# What is $f(r\cos(\theta), r\sin(\theta))$ equal to $1$ in this double integral?

Use a double integral to find the area of the region. The region inside the circle $(x − 2)^2 + y^2 = 4$ and outside the circle $x^2 + y^2 = 4$.

I understand how to get the limits of integrand for this region on $r$ and $\theta$, but when you set up the integral it is as follows.

The general form of a double integral in polar is:

$$\iint f(r\cos(\theta), r\sin(\theta)) r \, dr \, d\theta$$

However when evaluating this integral it becomes:

$$\iint r \, dr \, d\theta$$

$$f(r\cos(\theta), r\sin(\theta)) =1$$
You are integrating to find the area. If $f(x, y)$ is any constant, $f$, then $\iint f \,dA= f\iint dA= fA$ where $A$ is the area of the region of integration. In particular, since we are trying to find the area, we want to take $f= 1$: $\iint dA= A$.