Step 1: Understand the curves.
We want the area between $y^2=x$ (a sideways parabola) and $y=|x|$. In the first quadrant $y=|x|$ is just the line $y=x$, and that is where both curves enclose a region.
Step 2: Find where they meet.
On $y^2=x$ in the upper part, $y=\sqrt{x}$. Set $\sqrt{x}=x$. Squaring gives $x=x^2$, so $x(x-1)=0$, giving $x=0$ and $x=1$.
Step 3: Decide which curve is higher.
Between $x=0$ and $x=1$, pick $x=\tfrac{1}{4}$. Then $\sqrt{x}=0.5$ and $x=0.25$. So $\sqrt{x}$ is on top and the line $y=x$ is below.
Step 4: Write the area integral.
\[ A=\int_0^1(\sqrt{x}-x)\,dx. \]
Step 5: Integrate.
\[ \int\sqrt{x}\,dx=\frac{2}{3}x^{3/2},\qquad \int x\,dx=\frac{x^2}{2}. \] So \[ A=\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^1. \]
Step 6: Put in the limits.
At $x=1$: $\tfrac{2}{3}-\tfrac{1}{2}=\tfrac{4-3}{6}=\tfrac{1}{6}$. At $x=0$ it is $0$. So the area is $\tfrac{1}{6}$.
\[ \boxed{\dfrac{1}{6}} \]