# Find the area of the inner loop of $r = 2 + 3 \cos(\theta)$

I need some help with this problem. I started with graphing the polar equation:

The table of values generated by calculator are: $$0 \to 5$$, $$\pi/2 \to 2$$, $$\pi \to -1$$, and $$3\pi/2 \to 2$$

I know that the angle of $$-1$$ is $$\pi$$. So, that is my lower bound. Now, I need to determine the angle at the pole by letting $$r = 0$$.

$$r = 2 + 3\cos\theta$$

$$0 = 2 + 3\cos\Theta$$

$$-2 = 3\cos\Theta$$

$$\Theta =\cos^{-1}\left(\frac{-2}{3}\right)$$

Then, I inputted the following in my calculator: $$A = 2(\frac{1}{2})\int_{\cos^{-1}(\frac{-2}{3}) }^{\pi}(2+3\cos\Theta )^{2}d\Theta$$

I put $$2$$ in the front because the area to be solved is just the half of the inner loop. To get the whole area, I multiplied it to two. In doing so, it returned an answer of approximately $$0.44$$ units squared.

Is this correct?

The origin is given by $$r=0$$ and parametrized on the curve by $$\theta=\arccos(-\frac{2}{3})$$. The turning point in the inner circle is given by $$\theta=\pi$$. Thus, one can compute the area as follows
\begin{align} A &= 2 \int_{\arccos(-\frac{2}{3})}^\pi \int_0^{2+3\cos \theta} r \, dr d\theta = 2 \int_{\arccos(-\frac{2}{3})}^\pi \frac{1}{2} (2+3\cos \theta)^2 \, d\theta \\ &= -3 \sqrt{5} + \frac{17}{2} \arccos{\frac{2}{3}} \approx 0.4409 \end{align}