How do I find the area of the region in one loop of the curve that has polar equation r^2 = 3 cos^2(theta) - 9 sin^2(theta) ???
Polar integration is $\displaystyle \frac{1}{2}\int_{\alpha}^{\beta}[f({\theta})]^{2}d{\theta}=\frac{1}{2}\int_{\alpha}^{\beta}r^{2 }d{\theta}$
You already have r^2.
The area of half a loop in the first quadrant is $\displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{6}}\left[3cos^{2}{\theta}-9sin^{2}{\theta}\right]d{\theta}$
Integrate and multiply by 2.