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.

Short 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-by-step solution

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.

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 \(a^2/4 \int_0^\pi (1 + \cos(2n\theta)) d\theta\).

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 \(\frac{\pi a^2}{2}n\).

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 \(\frac{\pi a^2}{2}n\).