How would one find the area between two polar curves and ones which overlap?

Question

Please as simple as possible, I am in calculus II. Am wondering how to solve these two problems:

1. Find the area $A$ of the region $R$ outside of the curve $r = \cos(2\theta)$ and inside the curve $r = 1$.

Quite lost on how to solve, could really use steps and clarifications. What gets me is how do we know when writing out the computations which is the inner and which is the outer and how does this affect the formula structure? Such as what if the first problem were to find area inside the curve $r = \cos(2\theta)$ and outside the $r = 1$.

Thank you

Answer 1

The graph of $r=\cos 2\theta$ is a 4-petal curve. The graph of a $r=1$ is a unit circle. Now, the total area of the $4-$petal curve is \begin{align} A_{pc}&=8\int_{0}^{\pi/4}\frac{1}{2}\cos^2 2\theta d\theta\\ &=4\int_{0}^{\pi/4}\frac{1+\cos 4\theta}{2}d\theta\\ &=2\left[\theta+\frac{\sin 4\theta}{8}\right]_{0}^{\pi/4}\\ &=\frac{\pi}{8}. \end{align}

The area of the circle $r=1$ is $A_c=\pi(1)^2=\pi$. Thus, the required area is $$A=A_c-A_{pc}=\pi-\frac{\pi}{8}=\frac{7\pi}{8}.$$

Answer 2

Since $\cos (2\theta) \le 1$, it follows that $\cos(2\theta)\le r \le 1$. Now you can calculate the area between the curves by finding the area under $\cos (2 \theta)$ and subtracting it from the area of the circle of radius $r=1$, the first area is $$\frac 12 \int_0^{2\pi} \cos (2\theta)^2 \mathrm{d}\theta=\frac {\pi}{ 2}$$ The area of the circle is $\pi$ therefore, the total area is $\pi -\frac {\pi}{2}=\frac{\pi}{2}$