Originally Posted by

**skeeter** note the attached graph ...

$\displaystyle r = \sin(2t)$ in red from $\displaystyle 0$ to $\displaystyle \frac{\pi}{8}$

$\displaystyle r = \cos(2t)$ in blue from $\displaystyle \frac{\pi}{8}$ to $\displaystyle \frac{\pi}{4}$

area swept out by $\displaystyle r = \sin(2t)$ from $\displaystyle 0$ to $\displaystyle \frac{\pi}{8}$ equals the area swept out by $\displaystyle \cos(2t)$ from $\displaystyle \frac{\pi}{8}$ to $\displaystyle \frac{\pi}{4}$

so, __taking advantage of the symmetry of the graph__, 16 times one of those single areas is the desired area defined by the problem.

$\displaystyle \displaystyle 16 \int_0^{\frac{\pi}{8}} \frac{\sin^2(2t)}{2} \, dt = 16 \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\cos^2(2t)}{2} \, dt$

either one of the above will calculate the total area.