Answer
Polar coordinates: $\int_{0}^{2\pi}(\int_{2}^{5}f(r\cos\theta,r\sin\theta)\,r\,dr)d\theta$
Work Step by Step
It would be better to represent the region in polar coordinates as the region has a consistent radius.
Since the region varies radially from 2 to 5 with $\theta$ ranging from $0\leq\theta\leq2\pi$, we have:
$$\iint_{R}f(x,y)dA=\int_{0}^{2\pi}\bigg(\int_{2}^{5}f(r\cos\theta,r\sin\theta)\,r\,dr\bigg)d\theta$$
