91Ó°ÊÓ

The disc method is a powerful tool for finding the volume of a solid of revolution. This technique is particularly useful when a region is rotated about an axis, typically the x-axis or y-axis.
When you rotate a region about the x-axis, imagine slicing the solid into many thin discs. Each disc has a small thickness, denoted by \(dx\), and a radius given by \(f(x)\), the function describing the curve.

• The volume of one thin disc is \(\pi [f(x)]^2 dx\).
• To find the total volume, you integrate over the interval covering the discs: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \(a\) and \(b\) are the bounds of the region.

Using the disc method allows you to set up an integral that, once evaluated, gives you the total volume of the solid. In our example, we rotated the region defined by \(y = e^x\) between \(x = 0\) and \(x = \ln 3\), leading to the integral \(\pi \int_{0}^{\ln 3} e^{2x} \, dx\).

In calculus, finding the antiderivative is essential for solving integrals, especially when calculating areas, volumes, and other quantities. An antiderivative, also known as the indefinite integral, reverses differentiation.
This means that if you differentiate an antiderivative, you'll get back to the original function.

• The symbol \(\int f(x) \, dx\) represents the antiderivative of \(f(x)\).
• Remember to add the constant of integration \(C\), since there are infinitely many antiderivatives differing by a constant.

For the function \(e^{2x}\), its antiderivative is \(\frac{1}{2} e^{2x} + C\).
This formula results from recognizing that the derivative of \(\frac{1}{2} e^{2x}\) provides us back \(e^{2x}\).
In our exercise, this understanding allowed us to integrate \(e^{2x}\) correctly over the interval and solve for the total volume.

The definite integral is an extension of the antiderivative. It sums up the infinitesimal changes across an interval to find a total value. Typically, it gives us areas under curves, volumes, or other cumulative quantities.
The definite integral is written as \(\int_{a}^{b} f(x) \, dx\), where \(a\) is the lower limit and \(b\) is the upper limit.

• To solve it, find the antiderivative of the function \(f(x)\).
• Then, evaluate this antiderivative at the upper and lower limits, subtracting the two results.

This process yields a specific value, not a formula with a constant like the indefinite integral. In our example, solving the definite integral \(\pi \int_{0}^{\ln 3} e^{2x} \, dx\) produced the solid's volume after evaluating \(\frac{1}{2} e^{2x}\) at \(x = \ln 3\) and \(x = 0\).
This calculation provided the precise volume \(4\pi\).

Exponential functions are crucial in understanding growth patterns, decay models, and more in mathematics. They take the form \(y = ab^{x}\), where \(a\) and \(b\) are constants, and \(b\) is typically the base of the function, such as Euler's number \(e\).
The function \(y = e^x\) represents continuous growth, and it has unique properties that simplify calculus operations:

• The derivative of \(e^x\) is itself, \(e^x\).
• The integral of \(e^x\) is also \(e^x + C\).

In the given exercise, \(y = e^x\) was part of the region for which we calculated the volume of revolution. Its simple derivation and integration properties made it straightforward to set up and solve the integral for the volume. Furthermore, when transformed to \(e^{2x}\), as required by the disc method, we could apply similar calculus strategies to find the antiderivative and evaluate the definite integral easily.