SOLUTION From the sketch of the curve in Figure 8,
we see that a loop is given by the region \[
D = \{(r, \theta) | -\pi/4 \le \theta \le \pi/4, 0 \le r \le \cos
2\theta\}
\] So the area is \[
A(D) = \iint_D dA = \int_{-\pi/4}^{\pi/4} \int_0^{\cos 2\theta} r dr
d\theta = \int_{-\pi/4}^{\pi/4} \left[ \frac{r^2}{2} \right]_0^{\cos
2\theta} d\theta
\] \[
= \frac{1}{2} \int_{-\pi/4}^{\pi/4} \cos^2 2\theta d\theta = \frac{1}{4}
\int_{-\pi/4}^{\pi/4} (1 + \cos 4\theta) d\theta = \frac{1}{4} \left[
\theta + \frac{1}{4}\sin 4\theta \right]_{-\pi/4}^{\pi/4} =
\frac{\pi}{8}
\]
