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

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

• Why $b=5\pi/6$? That does not satisfy $2+2\sin b = 1$. Commented May 30, 2022 at 1:29
• Whether the $\sin$ is negative should not matter much. If the question asks for the area outside the circles $r=2$ or $r=3$, the steps to determine intersection points are similar. Commented May 30, 2022 at 2:01
• @peterwhy Ok that's for the lower bound right? Now what about the upper one how should I calculate it ? Commented May 30, 2022 at 2:06
• The intersections of the cardioid and the circle by definition satisfies $2+2\sin\theta = 1$, if you pick the lower bound $a=-\pi/6$ (the purple intersection to the lower-right), then the upper bound would be the next intersection $b=7\pi/6$ (the green intersection to the lower-left). Commented May 30, 2022 at 2:12
• @peterwhy okay I think I got everything now but there is only one question how is the lower-right intersection at $-\pi/6$ ? Shouldn't it be $11\pi/6$? Or $-5\pi/6$. Commented May 30, 2022 at 2:18

You can plot the graph to guess the values. I think you should integrate from $$-\dfrac{\pi}{6}$$ to $$\dfrac{7\pi}{6}$$.
• the numbers appear because they are only values of $\theta$ s.t $\sin(\theta) = -\frac{1}{2}$.
• When I plotted the graph I found it starting at $\pi/6$ and ending at $7\pi/6$, so why did you pick $-\pi/6$ Commented May 30, 2022 at 1:49
• @AhmedRamsey Why from $\pi/6$? That does not satisfy $2+2\sin \theta = 1$. Commented May 30, 2022 at 1:51
• I bet you mistook somewhere, in drawing the graph or interpreting it. An intersection is certainly on the line $\theta = -\frac{\pi}{6}$. And $\theta=\frac{\pi}{6}$ cannot be the intersection.