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

Aug 30, 2014

The areas of both regions are $\frac{\pi}{2}$.

The graph of $r = \sin \left(2 \theta\right)$, $0 \le q \theta < 2 \pi$ looks like this: Since the area element in polar coordinates is $r \mathrm{dr} d \theta$, we can find the area of the four leaves above by
$A = {\int}_{0}^{2 \pi} {\int}_{0}^{\sin \left(2 \theta\right)} r \mathrm{dr} d \theta$.

Let us evaluate the inside integral first,
A=int_0^{2pi}[{r^2}/2]_0^{sin(2 theta)}d theta =int_0^{2pi}{sin^2(2 theta)}/2 d theta

By the double-angle idenitity ${\sin}^{2} \left(2 \theta\right) = \frac{1 - \cos \left(4 \theta\right)}{2}$,
A=1/4int_0^{2pi}[1-cos(4 theta)]d theta =1/4[theta-{sin(4 theta)}/{4}]_0^{2pi} =1/4[2pi-{sin(8pi)}/4-(0-{sin(0)}/{4})]=1/4(2pi)={pi}/2

Hence, the area is $\frac{\pi}{2}$.

For $r = \cos \left(2 \theta\right)$, the area can be found by
$A = {\int}_{0}^{2 \pi} {\int}_{0}^{\cos \left(2 \theta\right)} r \mathrm{dr} d \theta$