This problem involves finding the volume of a solid formed by square cross-sections taken perpendicular to the $x$-axis.

Step 1: Analyze the Problem

The solid's base is the region in the first quadrant bounded by the curves: $1x = y^2, \quad 2x = y^2, \quad x = 4.$ From these equations:

• $x = y^2$ and $y^2 = 2x$ give the bounds for $y^2$.
• The region is further restricted by $x = 4$, setting the upper limit.

Step 2: Set up the Geometry of the Squares

The side length of each square is determined by the distance between the curves: $\text{Side length} = \sqrt{2x} - \sqrt{x}.$ The area of the square is then: $\text{Area} = \left(\sqrt{2x} - \sqrt{x}\right)^2.$

Step 3: Integrate to Find the Volume

The volume is the integral of the cross-sectional area along the $x$-axis, from $x = 0$ (where the curves intersect) to $x = 4$: $V = \int_{0}^{4} \left(\sqrt{2x} - \sqrt{x}\right)^2 \, dx.$

Step 4: Expand the Square

Expanding $\left(\sqrt{2x} - \sqrt{x}\right)^2$: $\left(\sqrt{2x} - \sqrt{x}\right)^2 = 2x - 2x^{3/2} + x.$

Thus, the volume integral becomes: $V = \int_{0}^{4} \left(2x - 2x^{3/2} + x\right) \, dx.$

Step 5: Integrate Term by Term

Integrate each term:

1. $\int 2x \, dx = x^2$,
2. $\int -2x^{3/2} \, dx = -\frac{4}{5}x^{5/2}$,
3. $\int x \, dx = \frac{1}{2}x^2$.

Combining these: $V = \left[x^2 - \frac{4}{5}x^{5/2} + \frac{1}{2}x^2\right]_{0}^{4}.$

Step 6: Evaluate the Integral

At $x = 4$: $x^2 = 16, \quad \frac{4}{5}x^{5/2} = \frac{4}{5} \cdot 32 = \frac{128}{5}, \quad \frac{1}{2}x^2 = 8.$ Thus: $V = 16 + 8 - \frac{128}{5} = 24 - \frac{128}{5} = \frac{120}{5} - \frac{128}{5} = -\frac{8}{5}.$

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