Area inside the cardioid $r=2+2\sin\theta$ and outside the circle $r=1$

Question

Simply I saw a friend asking about the area inside the cardoid $r=2+2\sin \theta$ and outside the circle $r=1$ and I couldn't help. I know that the area is equal to $$ \int_{a}^{b} \frac{1}{2}((2+2\sin \theta)^2-1) \,d\theta $$ But when I tried to solve the equation $$ 2+2\sin \theta =1 $$ I found $\sin \theta =-\frac{1}{2}$ Which means $\theta=-\pi/6$.

Now I'm not quite sure should $a$ be equal to $-\pi/6$ or $7\pi/6$?, and for $b$ should it be $5\pi/6$.

Its so confusing for me because I didn't expect negative sin.

That's what I got when I tried to plot the graph

Answer 1

You can plot the graph to guess the values. I think you should integrate from $-\dfrac{\pi}{6}$ to $\dfrac{7\pi}{6}$.