Find the area enclosed by the curve $r=2+3\cos \theta$.

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.

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