Find the area limited by 4 curves, using change of variables I'm trying to show that the area bound by the curves $r^2= 3\cos(2\theta)$, $r^2= 4\cos(2\theta)$, $r^2= 3\sin(2\theta)$,  $r^2= 4\sin(2\theta)$ in the first quadrant is equal to 
$$A= \frac{10 - 7\sqrt{2}}{4} $$
using a change of variables. Most of my tries have ended with integrals that don't have a solution. The closest I got was using 
$$\cos(2\theta)=r^2/u $$
$$\sin(2\theta)=r^2/v $$
But unfortunately that hasn't worked either, because the indefinite integral does not have a solution. Any ideas? 
Any suggestions are much appreciated!

 A: I don't think you need substitution for this problem, and I think you are missing a factor of 2 possibly because you didn't consider quadrant III.
We start with
$$
A=\iint \frac12\cdot 1_A\,\mathrm{d}(r^2)\,\mathrm{d}\theta
$$
Clearly $2\theta$ is in the first quadrant, to get $\sin(2\theta)\geq 0$ and $\cos(2\theta)\geq 0$.  So with a bit more thought,
$$
A=\int_0^{\pi/4}\int_{r^2=3\max(\sin2\theta,\cos2\theta)}^{r^2=4\min(\sin2\theta,\cos2\theta)} 1_{3\max(\sin2\theta,\cos2\theta)\leq 4\min(\sin2\theta,\cos2\theta)}\,\mathrm{d}(r^2)\,\mathrm{d}\theta
$$
since the quadrant III region is just the rotation of the quadrant I region about the origin.
Examining $0\leq3\max(\sin2\theta,\cos2\theta)\leq4\min(\sin2\theta,\cos2\theta)$ more closely gives $2\theta\in[\arctan(3/4),\arctan(4/3)]$, so
\begin{align*}
A&=\int_{2\theta=\arctan(3/4)}^{2\theta=\pi/4}\int_{r^2=3\cos2\theta}^{r^2=4\sin2\theta} \,\mathrm{d}(r^2)\,\mathrm{d}\theta+
\int_{2\theta=\pi/4}^{2\theta=\arctan(4/3)}\int_{r^2=3\sin2\theta}^{r^2=4\cos2\theta} \,\mathrm{d}(r^2)\,\mathrm{d}\theta\\
&=2\int_{2\theta=\arctan(3/4)}^{2\theta=\pi/4}(4\sin2\theta-3\cos2\theta)\,\mathrm{d}\theta\\
&=\left[-4\cos2\theta-3\sin2\theta\right]_{2\theta=\arctan(3/4)}^{2\theta=\pi/4}\\
&=\left[-4\cdot\frac1{\sqrt2}-3\cdot\frac1{\sqrt2}\right]-\left[-4\cdot\frac45-3\cdot\frac35\right]\\
&=\frac{10-7\sqrt2}2
\end{align*}
