Problem 19 - Calculus Early Transcendentals

Question

\(17-21\) Find the area of the region enclosed by one loop of the curve. $$ r=3 \cos 5 \theta $$

Step-by-Step Solution

Answer

The area of one loop of the curve is \(\frac{9\pi}{20}\).

Step 1: Understand the Problem

We need to find the area of one loop of the polar curve given by the equation \( r = 3 \cos 5\theta \). The curve is symmetric and the cosine function completes one loop every time \( 5\theta \) goes through a \( 2\pi \) range, i.e., \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{5} \).

Step 2: Calculate Bounds for One Loop

Because each loop completes when \( \theta \) moves from \( 0 \) to \( \frac{\pi}{5} \), these will be our integration limits to find the area of one loop of the curve.

Step 3: Set Up Integral for Area

The area \( A \) enclosed by a curve in polar coordinates is given by \( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). Substitute \( r = 3\cos(5\theta) \) and the bounds \( 0 \) to \( \frac{\pi}{5} \): \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} (3\cos(5\theta))^2 \, d\theta \].

Step 4: Simplify Integral

Substitute and simplify: \( A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} 9 \cos^2(5\theta) \, d\theta \). This further simplifies to: \( A = \frac{9}{2} \int_{0}^{\frac{\pi}{5}} \cos^2(5\theta) \, d\theta \).

Step 5: Apply Trigonometric Identity

Use the identity \( \cos^2(x) = \frac{1+\cos(2x)}{2} \): \[ A = \frac{9}{2} \int_{0}^{\frac{\pi}{5}} \frac{1 + \cos(10\theta)}{2} \, d\theta \]. This simplifies the integral to: \[ A = \frac{9}{4} \int_{0}^{\frac{\pi}{5}} (1 + \cos(10\theta)) \, d\theta \].

Step 6: Integrate

Separate the integral: \[ A = \frac{9}{4} \left( \int_{0}^{\frac{\pi}{5}} 1 \, d\theta + \int_{0}^{\frac{\pi}{5}} \cos(10\theta) \, d\theta \right) \]. The first integral is straightforward: \[ \int_{0}^{\frac{\pi}{5}} 1 \, d\theta = \frac{\pi}{5} \]. For the second: \[ \int_{0}^{\frac{\pi}{5}} \cos(10\theta) \, d\theta = \left[ \frac{\sin(10\theta)}{10} \right]_{0}^{\frac{\pi}{5}} = 0 \].

Step 7: Calculate Area

Combine the results of the separate integrals: \[ A = \frac{9}{4} \left( \frac{\pi}{5} + 0 \right) = \frac{9\pi}{20} \].

Key Concepts

Polar Coordinates

Polar coordinates provide a way to specify the position of a point in the plane. Unlike the traditional Cartesian coordinates, which use a horizontal and a vertical axis, polar coordinates use a radius and an angle to define a point's location.
In polar coordinates:

• The "radius" (r) is the distance from the origin (center point) to the point.
• The "angle" (\(\theta\)) is made from a reference direction, usually the positive x-axis.

In the context of the given problem, the polar equation is \(r = 3 \cos(5\theta)\). This equation describes a curve composed of multiple loops depending on the cosine function, which can be more efficient to describe using polar coordinates instead of Cartesian coordinates. Understanding how this function behaves across different values of \(\theta\) allows us to conveniently calculate properties like area.

Trigonometric Identities

Trigonometric identities are essential tools in simplifying expressions and solving equations involving trigonometric functions like sine and cosine. They allow us to rewrite expressions to make them easier to work with, which is critical in integration and other calculations involving these functions.
One key identity used in this exercise is the double angle identity for cosine:

• \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\)

This identity is crucial for converting the \(\cos^2\) term into a more straightforward expression that can be integrated easily. For the curve \(r = 3 \cos (5\theta)\), simplifying the integral involves replacing \(\cos^2(5\theta)\) with \(\frac{1 + \cos(10\theta)}{2}\), making the integration process manageable.

Integral Calculus

Integral calculus helps us find quantities that are accumulated over an interval, such as area, volume, and total change. The definite integral, in particular, is used to calculate the area under a curve, which is the focus of this problem.
For polar curves, the area \(A\) enclosed by a curve \(r(\theta)\) from \(\theta=\alpha\) to \(\theta=\beta\) is given by the formula:

• \(A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\)

Utilizing this formula involves setting up an integral specific to the curve in question. In this exercise, \(r = 3\cos(5\theta)\) and the limits are from \(0\) to \(\frac{\pi}{5}\), representing one loop of the curve. After substituting and simplifying using trigonometric identities, the integral can be evaluated to find the area under the polar curve. This method of integration over a polar region allows us to precisely determine the size of each loop, which would be more complicated in Cartesian coordinates.