Find the area enclosed by one loop of the lemniscate with equation $r^2 = 81cos(2\theta)$. Choose your limits carefully.
When I was solving this problem, my intuition was to take the integral from $0$ to $2\pi$ and multiply it by 4. When calculating it out, I got $162$, but my autograder is marking this as incorrect. Was I approaching the problem incorrectly? Or perhaps I made an arithmetic error, in which case would anyone be so kind as to offer an alternate solution?
I was also given $r_0 = 9$, but I'm not sure how this factors into solving the problem.
$\endgroup$1 Answer
$\begingroup$A single loop is defined by the angles between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ and the enclosed area equals$$ \int_{-\pi/4}^{\pi/4}\frac{1}{2}r^2\,d\theta = \frac{81}{2}\int_{-\pi/4}^{\pi/4}\cos(2\theta)d\theta=\frac{81}{4}\int_{-\pi/2}^{\pi/2}\cos(\theta)d\theta=\frac{81}{2}. $$
$\endgroup$
