# 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.

• I don't know if your integral is correct for polar coordinates. – Zach L Jun 27 '15 at 17:09
• Also, it appears that there are three areas that could count as being contained by the two circles. It appears that you are concerned with the middle area, whereas @imranfat is concerned with the rightmost area. For this area the bounds would be the same for both functions, but for the middle area, the bounds are actually different for both functions. For r =1, your bounds would be -pi/3 to +pi/3 and for r = 2cos$\theta$ your bounds would be pi/3 to 2pi/3. – Zach L Jun 27 '15 at 17:16

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.
• Wouldn't a single integral cover both of your last two integrals. Namely the integral from $\pi/3$ to $2\pi/3$ of $(2\cos{\theta})^2/2$? We could also just use twice the integral from $\pi/3 to \pi/2$ of $(2\cos{\theta})^2/2$? – Zach L Jun 27 '15 at 17:24
• @ZachL: There are many ways to do fewer than three integrals, including just one from $-\frac{\pi}2$ to $\frac{\pi}2$ of $\min(1,2\cos\theta)$. I wanted to keep it conceptually simple. Now that I look at what I wrote, I did overstate my case: you are correct. – Rory Daulton Jun 27 '15 at 17:27