Question

Find the volume of the solid generated by revolving the region bounded by \(y=e^{x}, y=0, x=0\), and \(x=\ln 3\) about the \(x\) -axis.

Short answer

The volume is \(4\pi\).

Step-by-step solution

Set up the integral for the solid of revolution

The volume of a solid of revolution generated by a function, in general, can be found using the formula for discs: \( V = \pi \int_{a}^{b} [f(x)]^2 \; dx \). In this case, the function is \(y = e^x\). The bounds are from \(x = 0\) to \(x = \ln 3\), because the region is bounded by these values.

Substitute the function into the volume formula

Given the function \(y = e^x\), we substitute it into the volume formula: \[ V = \pi \int_{0}^{\ln 3} (e^x)^2 \; dx \]Simplifying \((e^x)^2\) gives us \(e^{2x}\), so:\[ V = \pi \int_{0}^{\ln 3} e^{2x} \; dx \]

Integrate the function

To integrate \(e^{2x}\), recognize that the integral of \(e^{ax}\) is \(\frac{1}{a}e^{ax}\). Thus:\[ \int e^{2x} \; dx = \frac{1}{2} e^{2x} + C \]

Apply the definite integral

Compute the definite integral from 0 to \(\ln 3\):\[ \pi \left[ \frac{1}{2}e^{2x} \right]_{0}^{\ln 3} = \pi \left( \frac{1}{2}e^{2(\ln 3)} - \frac{1}{2}e^{0} \right) \]

Simplify the expression

Use properties of exponents \(e^{2(\ln 3)} = (e^{\ln 3})^2 = 3^2 = 9\) and \(e^0 = 1\). Then simplify:\[ \pi \left( \frac{1}{2} \times 9 - \frac{1}{2} \times 1 \right) = \pi \left( \frac{9}{2} - \frac{1}{2} \right) \]\[ V = \pi \cdot \frac{8}{2} = 4\pi \]

Conclude the solution

Thus, the volume of the solid generated by revolving the given region about the \(x\)-axis is \(4\pi\).

Key concepts

Solids of Revolution

Solids of revolution are intriguing geometrical shapes that occur when a two-dimensional area is revolved around an axis, producing a three-dimensional object. This concept is notably practical in calculus for determining the volume of these objects. In our exercise, the solid is created by rotating the region bounded by the curve \( y = e^{x} \), the \( x \)-axis, and vertical lines \( x = 0 \) and \( x = \ln 3 \) around the \( x \)-axis.

The idea is to imagine a slice of the area as a thin disc or circle perpendicular to the axis of rotation. The volume of each disc can be determined by calculating its area \( \pi [f(x)]^2 \) and then finding the sum of these small volumes from one boundary to the other, which is accomplished using integration. This is why the concept of solids of revolution directly relates to the integration techniques used in calculus.

Understanding solids of revolution requires visualizing the geometry involved and applying integration principles to represent the volume accurately. By learning this concept, students can apply similar principles to various other functions and boundaries, mastering the technique to solve complex geometry problems in calculus.

Definite Integral

The definite integral is a core concept in calculus, crucial for calculating the accumulated value over a certain interval. In the context of finding the volume of a solid of revolution, the definite integral helps compute the total volume by summing an infinite number of infinitely thin discs or washers. The integral encompasses two main elements: the limits of integration and the function itself.

In our previous exercise, the limits of integration are from \( x = 0 \) to \( x = \ln 3 \). The specific function we are integrating in this situation is derived from squaring the exponential function \( e^{x} \), giving \( (e^x)^2 = e^{2x} \).

The precision of the definite integral comes from its ability to consider every infinitesimal segment within these limits. Thus, it gives the exact volume of the entire solid when summed. By setting up and solving the definite integral of \( e^{2x} \) over the specified bounds, students learn the procedure to calculate such areas numerically, which would be impossible with simple arithmetic.

Exponential Functions

Exponential functions are an essential part of mathematics and occur frequently in various scientific fields. These functions are characterized by their rapid growth or decay, represented typically in the form \( y = a \cdot e^{bx} \). In our exercise, the specific function given is \( y = e^{x} \). Notably, this function describes one of the simplest yet most important forms where \( a = 1 \) and \( b = 1 \).

In the calculus problem discussed, \( e^{x} \) forms part of the region bounded that is to be revolved around the \( x \)-axis to form a solid. Understanding how exponential functions behave aids in visualizing how the solid appears when formed.

The exponential nature affects how rapidly or slowly the solid grows as the function \( e^{x} \) moves across different \( x \)-values. The geometric area under the curve and speed of growth also influence the total volume of the solid calculated in our integration steps. Recognizing the patterns in exponential growth can profoundly impact students’ abilities to predict and solve similar problems involving exponential models in different applications.