91Ó°ÊÓ

The Disk Method is a technique used to find the volume of a solid of revolution. Imagine slicing the solid into thin, circular disks perpendicular to the axis of revolution, resembling a stack of coins or CDs. Each disk has a small thickness (often represented as \( dx \) or \( dy \)) and a radius determined by the function being revolved.

In our problem, we revolve the region under the curve \( y = \frac{1}{x} \) around the \( x \)-axis. The formula to find the volume \( V \) using the Disk Method is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

The function \( f(x) = \frac{1}{x} \) represents the radius of each disk. The bounds \( b \) and \( 2 \) specify the region of interest for our problem. In essence, the Disk Method simplifies the calculation by summing the volumes of these infinitely thin disks, accounting for the entire solid.

The definite integral calculates the exact accumulated value of a function over a specified interval. In contexts like the Disk Method, it's used to find the total volume of a solid. The definite integral has limits (or bounds) which define the range over which the function is evaluated.

Mathematically, the definite integral is expressed as:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \( a \) and \( b \) are the lower and upper bounds, respectively. In this exercise, the bounds are \( b \) and \( 2 \).

After integrating the function \( \left( \frac{1}{x} \right)^2 \), the definite integral is evaluated at these bounds. This results in a precise number that reflects the total accumulation, allowing us to solve for the unknown variable, \( b \). The definite integral is crucial for calculating the volume here, by accounting for the continuous accumulation across the interval.

Integration is the process of finding a function that describes the area under a curve. It is the reverse operation of differentiation. When you integrate a function over an interval, you are essentially summing up infinitely small quantities to find the total.

In our exercise, we need to integrate \( \left( \frac{1}{x} \right)^2 \) across the interval from \( b \) to \( 2 \). The general form of the integral of \( x^{-n} \) is computed as follows:

• \( \int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} \)

For our specific function, \( n = 2 \), this simplifies to:

• \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \)

By calculating the antiderivative, we determine how the function changes over the given interval. Integration, thus, helps us find the function's behavior over a range, and in this context, it plays a key role in determining the volume of the solid generated by the revolving region.