91Ó°ÊÓ

The shell method is an insightful way to calculate the volume of a solid of revolution. It is particularly useful when the solid is rotated around a vertical line, like the y-axis in our case. The main concept involves envisioning the solid as made up of thin cylindrical shells. Each shell contributes a small volume, and by adding up (integrating) the volumes of all these shells, we can find the total volume.
The formula to find the volume using this method is:

• \( V = \int_{y_1}^{y_2} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dy \)

In this situation:

• The radius of each shell is given by the distance from the shell to the axis of rotation, which is just \(y\).
• The height is represented by the function opposite to the y-axis, \(x = \frac{4}{y^2}\).
• Thus, our integral becomes \( V = \int_{1}^{2} 2\pi y \cdot \frac{4}{y^2} \, dy = \int_{1}^{2} \frac{8\pi}{y} \, dy \).

So, the integral is evaluated to find the total volume of the solid generated.

Calculus is known for its powerful techniques to solve integrals, which describe accumulated quantities, such as areas, volumes, and other physical properties. A key aspect of using these techniques effectively is to properly set up the integral.
For our problem, we utilize a technique to integrate a function of the form \( \frac{1}{y} \). This leads to a natural logarithm result. Specifically, the integral \( \int \frac{1}{y} \, dy \) results in \( \ln|y| + C \), where \( C \) is a constant that gets eliminated as we evaluate over bounds:

• For the volume integral, we calculated \( 8\pi \int_{1}^{2} \frac{1}{y} \, dy = 8\pi (\ln 2 - \ln 1) \)
• Remember \( \ln 1 = 0 \), so the volume formula simplifies further to \( 8\pi \ln 2 \).

Practicing these techniques builds an understanding of transforming mathematical expressions and extracting meaningful physical interpretations from them.

The centroid of a solid is the average position of all points in the body, often considered as a balancing point. For solids of revolution, the calculation of the centroid is simplified due to symmetry. Often, one of the coordinates is zero, as in our solid of revolution around the y-axis where \( \bar{y} = 0 \).
To find \( \bar{x} \), we set up the integral for the first moment about the axis, which considers both the area and the distance of each area piece from the axis. The formula for \( \bar{x} \) in our context is:

• \[ \bar{x} = \frac{1}{V} \int_{1}^{2} 2\pi x^2 y \, dy \]

Replacing \( x^2 \) as \( \left(\frac{4}{y^2}\right)^2 = \frac{16}{y^4} \):

• \[ \bar{x} = \frac{16}{\ln 2} \int_{1}^{2} \frac{1}{y^3} \, dy \]

When evaluated, this leads to \( \bar{x} = \frac{6}{\ln 2} \).
The collection of these coordinates, \((\bar{x}, \bar{y}) = \left(\frac{6}{\ln 2}, 0\right)\), represents the centroid of the solid, showing where it evenly balances along both axes.