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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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up vote 4 down vote accepted

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

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@WillieWong: you are right. For arccos(−2/3)≤θ≤π,r is negative , so the domain is restricted to 0 to arccos(−2/3) for the graph in 1st and 2nd quadrant. There is symmetry about x-axis, that's why there is a multiplication factor of 2 initially. – Aang Jul 10 '12 at 12:08
Okay, a rough glance looks okay now. +1. I assume the OP accepted the answer because he/she knows how to compute $\sin \arccos (-2/3)$ by the Pythagorean theorem. – Willie Wong Jul 10 '12 at 13:50

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