Area of region inside circle and Lemniscate?

Question

What is the area of the region inside the circle $r=1$ and inside the lemniscate $r^2=2\cos(2\theta)$?

I can’t quite figure this one out. I was able to graph it easily and find the circle is one all the way around and the Lemniscate has tip petals on the x axis. How would you find the area if it is asking inside both figures? I know how to find out the answer if it is asking outside and inside. I found the integral range to be from $\pi/12$ to $\pi/4$. My answer after using the area for a region $1/2(f(\theta)^2-g(\theta)^2)$ is $\sqrt{3}-\pi/3$.

Answer 1

$ \theta = \frac{\pi}{12} \ $ is not the right value for their point of intersection in the first quadrant.

Equation of curves are $ \ r = 1$ and $r^2 = 2 \cos 2\theta$.

We need to find area that is inside the circle and the lemniscate (shaded area in the diagram). As you can see there is symmetry in all four quadrants so we can find area bound in first quadrant and multiply by $4$.

If we take intersection in first quadrant, $2 \cos2\theta = 1 \implies \theta = \frac{\pi}{6}$.

Also note that curve $r = \sqrt{2 \cos 2\theta}$ in first quadrant forms for $0 \leq \theta \leq \frac{\pi}{4}$.

For $0 \leq \theta \leq \frac{\pi}{6}$, $r$ is bound above by the circle and for $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{4}, $ r is bound above by the lemniscate.

So the area should be,

$A = 4 \: \bigg[\displaystyle \int_0^{\pi/6} \int_0^1 \: r \ dr \ d\theta + \int_{\pi/6}^{\pi/4} \int_0^{\sqrt{2\cos2\theta}} \: r \ dr \ d\theta \ \bigg]$