I need to calculate the area using a double integral (converting to polar system), which are bounded by the following curves: $(x^2+y^2)^2=2a^2(x^2-y^2)$ and $x^2+y^2 \ge a^2$. So I plotted both of these curves, assuming that a = 1. I converted it to the polar system and found $r=\pm\sqrt{2a^2cos2\theta}$. But I can't find $\theta$ using algebraic method. On the graph, I see that it can be equal to ${\pi}\over{6}$. Then my solution could be like this(the answer is correct):
$$ S=\iint\limits_{S} \,dx \,dy=4\int_{0}^{\pi\over6}d\theta\int_{a}^{\sqrt{2a^2cos2\theta}}rdr$$ I tried to find intersection point by doing this: $\sqrt{2a^2(x^2-y^2)}=a^2$. But i think that's wrong.
So basically my question is, how can i find this $\pi\over6$ by calculating it? How do I find this intersection point?
COMMENT AS A HINT.-Calculation of some points in the figure of circle and lemniscata is easy and you know the polar coordinates, say $r=f_1(\theta)$ and $r=f_2(\theta)$ for each curve. A way you can finish is the following:
►Calculate the angles $\theta$ corresponding to points $A,B,C$ in the attached figure; let these $\alpha,\beta,0$.
►►The shadow area, $A_1$, is given by $$\frac12\int_{\alpha}^{\beta}(f_1(\theta))^2d\theta-\frac12\int_{\alpha}^{\beta}(f_2(\theta))^2d\theta$$
►►►You can finish multiplying by $4$ the area given by
$$\frac{\pi}{4}+A_1+\frac12\int_{0}^{\alpha}(f_1(\theta))^2d\theta-\frac{\sqrt{-2+\sqrt5}}{2}$$