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!