Find the area of the small loop of the limacon (graph):
$$r = 1+2\cos(\theta)$$
What I tried:
Set $r=0$ to get $\theta = 2\pi/3, 4\pi/3$.
Then
$$A = \frac 1 2 \int_{2\pi/3}^{4\pi/3} r^2 d\theta$$
Is that right? Why do we set $r=0$?
My book seems to use symmetry. The answer given is:
$$A = \int_{2\pi/3}^{\pi} r^2 d\theta$$
so I'm not sure on what that symmetry is based. Is it based on the one above? Or possibly
$$A = \frac 1 2 \int_{\pi/3}^{\pi} r^2 d\theta$$
?
Best Answer
Never mind. I got it. We set $r=0$ because those give the $\theta$'s where the small loop starts and ends. This can be seen by retracting the polar graph based on the Cartesian graph of
$$y = 1+2\cos(x)$$
The small loop begins at $2\pi/3$ and ends at $4\pi/3$.