Please explain why for $r=1-a\cos^2(3\theta)$ the leaves have the same size only in the case $a=1$ and $a=2$.
Does anyone have an answer to this please?
The reason can be seen in a graph of $r$ versus $\theta$.
The leaves are created by the intervals $[u,v]$ on which $r(u)=r(v)=0$ and for which r is non-zero anywhere else on $[u,v]$.
In the case of $a=1$, you have this graph:
On this we can see that all the loops will be identical. Similarly with $a=2$; the graph looks like this:
We can see the symmetry in this graph: the loops created by the intervals of $\theta$ on which $r>0$ are identical to those created by the intervals of $\theta$ on which $r<0$.
Once we increase $a$ beyond 2, however, the graph of $r$ versus $\theta$ is no longer balanced about the horizontal axis: for $a>2$, the graph extends farther below the horizontal axis than above; as a result, there are two different size leaves. Here is $r$ versus $\theta$ for $a=3$, for example:
The intervals where $r$ is positive create smaller leaves than those for which $r$ is negative: the maximum $r$ is always 1, and the minimum is $1-a$. Thus the polar curve will have leaves of length 1, and leaves of length a-1.
This situation will hold for all $a>1, a\ne 2$: there will be two sizes of leaves. For $a<1$, there are no leaves, since $r$ is strictly positive.