# Explain why for $r=1-a \cos^2(3\theta)$ the leaves have the same size only in the case $a=1$ and $a=2$

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?

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When you say size do you mean volume? Because compared individually, they definitely don't have the same size. –  Pedro Tamaroff Apr 8 '12 at 19:49

## 1 Answer

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.

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Thank you so much. –  Lev Apr 8 '12 at 20:29
Also can u please tell me how do i algebraically determine how many leaves a function like this have? –  Lev Apr 8 '12 at 21:29
Start by looking at the graph of $r$ versus $\theta$, with $0 \le \theta 2\pi$. Every time $r=0$ corresponds to a return to the origin, i.e., the end of one leaf and the start of another. By counting the number of such points, you can get an upper bound on the number of leaves. The reason this might not give you the exact number of leaves is because some leaves my be identical (this happens in the $a=1$ case). To check if this happens you need to compare $r(\theta)$ to $r(\theta+\pi)$: if $r(\theta)=-r(\theta+\pi)$ for any $0\le \theta < \pi$, then there will be repetition. –  Matthew Conroy Apr 8 '12 at 21:40
Thanks Matthew, you are a life saver. –  Lev Apr 8 '12 at 21:50