I was doing some work finding the areas of rose curves. The rose curve is a polar curve given by the equation
$$ r(\theta) = \cos{k\theta} $$ When $k$ is even, the area is $\pi/2$, and when $k$ is odd, the area is $\pi/4$. But I got stuck when thinking about the area of the rose curve when $k$ is rational and not an integer. The petals formed by such a rose curve all overlap each other, so what part of the curve would be considered the area? Would the area be the entire area surrounded by the curve, or would it be only the "inner" parts of the curve?
Furthermore, if $k$ is irrational, then the rose will have infinitely many petals and thus have infinite length. This means that the curve forms a dense set on the unit disc. So would the area of this curve be $\infty$, or would it be the area of a circle that encloses the curve?
EDIT: Maybe a better generalization of the question is; what is defined as the area of a self-intersecting figure?
$I_{even} = 2\pi / (\pi/k)\cdot \int_{-\pi/(2k)}^{\pi/(2k)} \cos^2(k\theta)/2\; d\theta = k\cdot [\theta/2+\sin(2k\theta)/4k] = k\cdot (\pi/(2k) + 0) = \pi/2$ for $k \in \mathbb{2Z}$. Also, note that $I_{odd} = 1/2\cdot I_{even} = \pi/4$ for $k$ odd. Note that the circle of radius $1$ contains this curve and so the area must be less than $\pi$. When $k$ is not an integer but rational, then the curve is not well-defined (think +$2\pi$). If irrational, then the curve is not closed, so how could the area even be defined?
Feel free to ask more questions if required.