Find the area of the region that lies inside $r=\sin(2\theta)$ and outside of $r=\cos(2\theta)$

Question

Find the area of the region that lies inside $r= \sin(2\theta)$ and outside of $r=\cos(2\theta)$ using polar coordinates.

Generally, I could use help setting up the integral in order to solve for the area of this region. I am having trouble understanding how to find the interval from a to b in order to integrate. I am also having trouble understanding how symmetry applies to the integral in this case.

Answer 1

First you find the intersections between $\sin(2 \theta)$ and $\cos(2\theta)$

$ \sin(2 \theta) = \cos(2 \theta) $

Implies $ 2 \theta = \dfrac{\pi}{4} \Rightarrow \theta = \dfrac{\pi}{8} $

Therefore, the required area is

$ \text{Area} = 4 (\frac{1}{2}) \displaystyle \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (\sin^2(2\theta) - \cos^2(2\theta) ) d\theta = -2 \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos(4 \theta) d\theta\\ =\displaystyle - \dfrac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos(u) du = -\dfrac{1}{2} \Bigg[ \sin(u) \Bigg]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = 1 $