Finding area inside one curve, outside the other, Why $2\pi$ and $\pi$ for the integral?

Question

How are they determining that you should use $2\pi$ and $\pi$?

Answer 1

To be inside the first and outside the second we need $4-4\sin \theta \gt 4$, which means $\sin \theta \lt 0$, which is true for $\pi \lt \theta \lt 2 \pi$

Answer 2

Note that if we write the equations as \begin{equation} r_1=4-4\sin\theta \end{equation} \begin{equation} r_2=4 \end{equation} we see that $r_1=r_2$ only if $\sin\theta=0.$ That happens at any multiple of $\pi$ but $\pi$ and $2\pi$ are obvious choices.

And the graph of $r_1=4-4\sin\theta$ lies outside the circle $r_2=4$ on the interval $(\pi,2\pi)$.