The problem describes a solid lying between two planes perpendicular to the $y$-axis, specifically at $y = 0$ and $y = 1$. The cross-sections perpendicular to the $y$-axis are circular disks, and the diameter of these disks runs from the $y$-axis to the parabola $x = \sqrt{3y^2}$.

Step-by-step solution:

1. Interpret the diameter of the circular disks:
   The diameter of each disk at a point $y$ is the distance from $x = 0$ (the $y$-axis) to the curve $x = \sqrt{3y^2}$. This simplifies to $x = \sqrt{3}y$.

   Therefore, the diameter of each circular cross-section is: $D(y) = \sqrt{3}y$

2. Find the radius of the disk:
   The radius $r(y)$ of the circular disk is half of the diameter: $r(y) = \frac{D(y)}{2} = \frac{\sqrt{3}y}{2}$

3. Find the area of each cross-section:
   The area $A(y)$ of a circular cross-section is given by the formula for the area of a circle: $A(y) = \pi \left( r(y) \right)^2 = \pi \left( \frac{\sqrt{3}y}{2} \right)^2 = \pi \frac{3y^2}{4}$

4. Set up the integral for the volume:
   The volume of the solid can be found by integrating the area of the cross-sections from $y = 0$ to $y = 1$: $V = \int_0^1 A(y) \, dy = \int_0^1 \pi \frac{3y^2}{4} \, dy$

5. Evaluate the integral: $V = \frac{3\pi}{4} \int_0^1 y^2 \, dy$ The integral of $y^2$ is: $\int y^2 \, dy = \frac{y^3}{3}$ Evaluating this from 0 to 1: $V = \frac{3\pi}{4} \left[ \frac{1^3}{3} - \frac{0^3}{3} \right] = \frac{3\pi}{4} \times \frac{1}{3} = \frac{\pi}{4}$

Final Answer:

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

Would you like any details or have any other questions about this solution?

Here are 5 related questions to further your understanding:

1. How do we set up integrals for volumes of solids with known cross-sectional areas?
2. Why is the radius of the disk half of the diameter in this problem?
3. How does changing the shape of the cross-section affect the volume integral?
4. What if the parabola was a different function, like $x = y^2$? How would that affect the volume?
5. Could we solve this problem using cylindrical shells instead of cross-sectional disks?

Tip: When dealing with volume integrals, always start by identifying the shape of the cross-sections and express their area as a function of the axis of integration.