Suppose we want to find the area between the circles $x^2+y^2=1$ and $x^2+y^2=2$ using integrals in polar coordinates.
I understand that the integral is $$\int_{1} ^{2} \int_{0} ^{2\pi}r dθdr$$ as the radius $r$ extends from $1$ to $2$ and goes one full circle. But,
If I wanted to write this integral in the form $$I=\int_{} ^{} \int_{U} ^{}f(x,y) d(x,y)$$
and then convert it to polar coordinates $$I=\int_{1} ^{2} \int_{0} ^{2\pi}f(rcosθ,rsinθ) rdθdr$$ how can I find $f(x,y)$ in order to do this?
Note that this double integral is for area, so the only thing that will occupy the integrand is your Jacobian. When converting from polar back to rectangular, the integrand $r$ becomes $1$.
EDIT: If you are confused about why this is, set up a Jacobian matrix with the conversions. Let's pretend we are in rectangular and going to polar to make my job easier.
We know $x=r\cos\theta$ and $y=r\sin\theta$. Now, when we convert to polar, we have to miltiply the integrand by the following determinant matrix called the Jacobian matrix since we have changed our variables.
$$J=\begin{vmatrix}x_r & x_v \\ y_r & y_v\end{vmatrix}=\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r$$
Thus, upon converting from polar back to rectangular, we must divide by this factor $r$. This is why $f(x,y)=1$
We then change the differentials back to $dx$ and $dy$.
Additionally, we rearrange the equations and change the bounds accordingly.