Answer
Area of one loop is $\frac{1}{2}$.
Work Step by Step
First, we find the limits of integration.
The lemniscate ${r^2} = \cos 2\theta $.
When $\theta=0$, we have $r=1$, so the rose starts at the rectangular coordinates $\left( {1,0} \right)$. When $\theta = \frac{\pi }{4}$, we have $r=0$. From Figure 18, we see that the lemniscate completed the upper-half of the shaded region for the interval $0 \le \theta \le \frac{\pi }{4}$. Thus, by symmetry the area enclosed by one loop is twice the area of the upper-half shaded region:
area of one loop = $2\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} \cos 2\theta {\rm{d}}\theta $
area of one loop = $\frac{1}{2}\sin 2\theta |_0^{\pi /4} = \frac{1}{2}$.
