91Ó°ÊÓ

The disk method is a powerful technique used in calculus to find the volume of a solid of revolution. To understand this method, imagine slicing the solid into thin circular disks. Each of these disks has a small thickness, which can be represented as a differential element. These disks have a radius equal to the function value at a particular point along the axis of rotation. The volume is then obtained by summing up the volumes of all these disks.
For instance, if a region is rotated around the x-axis, the volume of one disk can be given by: \( V = \pi \[ f(x) \]^{2} \Delta x \). Here, \Delta x \ is the thickness of the disk and \ f(x) \ is the radius. To find the total volume, you need to integrate this expression over the desired interval:

\[ V = \pi \int_{a}^{b} \[ f(x) \]^{2} dx \]
This formula changes slightly depending on the axis of rotation and function variables involved.
The disk method is particularly useful when the solid’s cross section perpendicular to the axis of rotation is a disk.

A definite integral is a fundamental concept in calculus, used to find areas under curves, among other applications. When calculating volumes of solids, the definite integral helps sum up infinitesimally small slices or disks over an interval.
In our example, the region bounded by \ y = x^{6}, y = 1, \ and the y-axis, rotated around the x-axis, results in a solid. To find its volume, we set up an integral with the appropriate boundaries and integrate. Specifically, our integral looks like:
\[ V = \pi \int_{0}^{1} \[ y^{1/6} \]^{2} dy \]
Here, the bounds are from 0 to 1 because that's the interval for y after rotation. After simplifying and integrating, we obtain the volume of the solid. The power rule of integration, which states \ \( \int x^{n} dx = \frac{x^{n+1}}{n+1} \), is particularly handy in this context.
After performing the integration and applying the boundaries, we get our final volume as:
\[ V = \frac{3\pi}{4} \]
This result shows the powerful use of definite integrals in determining volumes of solids of revolution.

An inverse function essentially reverses the operation of the original function. If \ f(x) \ is a function, \ f^{-1}(x) \ is its inverse. The roles of the dependent and independent variables are swapped. In the context of volumes of solids, inverse functions come in handy when setting up integrals.
For the exercise provided, the inverse function finds the relationship between \ x \ and \ y \ from \ y = x^{6} \. We solve for \ x \ in terms of \ y \ to get \ x = y^{1/6} \. Then, this inverse function is used in our integral setup as:
\[ V = \pi \int_{0}^{1} \[ y^{1/6} \]^{2} dy \]
By understanding and utilizing the inverse function, we can correctly express one variable in terms of another and proceed with our integration to find the volume of the solid. This capability is crucial for solving problems involving curves and regions rotated around an axis.