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The Disk Method is used to find the volume of a solid of revolution, where the solid is formed by rotating a region around an axis. This method involves slicing the solid perpendicular to the axis of rotation, forming disk-shaped cross-sections.
To calculate the volume using this method, we consider the circular disks. The formula to find the volume, depending on whether there is a hole or not, is:

• If there is no hole (pure disk): \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]Here, \( R(y) \) is the radius of the disk.
• If there is a central hole (washer): \[ V = \pi \int_{a}^{b} (R(y)^2 - r(y)^2) \, dy \]Here, \( R(y) \) is the outer radius, and \( r(y) \) is the inner radius.

For the given problem, the outer radius comes from the line \( x=2 \), while the inner radius is the curve \( y = \frac{x^2}{8} \), rewritten here as functions of \( y \) for integration around the \( y \)-axis. The calculation provides the volume after applying limits of integration from \( y=0 \) to \( y=2 \). This results in the volume \( 8\pi \) cubic units.

Unlike the Disk Method, the Shell Method involves revolving around the axis, where you consider cylindrical shells. It is often more straightforward for solids revolving around the vertical axis, typically the \( y \)-axis, when the region is described with functions of \( x \).
In this technique, the volume is calculated by summing up the volumes of thin cylindrical shells:

• Volume of shells: \[ V = 2\pi \int_{a}^{b} (x \cdot h(x)) \, dx \]
• Here, \( x \) is the radius of the shell, and \( h(x) \) is the height of the shell.

For our problem, \( h(x) \) is the difference between the functions \( 2-x \) and \( \frac{x^2}{8} \), representing the height of the region.
The integration occurs over the interval \( x=0 \) to \( x=4 \), which again results in the same volume \( 8\pi \) cubic units. Using this method can sometimes simplify the setup when dealing with the \( y \)-axis.

Finding intersection points of curves is crucial when dealing with solids of revolution. It helps define boundaries or limits of integration.
In the given exercise, the curves are \( y = \frac{x^2}{8} \) and \( y = 2-x \). By setting these two equations equal, you can solve for \( x \) to find where they intersect:

• Solve: \[ \frac{x^2}{8} = 2 - x \].
• The solution to this gives the x-values of intersection: \( x=2 \) and \( x=4 \).

These points are then plugged back into the original equations to find corresponding y-values, resulting in points \((2,0)\) and \((4,-2)\). These intersection points are not only critical in setting up the integrals but also ensure that the region \( R \) is appropriately defined.

Definite Integrals play an essential role in calculating the volumes of solids of revolution. They provide a way to sum infinitesimally small quantities over a specific interval.
For a solid of revolution, these integrals are used to add up the tiny volumes of disk sections or cylindrical shells. Here’s why they’re important:

• They allow calculation over a given interval, ensuring the entire region is considered.
• They take into account both the radius and height in forming volumes, adapting to the shape of curves.

In the exercise, for both the Disk and Shell Methods, definite integrals help to find exact volumes by analyzing how functions intersect and form boundaries. By evaluating these integrals, either through manual calculations or numerical methods, the solid’s volume is accurately determined as \( 8\pi \). This shows how mathematical theory translates into practical volume calculations.