find the area of the region that lies inside both curves. r = sin(2θ), r = cos(2θ)

Explanation:

Step1: Find intersection points

Set $\sin(2\theta)=\cos(2\theta)$. Then $\tan(2\theta) = 1$, so $2\theta=\frac{\pi}{4}+k\pi$, $\theta=\frac{\pi}{8}+\frac{k\pi}{2},k\in\mathbb{Z}$. In the range $[0, 2\pi]$, we consider the relevant intervals for the area - calculation.

Step2: Use the formula for the area in polar coordinates

The area $A$ between two polar curves $r_1(\theta)$ and $r_2(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}|r_1^{2}(\theta)-r_2^{2}(\theta)|d\theta$. Since the curves are symmetric, we can find the area of one - part and multiply by the number of symmetric parts. The area of the region inside both curves is given by: [ \begin{align*} A&=4\times\frac{1}{2}\int_{0}^{\frac{\pi}{8}}(\cos^{2}(2\theta)-\sin^{2}(2\theta))d\theta\ &= 2\int_{0}^{\frac{\pi}{8}}\cos(4\theta)d\theta \end{align*} ]

Step3: Integrate

Let $u = 4\theta$, $du=4d\theta$. When $\theta = 0$, $u = 0$; when $\theta=\frac{\pi}{8}$, $u=\frac{\pi}{2}$. Then $2\int_{0}^{\frac{\pi}{8}}\cos(4\theta)d\theta=\frac{2}{4}\int_{0}^{\frac{\pi}{2}}\cos(u)du$. [ \begin{align*} \frac{1}{2}[\sin(u)]_{0}^{\frac{\pi}{2}}&=\frac{1}{2}(1 - 0)\ &=\frac{1}{2} \end{align*} ]

Answer:

$\frac{1}{2}$