This is what I am pretty sure is right, and understand. You integrate from $0$ to $\frac {π}{4} $ radians, but where does you $r$, radius begin when you integrate? It must be some kind of function, going from the cartesian coordinate points $(2,0)$ to $(1,0)$, and then up to $(1,2)$.
$$\int_0^{\fracπ4} \int_?^{2cos(θ)}\ r \,drdθ$$
Other solutions that may or may not be correct, need answers.
$\int_{0}^{2π} \int_0^{2cos(θ)}\ r \,drdθ$ - $\int_{\fracπ4}^{2π} \int_0^{2cos(θ)}\ r \,drdθ$
If I interpret you correctly, you want to integrate over the red quarter circle, in terms of ordinary polar coordinates centered at the origin $(0,0)$. Then the lower limit for $r$ is given by starting at the line $x=r\cos \theta=1$, i.e., $r=1/\cos \theta$.