The area of a single petal/leaf of a rose curve

Question

The function given is $r=12\cos(3\theta)$, the graph of this function shows a $3$ petal/leaf rose. Now one way to find the area of a single petal is to do $\frac{1}{3}\int_{0}^{2π}\int_{0}^{12\cos(3\theta)} r drd\theta$ this gives the value $24\pi$
another way which should give the same value is $2 \int_{0}^{\pi/6}\int_{0}^{12\cos(3\theta)} r drd\theta$
but this equals $12\pi$.
What is wrong here?

Answer 1

The following animation (created with PSTricks) might be useful for finding the integration interval.

For the former, you have to change the interval of the outer integral as follows. \begin{align} A&=\frac{1}{3}\int_0^\pi\int_0^{12\cos 3\theta} r\,\mathrm{d}r\,\mathrm{d}\theta\\ &=12\pi \end{align}