Question

What is the area inside the polar curve $r=1$, but outside the polar curve $r=2\cos \theta$?

Answer 1

Solution

Here is the graph of the two curves. The shaded area, $A$, is the area of interest:

This is a symmetrical problems so we only need find the shaded area, $B$ and subtract twice this from that of a unit circle ($r=1$).

We can find the polar coordinate of the point of intersection in Q1 by simultaneously solving the polar equations:

$r=2\cos \theta$
$r=1$

From which we get:

$2\cos \theta =1\Rightarrow \cos \theta =\frac{1}{2}$
$\therefore \theta =\frac{\pi }{3}$

So we can easily calculate the area, $B$, which is that of the a circle sector $C$ and that bounded by the curve $r=2\cos \theta$ where $\theta \in (\frac{\pi }{3},\frac{\pi }{2})$

The area, $C$ is simply $\frac{1}{2}{r}^2\theta$:

$A_C=\frac{1}{2}\cdot 1\cdot \frac{\pi }{3}$
$=\frac{\pi }{6}$

And we calculate the area of the segment $B$, via Calculus using:

$A=\int \frac{1}{2}{r}^{2}d\theta$

Thus:

$A_D={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2}{\left(2\cos \theta \right)}^{2}d\theta$
$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2}4{\cos }^2\theta d\theta$
$=2{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$
$=2{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2}(\cos 2\theta +1)d\theta$
$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\cos 2\theta +1d\theta$
$={[\frac{1}{2}\sin 2\theta +\theta ]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}$
$=(\frac{1}{2}\sin \left(\pi \right)+\frac{\pi }{2})-(\frac{1}{2}\sin \left(\frac{2\pi }{3}\right)+\frac{\pi }{3})$
$=(0+\frac{\pi }{2})-(\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{\pi }{3})$
$=\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{3}$
$=\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

So then the total area we require, is given by:

 

$A_A=\pi {\left(1\right)}^2-2(A_C+A_D)$
$=\pi -2(\frac{\pi }{6}+\frac{\pi }{6}-\frac{\sqrt{3}}{4})$
$=\pi -2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$
$=\pi -\frac{2\pi }{3}+\frac{2\sqrt{3}}{4}$

$=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$