The problem describes a solid bounded by planes perpendicular to the $y$-axis at $y = 0$ and $y = 1$. The cross-sections perpendicular to the $y$-axis are circular disks, with diameters running from the $y$-axis to the parabola $x = \sqrt{3y^2}$.

Step-by-Step Approach:

1. Understanding the Cross-Sectional Area: The cross-sections are circular disks, where the diameter extends from the $y$-axis ($x = 0$) to the curve defined by $x = \sqrt{3y^2}$.

   Therefore, the radius $r$ of each disk at a given $y$ is half the length of the diameter: $r = \frac{\text{diameter}}{2} = \frac{\sqrt{3y^2}}{2} = \frac{\sqrt{3} y}{2}$

2. Area of a Disk: The area $A(y)$ of a circular cross-section is given by: $A(y) = \pi r^2 = \pi \left( \frac{\sqrt{3} y}{2} \right)^2 = \pi \times \frac{3y^2}{4} = \frac{3\pi y^2}{4}$

3. Volume of the Solid: To find the volume, we integrate the area of the cross-sections from $y = 0$ to $y = 1$: $V = \int_0^1 A(y) \, dy = \int_0^1 \frac{3\pi y^2}{4} \, dy$

4. Calculating the Integral: Now, solve the integral: $V = \frac{3\pi}{4} \int_0^1 y^2 \, dy = \frac{3\pi}{4} \left[ \frac{y^3}{3} \right]_0^1$ $V = \frac{3\pi}{4} \times \frac{1^3 - 0^3}{3} = \frac{3\pi}{4} \times \frac{1}{3} = \frac{\pi}{4}$

Final Answer:

The volume of the solid is $\frac{\pi}{4}$.

Would you like further details or have any questions?

Here are 5 related questions to expand this concept:

1. How would the volume change if the diameter ran from $x = -\sqrt{3y^2}$ to $x = \sqrt{3y^2}$?
2. What happens to the volume if the parabola is changed to $x = \sqrt{2y^2}$?
3. How would you find the volume if the shape was a square instead of a circular disk?
4. What is the physical significance of such solids in real-world applications?
5. How does the method change if the parabola is rotated or tilted?

Tip: When integrating for volume, always ensure that the limits of integration and cross-sectional area are correctly defined based on the given geometry.