Step 1: Picture the region.
The curve $y^2 = x$ is a parabola lying on its side, opening to the right. We also have the line $y = 4$ and the Y-axis, which is the line $x = 0$. We want the area trapped between these.
Step 2: Decide how to slice.
The region is bounded on the left by $x = 0$ and on the right by the parabola $x = y^2$. Since the left and right walls are easy to write in terms of $y$, it is simplest to integrate with respect to $y$.
Step 3: Find the $y$ limits.
The parabola meets the Y-axis at the origin, where $y = 0$. The top boundary is the line $y = 4$. So $y$ runs from $0$ to $4$.
Step 4: Write the width of a strip.
For a fixed $y$, the strip stretches from $x = 0$ on the left to $x = y^2$ on the right. So its width is $y^2 - 0 = y^2$.
Step 5: Set up the integral.
Adding all the thin strips gives \[ \text{Area} = \int_{0}^{4} y^2 \, dy. \]
Step 6: Evaluate.
\[ \int_{0}^{4} y^2\,dy = \left[\frac{y^3}{3}\right]_0^4 = \frac{64}{3}. \] So the area is $\frac{64}{3}$ square units. \[ \boxed{\text{Area} = \dfrac{64}{3}\ \text{sq. units}} \]