Originally Posted by

**ThePerfectHacker** Look at picture below.

Now it seems dangerous to compute the integral from [0,2pi] because we will count the inside loop twice.

Instead draw the limacon for 0<=theta <= pi

(Shown below).

That area is equal to,

INT[0,pi](1/2)r^2 d(theta)=4.71

Now draw the limacon for 0<=theta<=pi/3

(Shown below).

That area is equal to,

INT[0,pi/3](1/2)r^2 d(theta)=.2718

Then if you subtract the area in the first integration from this one you get the area without the half-loop.

That is, 4.43

By Symettry we have total area, 8.86

Note, if you find,

INT[0,2pi](1/2)r^2 d(theta) = 9.42

And if you subtract .2718 twice you get,

8.86 (because we counted the half-loops twice).