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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.

A solid lies between 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 = √(3y^2). Find the volume of the solid.

Calculate the volume of a solid bounded between y=0 and y=1 with cross-sectional disks having diameters from the y-axis to the parabola x=√(3y^2). This solution uses integral calculus and the disk method for volume calculation.

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