# 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.

#### Solutions Collecting From Web of "Find the area enclosed by the curve $r=2+3\cos \theta$."

Area of curve, $$A=2\int_0^{\arccos(-2/3)}\frac{r^2}{2}d\phi$$ $$\implies A=2\int_0^{\arccos(-2/3)} \frac{(2+3\cos\phi)^2}{2}d\phi$$ $$=2\int_0^{\arccos(-2/3)} \frac{4+9\cos^2\phi+12\cos\phi }{2}d\phi$$ $$=\int_0^{\arccos(-2/3)} (4+9\cos^2\phi+12\cos\phi) d\phi$$ $$=3\sqrt{5}+\frac{17}{2}\cos^{-1}(\frac{-2}{3}).$$