The question states:
The graph of $2\cos(3\theta)$ has three petals (also called "leaves" or "lobes").
The intersection of one of those petals with the circle $r = 1$ is shaded in the figure. Find the area of the shaded region.
I already know the formula, $$A = 1/2\int_a^b f(\theta)^{2}-g(\theta)^2d\theta$$ the problem obviously is finding the bounds. I set the functions equal to each other and solved:
$2\cos(3x) = 1$
$\cos(3x) = 1/2$
$3x = \arccos(1/2)$
$3x = \pi/3$
$x = \pi/9$
So wouldn't the functions intersect when $\theta = -\pi/9, \pi/9$?
Can I not just exploit symmetry and calculate the integral from $0$ to $\pi/9$ then multiply the entire thing by $2$?
And Finally, would $r = 1$ be the right most curve, and the $2\cos(3\theta)$ be the left most?
Best Answer
Yes, the curves intersect at $\theta = \pm \frac \pi 9$ and you can use symmetry to integrate from $0$ to the maximum value of $\theta$, then double the result. What you have missed is that the green area extends above the line $\theta=\frac \pi 9$. From $\theta=0$ to $\theta=\frac \pi 9$ you are right, the right curve is $r=2\cos (3 \theta)$ and the left curve is $r=1$. There is some area at $\theta$ larger than $\frac \pi 9$ where the left limit is $r=0$ and the right limit is $r=2\cos (3 \theta)$