the question is
Find the area enclosed by the curve:
$r=2+3\cos \theta$
Here's my steps:
since when $r=0$, $\cos \theta=0$ or $\cos\theta =\arccos(-2/3)$.
so the area of enclosed by the curve is 2*(the area bounded by $\theta=\arccos(-2/3)$ and $\theta=0$)
the answer on my book is $5\sqrt{5}+(17/2)*\arccos(-2/3)$
I have no idea why there is a $5\sqrt{5}$ since $\arccos(-2/3)=2.300523984$ on my calculator.