91Ó°ÊÓ

When it comes to calculating the volume of a solid of revolution, the disk method is a reliable technique. We use this method when the solid is formed by rotating a region around an axis, such as the x-axis in our exercise.

The idea is straightforward: imagine the solid being composed of many tiny disk-shaped slices stacked along the axis of rotation. Each disk is infinitesimally thin and has a small height, which we call \(dx\), if we're considering the x-axis.
This contributes a small volume, which can be represented by the formula \(V = \pi (R^2 - r^2) \, dx\), where \(R\) is the outer radius, and \(r\) is the inner radius.

In the specific problem, the outer function is \(y = e^{3x}\), and the inner function is \(y = e^x\). These functions determine the radii of our disks, allowing us to find the volume by integrating over the bounded region.

Integral evaluation is the process that transforms our setup expression into an actual numerical value, representing the volume of the rotated region.
The initial integral we need to solve is \[V = \pi \int_{0}^{1} (e^{6x} - e^{2x}) \, dx\] which arises from the application of the disk method.

To solve this, we use the Fundamental Theorem of Calculus, which helps us compute the definite integral and thus the volume.
This theorem states that by finding an antiderivative for the expression and then evaluating it at the bounds of integration \(x=0\) and \(x=1\), we can find the volume.

Typically, each component of the integrand is handled separately:

• Find an antiderivative of \(e^{6x}\), which is \(\frac{e^{6x}}{6}\)
• Find an antiderivative of \(e^{2x}\), which is \(\frac{e^{2x}}{2}\)

By subtracting these evaluated anti-derivatives at the upper and lower limits, we calculate the complete volume.

Exponential functions are particular types of functions where the variable appears in the exponent, such as \(e^{3x}\) or \(e^x\). These are commonly seen when dealing with rotations because they often describe growth, decay, or transformations easily depicted in practical physical problems.

One key feature of exponential functions like \(e^x\) is that their rate of change is proportional to their current value.
This allows them to model real-world phenomena effectively and leads to straightforward integration when they appear in calculus problems.

In our exercise, it’s important to recognize how the squared form \((e^{3x})^2 = e^{6x}\) and \((e^x)^2 = e^{2x}\) appear naturally when setting up the integrals for our volume calculation.
These transformations simplify the integration process and use the properties of exponents, making it easier to apply integral evaluation techniques.

A definite integral represents the signed area under a curve, within specific bounds. In the context of our problem, it tells us the volume of the solid formed by rotation.

The expression \(\pi \int_{0}^{1} (e^{6x} - e^{2x}) \, dx\) is an example of a definite integral. Here, the bounds \(0\) and \(1\) specify the region we examine along the x-axis.
Once evaluated, the definite integral applies the limits to the antiderivative, which is derived from the integrand.

This results in a concrete numerical value, which, multiplied by \(\pi\), gives the exact volume.

• The upper limit of integration corresponds to \(x=1\)
• The lower limit corresponds to \(x=0\)

By calculating the antiderivative

1. at the upper limit and subtracting its value at the lower limit,
2. the process concludes by simplifying the result to obtain the final volume.

This ensures a clear understanding of why precise boundaries are essential in determining real-world quantities.