Solution:

The area bounded by the parabola, $x^{2}=y$, and the line, $y=|x|$, can be represented as

The given area is symmetrical about $y$-axis.
$\therefore$ Area $\mathrm{OACO}=$ Area $\mathrm{ODBO}$
Point of intersection of the parabola, $x^{2}=y$, and line, $y=x$, is $A(1,1)$
Area of $\mathrm{OACO}=$ Area $\triangle O A B-$ Area OBACO
$\therefore$ Area of $\Delta \mathrm{OAB}=\frac{1}{2} \times \mathrm{OB} \times \mathrm{AB}=\frac{1}{2} \times 1 \times 1=\frac{1}{2}$
Area of $\mathrm{OBACO}=\int_{0}^{1} y d x=\int_{0}^{1} x^{2} d x=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}$
$\Rightarrow$ Area of $\mathrm{OACO}=$ Area of $\triangle \mathrm{OAB}-$ Area of $\mathrm{OBACO}$
$\begin{array}{l}
=\frac{1}{2}-\frac{1}{3} \\
=\frac{1}{6}
\end{array}$
As a result, the required area $=2\left[\frac{1}{6}\right]=\frac{1}{3}$ units.