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In here, there are two polar curves. I first graphed both of them. I also shaded the area that I am going to find.

enter image description here

I assume that $r = 3$ is at $\Theta = 0$. I am unsure about this, but I proceeded anyway. I did not know the $\Theta $ at the intersection, which will be my upper bound, so I solved it by equating both $r$'s:

\begin{align*} 3 \cos\Theta &= 1 + \cos\Theta \\ 2\cos\Theta &= 1 \\ \cos\Theta &= \frac{1}{2} \\ \Theta &= \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} . \end{align*}

After that, I inputted the following in my calculator to find the area. I multiplied it to two because I recognized that I am just getting the area of one half:

$$A = 2\frac{1}{2}\int_{0}^{\frac{\pi }{3}}\left[(3\cos\Theta )^{2}-(1+\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = \pi \,\text{units}^{2} $$

In my equation above, I also assumed that $r_{2}$ is $3\cos\Theta$ because it has the farthest circumference from the origin.

Is my solution correct?

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