To solve the given problem, we need to compute the volume of the solid where the cross-sections perpendicular to the $y$-axis are squares.

Problem Summary:

1. Base Region $R$: Defined by $y = \sqrt{x}$, $y = 2$, and $x = 0$ (the $y$-axis).
2. Cross-Sections: Squares with side length equal to the horizontal span of the region $R$ at a given $y$-value.
3. Volume Formula: $\text{Volume} = \int \text{Area}_{\text{square}} \, dy$ Since the area of a square is $s^2$, and the side length $s$ is given by $s = x$, we need to express $x$ in terms of $y$.

Step-by-Step Solution:

1. Express $x$ as a function of $y$: From $y = \sqrt{x}$, we square both sides: $x = y^2$

2. Determine the side length $s$: The side length of the square is given by the difference in $x$-coordinates across the region, which is $s = x = y^2$.

3. Area of Cross-Section: The area of each square is: $\text{Area} = s^2 = (y^2)^2 = y^4$

4. Set up the integral for the volume: The volume is the integral of the area from $y = 0$ to $y = 2$: $\text{Volume} = \int_0^2 y^4 \, dy$

5. Evaluate the integral: Compute $\int_0^2 y^4 \, dy$: $\int y^4 \, dy = \frac{y^5}{5} + C$ Apply the bounds $y = 0$ to $y = 2$: $\text{Volume} = \left[ \frac{y^5}{5} \right]_0^2 = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5}$

Final Answer:

The volume of the solid is: $\boxed{\frac{32}{5} \text{ cubic units}}$

Would you like me to elaborate on any step, or verify using another method?

Related Questions:

1. How would the volume change if the cross-sections were semicircles instead of squares?
2. What is the significance of the bounds $y = 0$ and $y = 2$ in this problem?
3. Can this method be extended to other shapes, such as triangles or rectangles?
4. How do we determine if this problem could also be solved in terms of $x$?
5. Why is it necessary to square $s = y^2$ to find the area?

Tip:

When setting up integrals for volume, always confirm that your variable of integration matches the axis perpendicular to the cross-sections.