How do you find the area of one petal of r=2cos3theta?

Jun 10, 2017

$A = \frac{\pi}{3}$

Explanation:

First, graph $r = 2 \cos \left(3 \theta\right)$ to get an idea of what the petals look like. It can be really helpful to draw concentric circles and radial angle lines on graph paper, so that you have a polar graph, like this:

Next, using either a graphing utility or this graph paper, plot the graph using convenient points.

$\left.\begin{matrix}\theta & r \\ - - & - - \\ 0 & 2 \\ \pm \frac{\pi}{12} & \pm \sqrt{2} \approx \pm 1.41 \\ \pm \frac{\pi}{8} & \pm \sqrt{2 - \sqrt{2}} \approx \pm 0.77 \\ \pm \frac{\pi}{6} & 0\end{matrix}\right.$

The area of a petal can be determined by an integral of the form

$A = \frac{1}{2} {\int}_{\alpha}^{\beta} r {\left(\theta\right)}^{2} d \theta$

Notice the petal in Quadrant I and IV does not extend past $\pm \frac{\pi}{6}$ and that it is perfectly split between the two quadrants. That implies that if we can find the are of just half a petal, then we can multiply the result by two and get the area of the entire petal.

Letting the interval of integration go from $\theta = 0$ to $\theta = \frac{\pi}{6}$ and doubling the entire integration gives

$A = 2 \times \frac{1}{2} {\int}_{0}^{\frac{\pi}{6}} 4 {\cos}^{2} \left(3 \theta\right) d \theta$

$A = 4 {\int}_{0}^{\frac{\pi}{6}} {\cos}^{2} \left(3 \theta\right) d \theta$

The half-angle identity for cosine says

${\cos}^{2} \left(u\right) = \frac{1 + \cos \left(2 u\right)}{2}$, so

${\cos}^{2} \left(3 \theta\right) = \frac{1 + \cos \left(6 \theta\right)}{2}$.

This changes our integral to

$A = 4 {\int}_{0}^{\frac{\pi}{6}} \frac{1 + \cos \left(6 \theta\right)}{2} d \theta$

$A = 2 {\int}_{0}^{\frac{\pi}{6}} \left(1 + \cos \left(6 \theta\right)\right) d \theta$

$A = 2 {\int}_{0}^{\frac{\pi}{6}} d \theta + 2 {\int}_{0}^{\frac{\pi}{6}} \cos \left(6 \theta\right) d \theta$

Using $u$-substitution, you can determine the integral of the cosine function.

$A = 2 {\left[\theta\right]}_{0}^{\frac{\pi}{6}} + 2 {\left[\frac{1}{6} \sin \left(6 \theta\right)\right]}_{0}^{\frac{\pi}{6}}$

$A = \frac{\pi}{3}$