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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 \( \frac{\pi}{4} \).

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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.

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.

Learn how to calculate the volume of a solid bounded by planes perpendicular to the y-axis with circular cross-sections. This problem involves the parabola x = √3y^2 and uses integral calculus to determine the volume of the solid between y = 0 and y = 1.

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