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When determining the volume of a solid with a specific cross-section, understanding the concept of cross-sectional area is crucial. Cross-sectional areas are slices of the solid, perpendicular to an axis, that help us visualize and calculate how the shape changes along its length. For the given problem, these cross-sections are circular disks.

Think of slicing a loaf of bread. Each slice represents a cross-section, and the area of each slice will depend on its shape, which can vary depending on its position along the loaf. Similarly, in this problem, each cross-section is a circular disk whose size changes from one end to the other.

• The shape of each cross-section is determined by the equation of the parabola.
• The area of each circular disk is dependent on its radius.

This understanding helps to form the basis for computing the total volume of the solid using integration.

In this exercise, each cross-section perpendicular to the y-axis is a circular disk. Understanding the properties of these disks is essential to calculating the volume.

A circular disk is defined by its radius, which is half the diameter. Here, the diameter runs from the y-axis to the parabola. Since the parabola is given by the equation \(x = \sqrt{5}y^2\), the diameter is \(x\), and so the radius \(r\) is \(\frac{x}{2}\).

• Each disk's area is calculated using the formula for the area of a circle: \(A = \pi r^2\).
• This step is vital as it allows the integration process to represent the changing size of the disks along the y-axis.

In simpler terms, by knowing the radius at any point, we can find the area of the disk at that point, leading us into the integration stage to sum up these areas into a volume.

The calculation of volume involves integrating cross-sectional areas of the disks from one end of the solid to the other. This brings us to the concept of a definite integral.

A definite integral is used to calculate the total accumulation of a quantity, here the volume, by integrating between two specified bounds, which are \(y = 0\) and \(y = 2\) in this problem. The integral essentially sums up the infinite number of circular disks' areas along the interval to find the total volume.

• The limits of integration provide the bounds from which to calculate: here, the bounds are dictated by the planes perpendicular to the y-axis.
• The process involves setting up the integral of \(\pi r^2\).

This integral not only solves for the volume but also provides a deeper understanding of how the shape and size of the solid change across its length.

In this exercise, the radius function forms the core of converting the geometric problem into an analyzable mathematical one. The radius function determines the size of each cross-sectional disk by expressing the radius in terms of \(y\), which then plays an integral role in finding the volume.

The radius \(r(y)\) is derived as half the diameter, given by the parabola \(x = \sqrt{5}y^2\). Therefore, the radius function is \(f(y) = \frac{\sqrt{5}y^2}{2}\). This is essential because:

• It enables the calculation of the area of each cross-sectional disk.
• It allows us to set up the integral to find the total volume.

With this function, we can mathematically express the changes in the cross-section sizes along the length of the solid, leading to precise computation of the total volume through integration.