Consider the disc $D$ is centred at $(\bar{r},0)$ with radius $c$.
Illustration of $D$ :
The inner integration can be easily dealt with by using polar coordinate. However the difficulty is the outermost integration. This is how far I got to:
Converting the integral into polar coordinate we get:
$$\int_{-\alpha}^{\alpha}\int_{\bar{r}\cos(\theta)-\sqrt{-\bar{r}\sin^2(\theta)+ c^2} }^{\bar{r}\cos(\theta)+\sqrt{-\bar{r}\sin^2(\theta)+ c^2} } \frac{1}{r} r\, d r \,d \theta \\ = \int_{-\alpha}^{\alpha}2\sqrt{-\bar{r}\sin^2(\theta)+ c^2}\,d\theta $$
Now the problem is how to integrate and define the bounds $\alpha$ in terms of $\bar{r}$ and $c$. It seems there should be a clever substitution trick.
The final result should be something like $$2 \pi\left({\bar{r}}-\sqrt{\bar{r}^2-c^2}\right)$$
[ref. Mechanics of Materials 9th Edition c2014 - Hibbeler, Table 6-1]