# Finding the angles of polar curves when calculating the area

I am trying to understand how to choose the angles when doing area calculations on polar curves.

For example, to find the area inner loop of this limacon, $$1+2\sin\theta$$, I can identify four angles that seem to be tangent to where the limacon intersects with $$0$$. They are $$\pi\over6$$, $$5\pi\over6$$, $$7\pi\over6$$ , and $$11\pi\over6$$. I am assuming that the integral has to start and stop with these angles, because $$0$$ is where the inner loop of the limacon starts and stops. How do I know which ones to use to find the area of the inner loop (as opposed to the outer loop)? My book says the area is actually $$\int_{7\pi\over6}^{3\pi\over2}(1+2\sin\theta)^2d\theta$$, but I don't understand where this comes from, especially the $$3\pi\over2$$, which doesn't seem to be a starting or stopping point of the inner loop. Note that the inner loop has an interval $$\theta\in[0,2\pi]$$ which tells that the inner loop is defined by $$\sin\theta=-\dfrac12$$ which gives us $$\theta_{min}=\dfrac{7\pi}{6}$$ and $$\theta_{max}=\dfrac{11\pi}{6}$$.

Therefore, the area of the inner loop is $$\int_{\theta_{min}}^{\theta_{max}}\dfrac12r^2\theta\ d\theta=\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}\dfrac12[1+2\sin\theta]^2=\pi-\dfrac{3\sqrt{3}}{2}$$

• Would you just explain why the inner loop has an interval of $θ∈[0,2π]$? – agblt Mar 12 '19 at 23:31
• @agblt Notice in the above diagram that the inner loop is only in $1^{st}$ and $2^{nd}$ quadrant. So, we need $\theta\in[0,2\pi]$ – Key Flex Mar 13 '19 at 0:28

I am writing this answer to not only answer my specific question, but to give a general explanation of how finding the area of the limacon works, because in the many resources I looked at, I felt this point was insufficiently explained to me as a beginner.

In regards to my specific question, only two of those points I mentioned are valid choices, they are $$7\pi\over6$$ and $$11\pi\over6$$, because they are solutions to a point on the inner loop, $$r=0$$, as $$1+2\sin\theta=0$$ is on the inner loop. The other points, $$\pi\over6$$ and $$5\pi\over6$$ are not valid choices because they are not solutions to $$1+2\sin\theta$$ at any point on the inner loop, (the only time they are solutions are when $$1+2\sin\theta=2$$, which only happens on the outer loop.)

A better question would have been (a) how to determine the starting and ending angles of the outer loop. A different way of phrasing the question is, (b) how do we know that $$7\pi\over6$$ and $$11\pi\over6$$ are the starting and ending angles of the inner loop and not the outer loop?

Part of what motivates this question is a different question: (c) Why must we indeed have to worry at all about $$7\pi\over6$$ and $$11\pi\over6$$ when calculating the outer loop? If the only reason why we chose those angles were because we set $$1+2\sin\theta=0$$ out of convenience, because $$r=0$$ is a convenient starting point that is easy to find, but that is the reason for the confusion, then why not choose a different point that is definitely not on the inner loop such as $$1+2\sin\theta=2$$ and start from there to find the outer loop?

The answer to (c) is that it is not possible to find only the area of the outer loop with any starting point other than $$0$$. Observe what happens if we start at angle $$\theta ={\pi\over6}$$ where $$r=2$$. As you can see, if we start from there, we are forced to traverse the inner loop to get back where we started from! The only way to avoid this is to start at $$r=0$$, where we will not have to traverse the inner loop: So it turns out that our strength is also our downfall, because in choosing $$0$$ to avoid traversing the inner loop, we also fall prey to confusion as $$0$$ is also a valid starting point for the inner loop, and this leads to the question of how we choose the right angles that are solutions for $$r=0$$ that will only traverse the outer loop and not the inner loop, hence questions (a) and (b).

The answer to these is that for polar curves, we can only go counter-clockwise, as the whole angle system goes in a counter-clockwise direction starting for $$0$$ and going counter-clockwise to $$2\pi$$, and further if necessary. Therefore, it is impossible to start at $$7\pi\over6$$ and go to $$11\pi\over6$$ in a clockwise direction to traverse the entire outer loop of the limacon. It also won't work to go from $$7\pi\over6$$ to $$11\pi\over6$$ in a counter-clockwise direction and skip the inner loop, because that would not traverse the entire outer loop of the limacon, only the bottom point at $$0$$. And it definitely won't work to go from $$11\pi\over6$$ to $$7\pi\over6$$ in a counter-clockwise direction, because we can't go backwards in numbers.

So our only solution is to find two angles that are solutions for $$r=0$$, one that starts on the right side, and a second one that is greater than it that will be our ending angle on the left side. So we are forced to choose a new angle that is either less than $$7\pi\over6$$ (yet is still a solution for $$r=0$$), or one that is greater than $${11\pi\over6}$$ (yet is still a solution for $$r=0$$). Although there are an infinite number of choices, the two obvious choices are either $$-{\pi\over6}$$ to $$7\pi\over6$$, or $${11\pi\over6}$$ to $$19\pi\over6$$.