In this case your intuition gives the right answer. Place a general net on the plane, such that each cell is a "rectangular" restriction on r and θ. Thus each cell is a tiny slice of annulus. Its width is the change in r and its arc length is r times the change in θ. The area of a cell is not drdθ, but rather rdrdθ, introducing an extra factor of r. This suggests an integrand of f×r. Let's prove this using the previous theorem.
Let T be a rectangular region with height 2π and width h, where h is a radius that completely contains S. We will compute the integral over T, as r runs from 0 to h and as θ runs from 0 to 2π.
The map onto the xy plane is given by:
x ← r×cos(θ)
y ← r×sin(θ)
Take the partial with respect to r, and the partial with respect to θ, and build the jacobian as follows.
cos(θ), sin(θ)
-r×sin(θ, +r×cos(θ)
The determinant is r(cos2(θ)+sin2(θ)), or simply r. The integrand is indeed f×r.
The next section presents a beautiful application of this theorem.