Problem 29 - University Calculus: Early Transcendentals

Question

Find the area enclosed by one leaf of the rose \(r=12 \cos 3 \theta\).

Step-by-Step Solution

Answer

The area of one petal is \( 12\pi \).

Step 1: Understand the Polar Equation

The given polar equation is \( r = 12 \cos 3\theta \). This is a polar equation for a rose curve. The form \( r = a \cos k\theta \) can have \( \frac{k}{2} \) petals if \( k \) is even and \( k \) petals if \( k \) is odd. Here, \( k = 3 \), so there are 3 petals.

Step 2: Determine Petal Symmetry and Bounds

For the cosine-based rose curves, the petals are symmetric around the polar axis (the x-axis in Cartesian coordinates). One of the petals starts at \( \theta = 0 \) and ends at \( \theta = \frac{\pi}{3} \). We will calculate the area of one petal over this interval.

Step 3: Use the Polar Area Formula

The formula for the area enclosed by a polar curve \( r = f(\theta) \) is: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \] For our problem, \( r = 12 \cos 3\theta \) and we need the bounds \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \).

Step 4: Setup the Integral

Substitute \( r = 12\cos3\theta \) into the area formula: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} (12\cos3\theta)^2 \, d\theta \] = \( \frac{1}{2} \times 144 \int_{0}^{\frac{\pi}{3}} \cos^2 3\theta \, d\theta \).

Step 5: Simplify the Integral

Use the trigonometric identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to transform the integral: \[ A = 72 \int_{0}^{\frac{\pi}{3}} \frac{1 + \cos 6\theta}{2} \, d\theta \] = \( 36 \int_{0}^{\frac{\pi}{3}} (1 + \cos 6\theta) \, d\theta \).

Step 6: Evaluate the Integral

Split and integrate term by term: \[ A = 36 \left( \int_{0}^{\frac{\pi}{3}} 1 \, d\theta + \int_{0}^{\frac{\pi}{3}} \cos 6\theta \, d\theta \right) \] The first integral is \( \int_{0}^{\frac{\pi}{3}} 1 \, d\theta = \frac{\pi}{3} \) and the second is \( \int_{0}^{\frac{\pi}{3}} \cos 6\theta \, d\theta = \frac{1}{6} \sin 6\theta \Big|_0^{\frac{\pi}{3}} = 0 \).

Step 7: Calculate the Total Area

Combine the results: \( A = 36 \left( \frac{\pi}{3} + 0 \right) = 36 \times \frac{\pi}{3} = 12\pi \). Thus, the area of one petal is \( 12\pi \).

Key Concepts

Rose Curve

A rose curve in polar coordinates is a fascinating and beautiful mathematical pattern. It is given by the equations \( r = a \cos(k\theta) \) or \( r = a \sin(k\theta) \). The parameter \( k \) determines the number of petals the rose will display. Specifically, through these formulas, if \( k \) is even, the curve will have \( 2k \) petals, and if \( k \) is odd, there will be \( k \) petals. This makes rose curves unique and visually striking, perfect for illustrating symmetry and complexity in trigonometric curves.

• The parameter \( a \) dictates the length or amplitude of each petal.
• In cosine-based rose curves, the petals are symmetric about the polar axis (the horizontal axis in Cartesian coordinates).
• Each petal can be traced over a specific interval of \( \theta \), often zero to a fraction of \( \pi \).

This symmetry and periodic behavior make rose curves a classic example in studying the polar coordinate system and their applications in mathematics, art, and nature.

Polar Area Formula

The polar area formula is an integral formula used to compute the area enclosed by a curve in polar coordinates. When given a polar curve, described by \( r = f(\theta) \), the area \( A \) can be found using the following integral: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \] This formula accounts for the sweeping motion of the radius \( r \) as the angle \( \theta \) varies from \( \alpha \) to \( \beta \). The formula integrates the area of the infinitesimal sectors that \( r \) sweeps out as \( \theta \) changes.

• It uses the square of \( r \) to calculate the sector's area, ensuring all parts of the petal are captured.
• The integration limits \( \alpha \) and \( \beta \) are chosen so that they cover just a single petal of the rose curve, ensuring the area is only calculated for one section.

The polar area formula is especially helpful for complex shapes like rose curves, where traditional Cartesian methods might be cumbersome.

Trigonometric Identities

Trigonometric identities are essential tools in simplifying and solving equations involving trigonometric functions. They are particularly useful in calculus when integrating expressions involving trigonometric terms. A common identity used in integral calculations is the power-reduction formula: \[ \cos^2 x = \frac{1 + \cos 2x}{2} \]This identity transforms difficult-to-integrate expressions like \( \cos^2 x \) into simpler forms, breaking them into constant and cosine terms that can be integrated more easily.

• The identity allows us to convert products of trigonometric functions into sums or differences, which can be directly integrated.
• Such transformations are frequently necessary in evaluating integrals in polar coordinates.

Understanding these identities is crucial as they simplify the problem-solving process, especially when working with polar area calculations and multiple-angle formulas.