Question

Multiple Choice Let \(R\) be the region enclosed by the graphs of \(y=e^{-x}, y=e^{x},\) and \(x=1 .\) Which of the following gives the volume of the solid generated when \(R\) is revolved about the \(x\) -axis? (a) $$\int_{0}^{1}\left(e^{x}-e^{-x}\right) d x$$ (b) $$\int_{0}^{1}\left(e^{2 x}-e^{-2 x}\right) d x$$ (c) $$\int_{0}^{1}\left(e^{x}-e^{-x}\right)^{2} d x$$ (d) $$\pi \int_{0}^{1}\left(e^{2 x}-e^{-2 x}\right) d x$$ (e)$$\pi \int_{0}^{1}\left(e^{x}-e^{-x}\right)^{2} d x$$

Short answer

The correct answer is (d) \(\pi \int_{0}^{1}(e^{2 x}-e^{-2 x}) d x\).

Step-by-step solution

Identify the correct integration interval and functions to revolve

The region \(R\) is described by two functions \(y = e^{x}\) and \(y = e^{-x}\) and the line \(x=1\), which implies that the region starts at \(x = 0\) where \(e^{x} = e^{-x}\) and ends at \(x=1\). This identifies the interval of the integral as [0,1]. As \(R\) is revolved around the \(x\)-axis, the outer function is \(e^{x}\) since it is above the \(x\)-axis, whereas the inner function that forms the hole is \(e^{-x}\).

Set up the integral using the Disk Method

The volume of the solid is given by the Disk Method as \(V = \pi \int_{a}^{b} ([f(x)]^{2} - [g(x)]^{2}) dx\), where \(f(x)\) is the outer function and \(g(x)\) is the inner function. In this problem, \(f(x) = e^{x}\) and \(g(x) = e^{-x}\). So, the integral becomes \(V = \pi \int_{0}^{1} [(e^{x})^{2} - (e^{-x})^{2}] dx\). After simplifying the integrand, the integral becomes \(V = \pi \int_{0}^{1} (e^{2x} - e^{-2x}) dx\).

Identify the answer

In the provided multiple choices, the integral \(V = \pi \int_{0}^{1} (e^{2x} - e^{-2x}) dx\) corresponds to option (d).

Key concepts

Disk Method

The Disk Method is a technique used to find the volume of a solid of revolution when a region in the plane is revolved around an axis. The name 'Disk Method' comes from the shape of the slices that are formed when the region is rotated; they resemble circular disks.

Imagine slicing the solid perpendicular to the axis of rotation into thin disks. Each disk's volume is approximately the volume of a cylinder, \( V_{disk} = \pi r^2 h \), where \( r \) is the radius of the disk, and \( h \) is its height (or thickness), which is infinitesimally small. In the Disk Method, \( r \) corresponds to the value of a function \( f(x) \) at a given point \( x \) along the axis of rotation.

When you integrate the area of these disks from one boundary of the region \( a \) to another \( b \) along the axis of revolution, you sum up the volumes of all the slices, thus obtaining the volume of the entire solid:
\[ V = \pi \int_{a}^{b} [f(x)]^2 dx \.
\]
However, if there's an inner function \( g(x) \), creating a hollow region (like in a washer), the volume is modified to account for the empty space:
\[ V = \pi \int_{a}^{b} ([f(x)]^2 - [g(x)]^2) dx \.
\]
This adjustment results in the formula used in the aforementioned exercise, enabling us to find the volume of more complex solids of revolution like the one generated by revolving the region \( R \) around the x-axis.

Definite Integral

The definite integral is a fundamental concept in calculus that quantifies the accumulation of quantities, such as areas under curves, total displacement given by a velocity function, or the volume in the context of solids of revolution.

The definite integral is represented by the notation \( \int_{a}^{b} f(x) dx \) and is defined as the net area between the curve \( y = f(x) \) and the x-axis, from \( x = a \) to \( x = b \). If the function lies above the x-axis in this interval, the definite integral represents the area of that region. If the function dips below the x-axis, those parts of the integral represent negative area, or the area is subtracted.

To evaluate a definite integral, we find the antiderivative of the function—known as the indefinite integral—and then apply the Fundamental Theorem of Calculus. This theorem allows us to calculate the integral by taking the difference of the antiderivative evaluated at the upper and lower limits of integration:
\[ F(b) - F(a) \.
\]
In the exercise, the definite integral is applied to find the volume of the solid by integrating the areas of the disks along the interval from 0 to 1. The integral's result gives us a numerical value representing the volume of the solid after revolving the region within the specified bounds, which in this case, matches to option (d) from the multiple choices.

Exponential Functions

Exponential functions are characterized by their base raised to a variable exponent. They are essential in modeling growth and decay processes, such as populations, radioactive decay, and interest in finance. The general form of an exponential function is \( f(x) = a \cdot b^x \), where \( a \) is a constant, \( b \) is the base, and \( x \) is the exponent.

Exponential functions have distinctive properties:

• The base \( b \) is a positive real number.
• For \( b > 1 \) the function reflects exponential growth, and for \( 0 < b < 1 \) it reflects exponential decay.
• The function is always positive, \( f(x) > 0 \) for any real \( x \) value.
• The rate of increase or decrease is proportional to the current function value, meaning it changes at a rate constant in percentage, not in absolute value.

In the context of our exercise, the functions \( y = e^{x} \) and \( y = e^{-x} \) are exponential functions with the natural base \( e \) (approximately equal to 2.71828). Here, the function \( e^{x} \) represents exponential growth, while \( e^{-x} \) represents exponential decay. When revolving these functions around the x-axis and applying the Disk Method, as illustrated in the step-by-step solution, we use the properties of these functions to determine the volume of the resulting solid by integrating their squares over the given interval, emphasizing the dynamic behavior of exponential functions in geometric applications.