Problem 7 - The Calculus with Analytic Geometry

Question

Find the area of the region enclosed by one loop of the graph of the given equation.\(r=3 \cos 2 \theta\)

Step-by-Step Solution

Answer

The area is \(\frac{9\pi}{4}\).

Step 1: Identify the given equation

The given equation is in polar coordinates: \(r = 3 \cos(2\theta)\).

Step 2: Determine the range for one loop

For the curve \(r = 3 \cos(2\theta)\), one loop is completed when the argument of the cosine function, \(2\theta\), completes a cycle from \(0\) to \(2\pi\). Hence, \(\theta\) ranges from \(0\) to \(\pi\).

Step 3: Set up the integral for the area in polar coordinates

The area enclosed by a polar curve \(r = f(\theta)\) from \(\theta = a\) to \(\theta = b\) is given by: \[A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 d\theta.\]

Step 4: Calculate the integral

Evaluate the integral: \[A = \frac{1}{2} \int_{0}^{\pi} [3 \cos(2\theta)]^2 d\theta.\] This simplifies to: \[A = \frac{9}{2} \int_{0}^{\pi} \cos^2(2\theta) d\theta.\] Convert \(\cos^2(2\theta)\) using the identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\): \[A = \frac{9}{2} \int_{0}^{\pi} \frac{1 + \cos(4\theta)}{2} d\theta.\] This simplifies to: \[A = \frac{9}{4} \int_{0}^{\pi} (1 + \cos(4\theta)) d\theta.\]

Step 5: Evaluate the integral parts

The integral can be split into two parts: \[A = \frac{9}{4} \left[ \int_{0}^{\pi} 1 \d\theta + \int_{0}^{\pi} \cos(4\theta) \d\theta \right].\] The first part is: \[\frac{9}{4} \left[ \theta \right]_{0}^{\pi} = \frac{9}{4} (\pi - 0) = \frac{9\pi}{4}.\] The second part is: \[\frac{9}{4} \left[ \frac{1}{4} \sin(4\theta) \right]_{0}^{\pi} = \frac{9}{16} \left[ \sin(4\theta) \right]_{0}^{\pi}.\] Since \(\sin(4\theta)\) is zero at both \(0\) and \(\pi\), this term adds nothing to the area.

Step 6: Combine results to find the total area

The total area is: \[A = \frac{9\pi}{4}.\]

Key Concepts

Polar Coordinates

Polar coordinates are a way to represent points on a plane. Instead of using the traditional Cartesian coordinates \((x, y)\), polar coordinates use a radius \(r\) and an angle \(\theta\). Here's a simple breakdown:

• \(r\): the distance from the origin to the point.
• Angle \(\theta\): the angle from the positive x-axis to the point.

This method is especially useful when dealing with curves and areas in circular regions. For example, the equation \(r = 3 \cos(2\theta)\) represents a curve plotted using polar coordinates.

Definite Integral

A definite integral allows us to calculate the area under a curve within a specified range. If you have a function \(f(x)\) and you want to find the area from \(x = a\) to \(x = b\), you would set up an integral like this:
\[ \int_{a}^{b} f(x) \, d\theta \]
In the context of polar coordinates, we have a similar formula to calculate the area enclosed by a curve \(r = f(\theta)\):
\[ A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 \, d\theta \]
This formula accounts for the radial nature of the area.

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions that are true for all values of the variables. They are very useful for simplifying expressions and solving integrals. One key identity used in this exercise is:
\[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \]
This identity helps transform \(\text{cosine squared}\) terms into a more integrable form. For example, in the exercise, we used it to convert \(\cos^2(2\theta)\) into:
\[ \cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2} \]
which simplifies our definite integral.

Cosine Function

The cosine function, represented as \(\cos(x)\), is one of the primary trigonometric functions. It calculates the horizontal coordinate of a point on the unit circle. Key properties include:

• Periodicity: \(\cos(x + 2\pi) = \cos(x)\).
• Symmetry: \(\cos(-x) = \cos(x)\).
• Ranges between -1 and 1.

In the given exercise, we use the cosine function with a doubled argument, \(2\theta\). This impacts the period and symmetry of the curve:
\[ r = 3 \cos(2\theta) \]
This formula describes a four-petalled rose curve in polar coordinates, each petal representing one loop as \(2\theta\) ranges from 0 to \pi.