How to determine polar coordinate visually I am studying calculating arc length on polar coordinates on khan academy and I encounter this question.

Let R be circular segment that lies inside the circle $r=4cos(\theta)$
  and the left of $r=sec(\theta)$. 
Calculate the perimeter of the shaded region.


I know how to use formula, and I familiar with polar coordinates. However, when I encountered this I set boundaries for my integral as $\pi/3$ and $5\pi/3$! Which is wrong. so after struggling a lot and realizing that I am doing things wrong, by solving this equation $4cos(\theta)=sec(\theta)$ I realized that I must set my boundaries as $\pi/3$ and $2\pi/3$. The lower boundary is apparent by the diagram, but the other one seems strange to me! 
In the hints there is this line:

The entire circle is traced out once from $\theta=0$ to $\theta=\pi$ thus the arc is traced from $\theta=\pi/3$ to $\theta=2\pi/3$

Unfortunately, I am not even able to understand what this hint even is about! 
So my questions are:
How can I determine boundaries visually from the diagram and what this hint talk about? the circle must be traced out from $0$ to $2\pi$, why it says $\pi$??
 A: We can calculate the perimeter using regular geometry.
Using the angle markings, we can deduce that the perimeter of the curved part is a third of the perimeter of the circle. Connecting the $\frac13 \pi$ and the $\frac53 \pi$ lines to the circle's center, we form two equilateral triangles. Their angles added together must be $\frac23 \pi$. Therefore, the perimeter of the curved part is $\frac23 \pi r = \frac43 \pi$. After drawing altitudes from the top to the bottom side on both of the equilateral triangles and using special 30-60-90 triangle rules, we see that half of the line is $\sqrt{3}$, so the whole line is $2\sqrt{3}$. Adding these together we get the perimeter as $\boxed{\frac43 \pi + 2\sqrt{3}}$.
A: I'm taking a different view here and will determine the perimeter by analysis in the complex plane. We will divide the problem in to two parts, the circular arc and the straight line. These can be readily shown to be described by
$$
r=4\cos\theta,\quad \theta\in[\pi/3,2\pi/3]\\
r=\sec\theta,\ \quad \theta\in[-\pi/3,2\pi/3]
$$
The general equation for the arc length in the complex plane is
$$s=\int |\dot z| du\color{gray}{=\int\sqrt{1+\left(\frac{dy}{dx} \right)^2}\ dx}$$
We start with the arc...
$$
z=4\cos\theta\ e^{i\theta}\\
\dot z=(-4\sin\theta+i\cos\theta)\ e^{i\theta}\\
|\dot z|=4\sqrt{\sin^2\theta+\cos^2\theta}=4\\
\int_{\pi/3}^{2\pi/3}4d\theta=\frac{4\pi}{3}
$$
The straight line can be much simpler than this, but here we are demonstrating a method. Thus,
$$
z=\sec\theta\ e^{i\theta}\\
\dot z=(\tan\theta \sec\theta+i\sec\theta)\ e^{i\theta}\\
|\dot z|=\sec\theta\sqrt{1+\tan^2\theta}=\sec^2\theta\\
\int_{-\pi/3}^{\pi/3}\sec^2\theta\ d\theta=\tan\theta\big|_{-\pi/3}^{\pi/3}=2\sqrt{3}
$$
Thus, the total perimeter is given by
$$\boxed{s=\frac{4 \pi}{3} + 2\sqrt{3}}$$
This result has been verified numerically.
