Originally Posted by

**shawsend** I'm pretty sure they both have the same loci of points. That is, if $\displaystyle \mathbb{S}=\left\{(r,\theta):r=2+\sin(\theta)\righ t\}$ and $\displaystyle \mathbb{G}=\left\{(r,\theta):r=-2+\sin(\theta)\right\}$, then $\displaystyle \mathbb{S}=\mathbb{G}$. I'm just a little unaccustomed to dealing with negative radii: when the radius is negative, the point in the polar plane is actually located at $\displaystyle (|r|,\theta+\pi)$. So it's just one limicon and can integrate it accordingly.