$\text{ The area (in sq. units) of the region }\mathrm{A}=\left\{(\mathrm{x},\mathrm{y}):\left|\mathrm{x}\right|+\left|\mathrm{y}\right|\le 1,2{\mathrm{y}}^2\ge \left|\mathrm{x}\right|\right\}\text{ is: }$

$\begin{array}{l}\left|x\right|+\left|y\right|\le 1 \text{ represent region of inside a square }\\ 2{\mathrm{y}}^2\ge \left|\mathrm{x}\right| \Rightarrow \left\{\begin{array}{l}2{\mathrm{y}}^2\ge \mathrm{x}\text{ if }\mathrm{x}\ge 0\\ \ge -\mathrm{x}\text{ if }\mathrm{x}<0\end{array}\right.\end{array}$

$\text{ which represents region outside/on the two parabolas }{\mathrm{y}}^2=\mathrm{x}/2\text{ and }{\mathrm{y}}^2=-\mathrm{x}/2$By symmetry,

$\begin{array}{c}\text{Require area}=4\times \left[Areaboundedin{Q}_1\right]\\ =4\times \left[Areaboundedbyx+y=1,y^2=\frac{x}{2}andy-axis\right]\\ =4\times \left[\underset{0}{\overset{\frac{1}{2}}{\int }}xdy+Areaof{\Delta }^{le}formedbyx+y=1,y=\frac{1}{2},y-axis\right]\\ =4\times \left[\underset{0}{\overset{\frac{1}{2}}{\int }}\left(2y^2\right)dy+\frac{1}{2}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\right]\\ =4\times \left[{\left[\frac{2y^3}{3}\right]}_0^{\frac{1}{2}}+\frac{1}{8}\right]\\ =4\times \left[\frac{2}{3}\times \frac{1}{8}+\frac{1}{8}\right]\\ =\frac{1}{3}+\frac{1}{2}\\ =\frac{5}{6}sq.units\end{array}$