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Bonus: Cylindrical and spherical coordinates
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 $r=\frac{1}{2}$.
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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*} |