The parabola $y^2 = x$ encloses an area between $x = 0$ and $x = 1$. Solving for $y$ yields $y = \sqrt{x}$. The area is calculated by integrating this function from $x = 0$ to $x = 1$: \[ \text{Area} = \int_0^1 \sqrt{x} \, dx. \] The antiderivative of $\sqrt{x}$ is $\frac{2}{3} x^{3/2}$. Evaluating this definite integral: \[ \text{Area} = \left[\frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2}) - \frac{2}{3} (0^{3/2}) = \frac{2}{3}. \] The enclosed area is $\frac{2}{3}$ square units.