Problem 6 - Thomas Calculus

Question

Find the areas of the regions in Exercises \(1-8\) Inside one leaf of the three-leaved rose \(r=\cos 3 \theta\)

Step-by-Step Solution

Answer

The area of one leaf is \( \frac{\pi}{12} \).

Step 1: Graph the Polar Equation

Begin by understanding that the equation of the rose curve is given by the polar equation \( r = \cos 3\theta \). The curve forms a three-leaved rose. By graphing this, you can visualize the three symmetrical petals that the curve presents.

Step 2: Identify Symmetrical Petal

Since the curve is symmetric and creates three petals, each petal represents approximatively one-third of the entire area. Focus on one petal, as solving for its area can then be multiplied by 1, if needed, since we only need the area inside one leaf.

Step 3: Determine Limits of Integration

The angle \( \theta \) for one petal of the three-leaved rose ranges from \( -\frac{\pi}{6} \) to \( \frac{\pi}{6} \). This is due to the periodic nature of the cosine function, and it is key to compute the integral within these limits.

Step 4: Setup Area Integral in Polar Coordinates

The area \( A \) inside one petal is found by the integral: \[A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (\cos 3\theta)^2 \, d\theta\]. This accounts for the symmetry by considering one petal between these limits.

Step 5: Solve Integral

Change variable by using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\), then: \[A = \frac{1}{2} \left( \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{1 + \cos 6\theta}{2} \, d\theta \right)\].Split the integral into two parts: \[= \frac{1}{2} \left(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{1}{2} \, d\theta + \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\cos 6\theta}{2} \, d\theta \right)\].

Step 6: Evaluate Integrals Separately

Calculate: \[\frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{1}{2} \, d\theta = \frac{1}{4} \cdot \left(\frac{\pi}{6} + \frac{\pi}{6} \right) = \frac{\pi}{12}\]For the second part, \(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos 6\theta \, d\theta\) evaluates to zero due to the cosine integral properties over its symmetrical cycle.

Step 7: Calculate the Final Area

Combine the results: \[A = \frac{1}{2} \cdot \left( \frac{\pi}{12} + 0 \right) = \frac{\pi}{12}\] Thus, the area of one leaf of the rose is \( \frac{\pi}{12} \).

Key Concepts

Integration in Polar Coordinates

When dealing with polar coordinates, integration takes a slightly different approach compared to Cartesian coordinates. In polar coordinates, a point in the plane is determined by a distance from the origin and an angle from the positive x-axis. This non-linear aspect makes polar integration unique.
This involves integrating over an angle instead of a typical x-y plane. The formula for the area of a region in polar coordinates is \( A = \frac{1}{2} \int r^2 \, d\theta \). Here, \( r \) is a function of \( \theta \), which represents the radial distance at a given angle.

• The \( \frac{1}{2} \) factor accounts for the way the polar area sweeps out an entire sector as it integrates.
• Integration limits, \( \theta_1 \) to \( \theta_2 \), are crucial as they define the arc over which you are measuring.

Understanding this form of integration allows for precise calculation of areas enclosed by polar curves.
It involves breaking down complex regions into manageable integral calculations.

Area of Polar Curves

Calculating the area enclosed by polar curves can sometimes seem daunting, but it’s quite direct once you grasp the concept. One common example is the rose curve, defined by an equation like \( r = \cos 3\theta \) in our exercise.
The area of a single leaf of the rose is the focus, which you find by integrating correctly over the designated interval. Each leaf reflects symmetry around the origin.
The general process involves:

• Graphing the curve to understand its shape and symmetry.
• Determining integration limits: for \( r=\cos 3\theta \), one leaf spans from \( -\frac{\pi}{6} \) to \( \frac{\pi}{6} \).
• Setting up the integral using the formula \( A = \frac{1}{2} \int r^2 \, d\theta \) for one leaf.

A typical step would include simplifying the integral with trigonometric identities such as \( \cos^2 x = \frac{1 + \cos 2x}{2} \). This simplifies calculations and accurately accounts for all parts within the interval.

Cosine Function Properties

The cosine function plays a pivotal role in understanding polar curves, especially those described by equations like \( r = \cos 3\theta \).
The cosine function is fundamental because:

• It has a periodic nature, repeating every \( 2\pi \), which helps in defining symmetrical curves like the rose with leaves at regular intervals.
• It’s an even function, meaning \( \cos(-\theta) = \cos(\theta) \), which allows for symmetry in polar graphs.
• Trigonometric identities, such as \( \cos^2 x = \frac{1 + \cos 2x}{2} \), are often used to simplify integrals by transforming squared cosine terms into manageable linear terms.

This even property of cosine is crucial when evaluating integrals over symmetric limits, often resulting in simpler zero results. In the exercise, this property ensures certain parts of the integral, like \( \int \cos 6\theta \, d\theta \), equate to zero, simplifying the area calculation considerably.