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Solution:
Let $I=\int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{x d y d x}{\sqrt{x^{2}+y^{2}}}$
Region of integration is: $\quad x \leq y \leq \sqrt{2-x^{2}}\\ 0 \leq x \leq 1$
Curves: $(i) y=x$ line
$\quad(i i) x=0, x=1 \quad$ lines parallel to the y axis
(iii) $y=\sqrt{2-x^{2}} \Rightarrow x^{2}+y^{2}=2$
Circle with centre $(0,0)$ and radius $\sqrt{2}$ .
Intersection of circle and $y=x$ line is $(1,1)$ in $1^{st}$ quadrant.
Divide the region into two parts as shown in fig.
After changing the order of integration:
$\begin{array}{ll}{\text { For one region: }} & {0 \leq x \leq y} \\ & {0 \leq y \leq 1}\end{array}$
$\begin{aligned} \text { For another region : } & 0 \leq x \leq \sqrt{2-y^{2}} \\ & 1 \leq y \leq \sqrt{2} \end{aligned}$
$\begin{aligned} \therefore \mathbf{I} &=\int_{0}^{1} \int_{0}^{y} \frac{x d x d y}{\sqrt{x^{2}+y^{2}}}+\int_{1}^{\sqrt{2}} \int_{0}^{\sqrt{2-y^{2}}} \frac{x d x d y}{\sqrt{x^{2}+y^{2}}} \\ &=\int_{0}^{1}\left[\sqrt{x^{2}+y^{2}}\right]_{0}^{y} d y+\int_{1}^{\sqrt{2}}\left[\sqrt{x^{2}+y^{2}}\right]_0^{\sqrt{2-y^{2}}} \\ &=\int_{0}^{1}(\sqrt{2} . y-y) d y+\int_{1}^{\sqrt{2}}(\sqrt{2}-1) d y \\ &=(\sqrt{2}-1)\left[\frac{y^{2}}{2}\right]_{0}^{1}+\left[\sqrt{2} y-\frac{y^{2}}{2}\right]_{1}^{\sqrt{2}} \end{aligned}$
$=1-\frac{1}{\sqrt{2}}$
$\therefore I=\frac{\sqrt{2}-1}{\sqrt{2}}$