find the area of the region. three petals of $r = \\cos(9\\theta)$

Explanation:

Step1: Recall area formula in polar - coordinates

The area $A$ of a polar curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. For the rose - curve $r = \cos(n\theta)$, the petals are symmetric. For $r=\cos(9\theta)$, the period of $\cos(9\theta)$ is $T=\frac{2\pi}{9}$. Each petal is formed over an interval of length $\frac{\pi}{9}$.

Step2: Determine the limits of integration for one petal

To find the limits of integration for one petal of $r = \cos(9\theta)$, we set $r = 0$, so $\cos(9\theta)=0$. Then $9\theta=\frac{\pi}{2}+k\pi$, $k\in\mathbb{Z}$, and $\theta=\frac{\pi}{18}+\frac{k\pi}{9}$. For one petal, we can take the interval $\theta\in[-\frac{\pi}{18},\frac{\pi}{18}]$.

Step3: Calculate the area of one petal

Using the area formula $A_1=\frac{1}{2}\int_{-\frac{\pi}{18}}^{\frac{\pi}{18}}\cos^{2}(9\theta)d\theta$. Since $\cos^{2}(x)=\frac{1 + \cos(2x)}{2}$, then $\cos^{2}(9\theta)=\frac{1+\cos(18\theta)}{2}$. So $A_1=\frac{1}{2}\int_{-\frac{\pi}{18}}^{\frac{\pi}{18}}\frac{1+\cos(18\theta)}{2}d\theta=\frac{1}{4}\int_{-\frac{\pi}{18}}^{\frac{\pi}{18}}(1 + \cos(18\theta))d\theta$. Integrating term - by - term: $\frac{1}{4}\left[\theta+\frac{\sin(18\theta)}{18}\right]_{-\frac{\pi}{18}}^{\frac{\pi}{18}}=\frac{1}{4}\left[\left(\frac{\pi}{18}+\frac{\sin(\pi)}{18}\right)-\left(-\frac{\pi}{18}+\frac{\sin(-\pi)}{18}\right)\right]=\frac{\pi}{72}$.

Step4: Calculate the area of three petals

Since the area of one petal is $A_1=\frac{\pi}{72}$, the area of three petals $A = 3A_1$. So $A=\frac{\pi}{24}$.

Answer:

$\frac{\pi}{24}$