Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=32$.

Given: ${\mathrm{x}}^2+{\mathrm{y}}^2=32\dots \dots \left(\mathrm{i}\right)$
Now, 
${\mathrm{x}}^2+{\mathrm{y}}^2=32\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=(\sqrt{32}{)}^2=(4\sqrt{2}{)}^2$
We can notice that it is a circle with centre (0, 0) and radius $4\sqrt{2}$
Given line is $\mathrm{y}=\mathrm{x}$.....(ii)
Solving equations(i) and (ii) for points of intersections,
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{x}}^2=32 [\mathrm{using} (\mathrm{ii}) \mathrm{in} (\mathrm{i}\left)\right] \\ \Rightarrow {\mathrm{x}}^2=16\\ \Rightarrow \mathrm{x}=\pm 4\\ \Rightarrow \mathrm{x}=4 (\mathrm{first} \mathrm{quadrant})\end{array}$
$\therefore$ Point of intersection is (4,4).

Now,
Required area = Area OMA + Area OMP 
= Area OMA + Area MPA
$\begin{array}{l}={\int }_0^4 \mathrm{x}\cdot \mathrm{dx}+{\int }_4^{4\sqrt{2}} \sqrt{(4\sqrt{2}{)}^2-{\mathrm{x}}^2}\cdot \mathrm{dx}\\ ={\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^4+\left[\frac{\mathrm{x}}{2}\sqrt{(4\sqrt{2}{)}^2-{\mathrm{x}}^2}\right.\\ {\left.+\frac{(4\sqrt{2}{)}^2}{2}{\sin }^{-1}\left(\frac{\mathrm{x}}{4\sqrt{2}}\right)\right]}_4^{4\sqrt{2}}\\ =\frac{16}{2}+\left[\left\{\frac{4\sqrt{2}}{2}\sqrt{(4\sqrt{2}{)}^2-(4\sqrt{2}{)}^2}\right.\right.\\ \left.\left.+\frac{32}{2}{\sin }^{-1}1\right\}-\left\{\frac{4}{2}\sqrt{(4\sqrt{2}{)}^2-(4{)}^2}+\frac{32}{2}{\sin }^{-1}\frac{1}{\sqrt{2}}\right\}\right]\\ \end{array}$
$\begin{array}{l}=8+\left(2\sqrt{2}\left(0\right)+16\times \frac{\mathrm{\pi }}{2}\right)\\ -\left(2\times 4+16\times \frac{\mathrm{\pi }}{4}\right)\\ =8+8\mathrm{\pi }-8-4\mathrm{\pi }\\ =4\mathrm{\pi }\text{ sq. units }\end{array}$
Therefore, the required area is $4\mathrm{\pi }{\text{ unit }}^2$