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The problem describes a solid between planes perpendicular to the \( y \)-axis at \( y = 0 \) and \( y = 2 \). The cross-sections of the solid are circular disks whose diameters run from the \( y \)-axis to the curve given by the equation \( x = \sqrt{17}y^2 \).

### Steps to solve:
1. **Understanding the cross-sections**: 
   The cross-sections perpendicular to the \( y \)-axis are circular disks. The diameter of each disk at a given value of \( y \) runs from the \( y \)-axis (\( x = 0 \)) to the curve \( x = \sqrt{17}y^2 \). So, the diameter of the disk at any \( y \) is:
   \[
   \text{Diameter} = \sqrt{17}y^2 - 0 = \sqrt{17}y^2.
   \]
   The radius of the disk is half the diameter, so the radius is:
   \[
   \text{Radius} = \frac{\sqrt{17}y^2}{2}.
   \]

2. **Area of a circular cross-section**: 
   The area \( A(y) \) of a circular cross-section at a given \( y \) is given by the formula for the area of a circle \( A = \pi r^2 \). Substituting the radius:
   \[
   A(y) = \pi \left( \frac{\sqrt{17}y^2}{2} \right)^2 = \pi \cdot \frac{17y^4}{4}.
   \]

3. **Volume of the solid**: 
   The volume of the solid can be found by integrating the area of the cross-sections along the \( y \)-axis from \( y = 0 \) to \( y = 2 \):
   \[
   V = \int_0^2 A(y) \, dy = \int_0^2 \pi \cdot \frac{17y^4}{4} \, dy.
   \]
   Simplifying the integral:
   \[
   V = \frac{17\pi}{4} \int_0^2 y^4 \, dy.
   \]
   Now, compute the integral of \( y^4 \):
   \[
   \int y^4 \, dy = \frac{y^5}{5}.
   \]
   Evaluating this from \( y = 0 \) to \( y = 2 \):
   \[
   \int_0^2 y^4 \, dy = \frac{(2)^5}{5} - \frac{(0)^5}{5} = \frac{32}{5}.
   \]

4. **Final volume**: 
   Substituting this back into the expression for the volume:
   \[
   V = \frac{17\pi}{4} \times \frac{32}{5} = \frac{17 \times 32 \times \pi}{20} = \frac{544\pi}{20} = 27.2\pi.
   \]

Thus, the volume of the solid is \( 27.2\pi \) cubic units.

Would you like further details or explanations on any part of the solution?

Here are 5 related questions for practice:
1. How would the volume change if the parabola equation was \( x = \sqrt{10}y^2 \)?
2. What if the cross-sections were squares instead of circular disks?
3. How can you use cylindrical coordinates to solve a problem with circular cross-sections?
4. What is the physical interpretation of cross-sectional volumes in real-world applications?
5. How would the limits of integration change if the region extended to \( y = 4 \)?

**Tip:** Always check the units of the result, especially in physical applications where the dimensions of objects must match!

A solid lies between planes perpendicular to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the y-axis are circular disks with diameters running from the y-axis to the parabola x = sqrt(17)y^2. Find the volume of the solid.

Learn how to find the volume of a solid with circular cross-sections, with diameters running from the y-axis to the parabola x = sqrt(17)y^2, using integration techniques. This problem covers concepts from solid geometry and calculus.

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