That is another thing. The integration domain is limited by the circle $\displaystyle (x-1)^2+y^2=1$ (equivalently $\displaystyle x^2+y^2-2x=0$) and $\displaystyle y=0$ (upper semidisk). Using polar coordinates $\displaystyle x=\rho\cos \theta,\;y=\rho\sin\theta$ we get $\displaystyle \rho^2-2\rho\cos\theta=0$ (equivalently $\displaystyle \rho(\rho-2\cos\theta)=0$). As a consequence the integration domain is $\displaystyle D\equiv \begin{Bmatrix} 0\leq \theta\leq \pi/2\\0\leq \rho\leq 2\cos\theta\end{matrix}.$