Area Between Polar Curves

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To get the area between the polar curve $r=f(\theta)$ and the polar curve $r=g(\theta)$ , we just subtract the area inside the inner curve from the area inside the outer curve. If $f(\theta) \ge g(\theta)$ , this means

$\frac{1}{2}\int_a^b f(\theta)^2 - g(\theta)^2 d\theta.$ Note that this is NOT

$\frac{1}{2}\int_a^b [f(\theta)-g(\theta)]^2 d\theta$ !! You first square and then subtract, not the other way around.

As with most ``area between two curves'' problems, the tricky thing is figuring out the beginning and ending angles. This is typically where $f(\theta)=g(\theta)$ .

In the following video, we compute the area inside the cardioid $r=1+\sin(\theta)$ and outside the circle $r=\frac{1}{2}$ .

Example: Find the area inside the circle

$r=2\cos(\theta)$ and outside the unit circle.
Solution: Here

$f(\theta) = 2\cos(\theta)$ and

$g(\theta)=1$ . These intersect when

$\cos(\theta)=1/2$ , i.e. at

$\theta = \pm \pi/3$ . That is,

$f(\theta)$ is only bigger than

$g(\theta)$ when

$-\pi/3 < \theta < \pi/3$ . Our area is then

$\begin{eqnarray*} \frac12 \int_{-\pi/3}^{\pi/3} [(2\cos(\theta))^2 - 1^2] \, d\theta &=& \frac12 \int_{-\pi/3}^{\pi/3} 1 + 2 \cos(2\theta) \, d\theta \cr &=& \frac12 (\theta + \sin(2\theta)) \, \Big |_{-\pi/3}^{\pi/3}\cr &=& \frac{\pi}{3} + \frac{\sqrt{3}}{2}. \end{eqnarray*}$