.
Solution: Consider how the graph is traced:
It starts at (3/2,0),
continues on to (1/2,Pi/2),
reaches the origin at 2Pi/3,
continues on to (-1/2,Pi),
reaches the origin a second time at 4Pi/3,
and finally returns to the initial point (3/2,2Pi),
Let A be the area inside the large loop and outside the small black loop
The polar integral
counts the area of the small loop twice. Thus, by symmetry:
