Problem 31 - Essential Calculus

Question

Conjecture Find the area of the region enclosed by \(r=a \cos (n \theta)\) for \(n=1,2,3, \ldots .\) Use the results to make a conjecture about the area enclosed by the function if \(n\) is even and if \(n\) is odd.

Step-by-Step Solution

Answer

The area enclosed by the function \(r=a \cos (n \theta)\), if \(n\) is odd, is \(\frac{\pi a^2}{4}\), and if \(n\) is even, the area is \(\frac{\pi a^2}{2}n\).

Step 1: Understanding the Polar Coordinate System

When given a function in polar coordinates, it represents a curve in a plane with an origin (also known as the pole) and a polar axis. For function \(r=a \cos (n \theta)\), there are several loops depending on the value of \(n\). If \(n\) is odd, there will be \(n\) loops. If \(n\) is even, there will be \(2n\) loops. This concept will be useful in predicting the area of the region enclosed by the function.

Step 2: Calculating the Area

Given the polar function, the area enclosed by the curve is given by \(\int \frac{1}{2} r^2 d\theta\). Since \(r = a \cos (n \theta)\), the formula becomes \(\int \frac{1}{2} (a \cos (n \theta))^2 d\theta = a^2/2 \int_0^\pi \cos^2(n\theta) d\theta\). Using the double angle identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\), the formula then becomes \(a^2/2 \int_0^\pi \frac{1 + \cos(2n\theta)}{2} d\theta\). Finally, this evaluates to \(a^2/4 \int_0^\pi (1 + \cos(2n\theta)) d\theta\).

Step 3: Evaluating the Integral

After evaluating the integral, we will end up with two terms: \(\frac{a^2}{4} [\pi + \frac{sin(2n\pi)}{2n}]\), where the second term will be zero since sine of a multiple of \(\pi\) is zero. So the area will be \(\frac{\pi a^2}{4}\), if \(n\) is odd. If \(n\) is even, there will be \(2n\) loops, hence the total area will be \(2n\) times that of when \(n\) is odd, which is \(\frac{\pi a^2}{2}n\).

Step 4: Making the Conjecture

From our calculations, we can make a conjecture that for the given polar function, when \(n\) is odd, the area enclosed by the curve is \(\frac{\pi a^2}{4}\), and when \(n\) is even, the area is \(\frac{\pi a^2}{2}n\).

Key Concepts

Polar Equations

Polar equations describe relationships between a radius and an angle, providing a way to represent curves in a plane. Unlike Cartesian equations that use a set of x and y coordinates, polar coordinates are expressed in terms of a radius \( r \) and an angle \( \theta \), with reference to the pole (the origin) and polar axis. In the equation \( r = a \cos(n\theta) \), \( n \) determines the number of loops the graph will make around the pole.

• If \( n \) is odd, the graph will feature \( n \) distinctive loops.
• If \( n \) is even, there will be \( 2n \) loops.

This understanding is crucial when predicting the shape and area enclosed by the polar function, as it directly influences the curve's lobes' frequency and symmetry around the pole.

Area Calculation

To calculate the area enclosed by a polar curve, we employ a specific integral formula. The essential formula for this is:\[\text{Area} = \int \frac{1}{2} r^2 d\theta\]For our function \( r = a \cos(n\theta) \), this becomes:\[\int \frac{1}{2} (a \cos(n\theta))^2 d\theta\]By simplifying this using trigonometric identities, namely the double angle identity: \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \), it expands to:\[\int \frac{1}{2} \left(a^2 (\frac{1+\cos(2n\theta)}{2})\right) d\theta\]These steps creatively simplify the integration, aiding in determining the area enclosed by complicated polar curves. This particular transformation is especially helpful because it converts a complex trigonometric expression into more manageable terms.

Integral Calculus

Integral calculus is crucial in finding the area under curves or enclosed by them, especially in polar coordinates. Once the integral is set up from the polar equation, we evaluate it over the appropriate interval, here from \( 0 \) to \( \pi \).

After evaluating \[\int_0^{\pi} (1 + \cos(2n\theta)) d\theta,\]we find two primary components: one involves a constant while the other diminishes because \( \sin(2n\pi) \) equals zero. Hence, if \( n \) is odd, our result reveals the area:\[\frac{\pi a^2}{4}\]For even \( n \), since the graph consists of \( 2n \) loops, the area becomes:\[\frac{\pi a^2}{2}n\]This demonstrates how integral calculus helps solve complex area problems by breaking down larger integrals into simpler parts. It exemplifies how recognizing periodic properties of trigonometric functions like sine and cosine simplifies what might seem intricate at first glance.