$R=3$ and $R=4-2sin(\theta)$
Find the area
I got the second part of the equation right ($1/2$ the integration of $\pi/6$ to $5\pi/6$ of $(4-2sin(x))^2)$ but he first part confuses me. The answer key stated that there was a $6\pi$ added out in front and they got $6\pi$ from $2/3\pi(3^2)$. where did these numbers come from?
$3=4-2\sin\theta\implies \theta=\arcsin\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6},\dfrac{5\pi}{6}$. Inside both curves, we have: $$A=\int^{5\pi/6}_{\pi/6}\dfrac{1}{2}\left((4-2\sin\theta)^2-3^2\right)d\theta$$ Now, $$\dfrac{9}{2}\int^{5\pi/6}_{\pi/6}d\theta=\dfrac{9}{2}\left(\dfrac{5\pi-\pi}{6}\right)=\\ \dfrac{9}{2}\times\dfrac{2\pi}{3}=3\pi$$ Hope this helps,