Explain finding the area of a region?

Question

How is the area of the region inside the lemniscate $r^2 = 6\cos(2\theta)$ and outside the circle $r = \sqrt3$ equal to $(3(\sqrt3) - \pi)$? Thank you for anyone that helps.

Answer 1

First, we solve for the point of intersection, specifically in the first quadrant, since we will use symmetry.

$$r^2=6\cos(2\theta)=3$$

$$\frac{1}{2}=\cos (2\theta)$$

In the first quadrant, cosine is $\frac{1}{2}$ at $\pi \over 3$. Thus

$$\theta=\frac{\pi}{6}.$$

Using our formula for polar area, we have $$\mathrm{Area}=4\cdot\frac{1}{2}\int_{0}^{\pi/6}(6\cos(2\theta)-3)d\theta$$

Answer 2

First, sketch a graph of the region so you know what you are dealing with. Here is a graph.

You want the regions at the left and the right that are inside the blue and outside the red. You see this is symmetric, so you can just take four times the area of the portion of the region in the first quadrant.

You see that $\theta$ starts at zero. To find the upper limit, solve your simultaneous equations for $\theta$. That is very easy. Your lower limit for $r$ is the expression in red and your upper limit is the expression in blue.

Now set that up, with the usual equation for an area in polar coordinates

$$A=\frac 12\int_{\theta_1}^{\theta_2}(r_2^2-r_1^2)\,d\theta$$

If you cannot finish from here, show us what work you have done and we can help you from there.