Problem 13 - The Calculus with Analytic Geometry

Question

Find the area of the intersection of the regions enclosed by the graphs of the two given equations.\(\left\\{\begin{array}{l}r=3 \sin 2 \theta \\ r=3 \cos 2 \theta\end{array}\right.\)

Step-by-Step Solution

Answer

The area of the intersection is \frac{9\pi}{8}.

Step 1: Understand the Polar Equations

Identify the given equations in polar coordinates: \[ r = 3 \sin 2\theta \] and \[ r = 3 \cos 2\theta \]. These represent rose curves.

Step 2: Determine the Intersection Points

Set the equations equal to find the points of intersection: \[ 3 \sin 2\theta = 3 \cos 2\theta \]. Simplify to get \[ \tan 2\theta = 1 \] which gives \theta = \frac{\pi}{8} and \theta = \frac{5\pi}{8}.

Step 3: Set Up the Integral for Area

The area of the intersection is given by the integral of the absolute difference of the functions divided by two within the limits of intersection: \[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} (3 \sin 2\theta)^2 d\theta \].

Step 4: Evaluate the Integral

Simplify \(3 \sin 2\theta\)^2 to get 9 \sin^2 2\theta. The integral becomes \[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} 9 \sin^2 2\theta d\theta \]. Use the identity \sin^2 x = \frac{1-\cos 2x}{2} to convert and integrate.

Step 5: Final Calculation

Completing the integral for \[ \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} \frac{9}{2} (1-\cos 4\theta) d\theta \] gives: \[ A = \frac{9}{4}\left[\theta - \frac{\sin 4\theta}{4}\right]_{\frac{\pi}{8}}^{\frac{5\pi}{8}} \]. Evaluate and simplify this to find the area \frac{9\pi}{8}.

Key Concepts

Understanding Polar Coordinates

In polar coordinates, any point on the plane is defined by a radius, denoted as r, and an angle, denoted as \( \theta \). This system is especially useful when dealing with circular or rotational symmetries. Unlike Cartesian coordinates \((x, y)\), polar coordinates can make equations involving spirals, circles, and other curves more straightforward.
One of the key aspects of polar coordinates to remember is that:

• The radius \( r \) is the distance from the origin to the point.
• The angle \( \theta \) is the angle formed with the positive x-axis.

For example, the equations \( r = 3 \, \text{sin} \, 2 \theta \) and \( r = 3 \, \text{cos} \, 2 \theta \) are examples of what are known as 'rose' curves, because their plots resemble petals of a rose.

Finding Intersection Points

To solve for the area of intersection between two curves, we first need to know where those curves intersect. In a polar coordinate system, we set the equations equal to each other and solve for \( \theta \).
Given:

• \( 3 \, \text{sin} \, 2\theta = 3 \, \text{cos} \, 2\theta \)

We simplify to get:

• \( \text{tan} \, 2\theta = 1 \)

Solving this returns the key angles:

• \( \theta = \frac{\pi}{8} \)
• \( \theta = \frac{5\pi}{8} \)

These angles represent the points where the two curves intersect, allowing us to set up the bounds for our integral accurately.

Using Definite Integrals for Area

We use definite integrals to find the area enclosed by curves in polar coordinates. The general formula for the swept area from angle \( \alpha \) to \( \beta \) is:
\[ A = \frac{1}{2} \, \int_{\alpha}^{\beta} r^2 \, d\theta \]
Here, the integral works by summing small wedges of areas. Each wedge has area proportional to \( r^2 \).
For our given functions \( r = 3 \, \text{sin} \, 2\theta \) and \( r = 3 \, \text{cos} \, 2\theta \), we calculate:
\[ A = \frac{1}{2} \, \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} (3 \, \text{sin} \, 2\theta)^2 \, d\theta \]Simplifying the expression under the integral will help us solve this step more effectively.

Applying Trigonometric Identities

Trigonometric identities simplify the process of integrating functions of trigonometric forms. For this problem, we use the Pythagorean identity:
\[ \text{sin}^2 x = \frac{1 - \text{cos} \, 2x}{2} \]
Applying it to our integral:
\[ A = \frac{1}{2} \, \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} 9 \, \text{sin}^2 \, 2\theta \, d\theta \]
Simplifies to:
\[ A = \frac{9}{4} \, \int_{\frac{\pi}{8}}^{\frac{5\pi}{8}} (1 - \text{cos} \, 4 \theta) \, d\theta \]
This turns our integral into an easier form. Evaluating this integral, we get:
\[ A = \frac{9}{4} \left[ \theta - \frac{\text{sin} \, 4\theta}{4} \right]_{\frac{\pi}{8}}^{\frac{5\pi}{8}} \], which simplifies to a final area calculation of:
\[ A = \frac{9\pi}{8} \].