Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$. 
Calculate
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$
where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.

I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{2}, \frac{1}{2}\right)$. And the determined the regions
\begin{split}
D & =\left\{(x,y)\in \mathbb{R}^{2}: x^{2}+y^{2}, y \le x\} = \{(r,\theta)\in \mathbb{R}^{2}: 0 \le r \le 1, 0 \le \theta \le \frac{\pi}{4} \right\}\\
B & =\left\{(x,y)\in \mathbb{R}^{2}: 0 \le x \le \frac{1}{\sqrt{2}}, 0 \le y \le x\right\}\\
C & =\left\{(x,y)\in \mathbb{R}^{2}: \frac{1}{\sqrt{2}} \le x \le 1,  \frac{1}{\sqrt{2}} \le y \le 1-x \right\}
\end{split}
Then,
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy = \iint_{D} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{B} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{C} (x^{2}+y^{2})^{-3/2} \,dx\,dy$$
Is this right? Or is there any other easiest way?

In polar coordinates, the integration region is enclosed by $\theta \in (0,\frac\pi4)$ and $ r\in (\frac1{ \sin \theta + \cos\theta},1)$. Then, the integral becomes
$$I=\int_0^{\pi4}\int_{\frac1{\sin \theta + \cos\theta}}^1 \frac1{r^2}drd\theta= \int_0^{\pi4}(\sin \theta +\cos \theta -1 )d\theta=1-\frac\pi4
$$
