Question

Find the area of the region enclosed by one loop of the curve. r = 7 cos(11θ)

Answer 1

First compute the period of the curve:
$\cos(11x)=\cos(11(x+T))\\=\cos(11x+11T )$
The above equation implies that:
$11x=11x+11T+2k\pi$
The smallest solution is: $T= \frac{2\pi}{11}$
First compute the following area:
$\int\limits^{\pi/22}_{-\pi/22} {7\cos(11x)} \, dx = \frac{14}{11}$
Then compute the following area:
$\left\vert\int\limits^{\pi/22+\pi/11}_{\pi/22} {7\cos(11x)} \, dx\right\vert=\dfrac{14}{11}$
Adding the two values we get the total area:
$\frac{14}{11}+\frac{14}{11}=\frac{28}{11}$