, \pi/2]\).

Step 3: Evaluate the Integral

We can solve the integrals using the Pythagorean identity (\(\sin^2 \theta + \cos^2 \theta = 1\)) and the integral of \(sin^2\) and \(cos^2\) (which is \(\theta/2 + \sin(2\theta)/4\)). Thus \(\int_{0}^{\pi/4} [(a \cos \theta)^2 - (a \sin \theta)^2] d\theta = a^2/8 [-2\theta + \sin(2\theta)+2\theta]\) evaluated between \(0\) and \(\pi/4\), and \(\int_{\pi/4}^{\pi/2} [(a \sin \theta)^2 - (a \cos \theta)^2] d\theta = a^2/8 [2\theta - \sin(2\theta)-2\theta]\) evaluated between \(\pi/4\) and \(\pi/2\). Adding both results will give us our final answer.

Key Concepts

Integral Setup for Polar Regions

When you're tasked with finding the area of a region defined in polar coordinates, you start by visualizing the shapes represented by the equations. In polar coordinates, the area can be determined by an integral that takes the form \[\begin{equation}\frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta, \end{equation}\] In the given exercise, we are dealing with two circles defined by their polar equations. The process of setting up the integral involves identifying the limits of integration, which corresponds to the points where the two curves intersect, or the boundaries of the region. Once these limits are determined, you can create an integral for each portion of the area you're interested in. In some cases, like with our example where we have two regions (from 0 to \pi/4 and from \pi/4 to \pi/2), you will have separate integrals that will be summed to find the total area.

Evaluating Trigonometric Integrals

Evaluating trigonometric integrals is a common task in calculus, especially when dealing with polar coordinates. It involves integrating powers of sine and cosine over a certain interval. In the context of our problem, after setting up the correct integrals, we must evaluate them which entails using trigonometric identities and known integral forms. For instance, integrals containing \sin^2\theta or \cos^2\theta can be calculated using the half-angle identities, making the otherwise challenging integration manageable. To solve these, you can employ techniques such as substitution or apply known integrals like \[\begin{equation}\int \sin^2\theta d\theta = \frac{\theta}{2} + \frac{\sin(2\theta)}{4}\text{ and }\int \cos^2\theta d\theta = \frac{\theta}{2} - \frac{\sin(2\theta)}{4}.\end{equation}\] These will lead to the result when applied within the limits of integration for our specific problem.

Intersection of Polar Curves

Understanding the intersection of polar curves is crucial in solving many calculus problems involving areas and lengths of polar curves. In our exercise, we're looking at the common interior of two circles, one described by \[\begin{equation} r = a \cos \theta, \end{equation}\] and the other by \[\begin{equation} r = a \sin \theta. \end{equation}\] These two curves intersect at points where the values of r given by both equations are the same, which leads us to find solutions to the equation \[\begin{equation} a \cos \theta = a \sin \theta. \end{equation}\] This happens at \theta = \pi/4 and 5\pi/4; however, we only consider the interval from 0 to \pi as a > 0 and due to the curve symmetries. Recognizing the intersection points allows us to determine the bounds for our integrals.

Pythagorean Identity Application

The Pythagorean identity is an essential tool in trigonometry and calculus, particularly when dealing with problems involving sine and cosine. It states that for any angle \theta, \[\begin{equation} \sin^2 \theta + \cos^2 \theta = 1. \end{equation}\] In solving our exercise's integrals, this identity allows us to simplify the integrands by recognizing that \[\begin{equation} \cos^2 \theta - \sin^2 \theta \end{equation}\] is equivalent to \[\begin{equation} 1 - 2\sin^2 \theta \end{equation}\] or \[\begin{equation} 2\cos^2 \theta - 1. \end{equation}\] By applying the Pythagorean identity, we can transform the original integral into a more solvable form, using the known integral values of \sin^2\theta and \cos^2\theta. Further, these transformations are pivotal in calculating the desired area enclosed between the polar curves in the problem.