Q:

Circles limits of integration with polar coordinates

Footnote: Got caught up thinking it asked for a 'mutual region' in both functions, while the question actually asked for area of the second function not covered by the first function. 

I have two circles: 

Radius 1 with center (0,0), i.e. $r=1$
Radius 1 with center (1,0), i.e. $r=2cos(\theta)$

Hence the area equals in both cicles equals:
$$A= \int_{-\pi/3}^{\pi/3} \int_{1}^{2cos(\theta)}r \space drd\theta$$
When finding the area that is enclosed by both circles, the limits of integration according to the answer is $-\pi/3$ and $\pi/3$. 
While I recognise that this is the intersection between the functions, why do I not integrate between $-\pi/2$ and $\pi/2$? It would appear that there is an region not 'covered' by these bounds of integration between the y-axis and these limits. I keep visualising that I only 'sweep' a sector of the area by doing this, missing the bits that extend beyond the limits.

A:

You say the integral is "the area that is enclosed by both circles."
Actually, the integral is correct for the area in the circle with center $(1,0)$ and outside the circle with center $(0,0)$. In my diagram, it is the pure-blue area.

For that crescent-shaped area, the angle from the origin is indeed $60°=\frac{\pi}3$, and the distance from the origin ranges from the inner circle $r=1$ to the other circle $r=2\cos\theta$. That explains the bounds on both $r$ and $\theta$.
If you actually did want the area inside both circles and wanted to do integration from the origin, you would need three integrals: one from $-\frac{\pi}3$ to $\frac{\pi}3$ to cover $r$ from $0$ to $1$, and another from $-\frac{\pi}2$ to $-\frac{\pi}3$ to cover $r$ from $0$ to $2\cos\theta$, and a third from $\frac{\pi}3$ to $\frac{\pi}2$ to cover $r$ from $0$ to $2\cos\theta$. But this is not what your given integral apparently wants.

