Area of region inside circle and Lemniscate?

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 $$\sqrt{3}-\pi/3$$.

• why don't you use the description of the circle in polar coordinates and compute the integral between both curves? Commented May 24, 2021 at 19:26
• I have tried that but don’t know if my answer would be correct. Commented May 24, 2021 at 19:28
• What have you tried? If You saw the region already, did you try and find their intersection points? Commented May 24, 2021 at 19:28
• @user242559 edit your answer with what you got so far. it's good practice in this site. Commented May 24, 2021 at 19:29
• i suggest you use polar coordinates. Also I believe the radius of the circle should be $\sqrt{2}$ not $1$. you can integrate with respect to $r^2$ instead of $r$ to simplify. Commented May 24, 2021 at 19:39

$$\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,

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

• You can alternatively write each of the integrals as single integral using formula of area as $\displaystyle \frac{1}{2} \int f(\theta)^2 \ d\theta$. Commented May 24, 2021 at 20:32