How to get the area between these $2$ functions? I have a function:
$(a)$ $r = 4\cos(2\theta)$
$(b)$ $r = 4\sin(2\theta)$.
I need at least a set up for the integral that will yield the area inside the rose (a) but outside the rose $(b).$
I cant seem to figure out which strip to use because the boundaries are confusing me. Any help?
 A: First of all it's good to graph the functions $(a)$ (red) and $(b)$ (blue):

The second step is to fund points of intersection, so we find all solutions of
$$\begin{cases}
r=|4\cos(2\theta)|\\
r=|4\sin(2\theta)|\\
\end{cases}$$
We are taking absolute values because $r$ may be negative and this might complicate things. The solutions are:
$$r=2\sqrt2,\quad\theta\in\left\{\pm\tfrac18\pi,\pm\tfrac38\pi,\pm\tfrac58\pi,\pm\tfrac78\pi\right\}$$
It's enough to find area between $\theta=0$ and $\theta=\tfrac{\pi}{8}$ and multiply it by $8$:
$$S=8\cdot\left(
\frac12\int_{0}^{\pi/8}r_a^2\,d\theta-
\frac12\int_{0}^{\pi/8}r_b^2\,d\theta
\right)$$
where $r_a=4\cos(2\theta),r_b=4\sin(2\theta)$. Simplifying:
$$S=8\cdot16\cdot\left(
\frac12\int_{0}^{\pi/8}(\cos(2\theta))^2\,d\theta-
\frac12\int_{0}^{\pi/8}(\sin(2\theta))^2\,d\theta
\right)$$
$$=64\int_{0}^{\pi/8}(\cos(2\theta)^2-\sin(2\theta)^2)\,d\theta$$
$$=64\int_{0}^{\pi/8}\cos(4\theta)\,d\theta$$
$$=64/4\big[\sin(4\theta)\big]_0^{\pi/8}=16$$
A: Note that in the first quadrant the two curves intersect at $\sin(2\theta)=\cos(2\theta) \Rightarrow \theta=\frac{\pi}{8}$.
In this quadrant:
the curve (a) $r=4\cos(2\theta)$ starts from $(1,0)$ for $\theta=0$  and the curve (b) $r=4\sin (2\theta)$ starts at $(0,0)$. So the area inside (a) but not (b) is
$$
 A_1=A_a-A_b=\frac{1}{2}\int_{0}^{\frac{\pi}{8}}\left(4\cos(2\theta)\right)^{\;2} d \theta-\frac{1}{2}\int_{0}^{\frac{\pi}{8}}\left(4\sin(2\theta)\right)^{\;2} d \theta=2
$$

where ( see the figure) $A_a$ is the area delimited by  the blue arc $AP$ ( rose (a)), the segment $PO$,and the segment$OA$, and $A_b$ is the area delimited by the red arc $OP$  (rose (b)) and the segment $PO$.
Use the symmetry to find the entire area inside (a) but not (b) $A=8A_1=16$.
