How do you find the area of the region bounded by the polar curves r^2=cos(2theta) and r^2=sin(2theta) ?

Nov 2, 2014

Let us look at the region.

(Note: ${r}^{2} = \cos 2 \theta$ in purple, and ${r}^{2} = \sin 2 \theta$ in blue)

Since there two identical region, we will find a half of one region then multiply by $4$. The combine area $A$ can be found by

$A = 4 {\int}_{0}^{\frac{\pi}{8}} {\int}_{0}^{\sqrt{2 \sin 2 \theta}} r \mathrm{dr} d \theta$

$= 4 {\int}_{0}^{\frac{\pi}{8}} {\left[{r}^{2} / 2\right]}_{0}^{\sqrt{\sin 2 \theta}} d \theta$

$= 2 {\int}_{0}^{\frac{\pi}{8}} \sin 2 \theta d \theta$

$= 2 {\left[\frac{- \cos 2 \theta}{2}\right]}_{0}^{\frac{\pi}{8}}$

$= - \cos \left(\frac{\pi}{4}\right) + \cos \left(0\right)$

$= - \frac{\sqrt{2}}{2} + 1 = \frac{2 - \sqrt{2}}{2}$

I hope that this was helpful.