Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-10, sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\)-axis. $$ y=e^{x}, y=\frac{e}{x}, y=0, \text { between } x=0 \text { and } x=3 $$

Step-by-Step Solution

Answer

The volume of the solid is \( \pi \left( \frac{1}{2}(e^6 - 1) - \frac{2e^2}{3} \right) \).

Step 1: Sketch the Region

To begin, we need to graph the equations to find the region \( R \): \( y = e^x \), \( y = \frac{e}{x} \), and \( y = 0 \). The region is bounded by these functions between \( x=0 \) and \( x=3 \). Plot these curves over the given interval. Note how they cross and form a closed region.

Step 2: Identify a Typical Vertical Slice

Typical vertical slices will be rectangles within the region \( R \). These slices have a width \( \Delta x \) and height determined by the vertical distance between the curves \( y = e^x \) and \( y = \frac{e}{x} \). These are adjusted based on \( y = 0 \) as the bottom boundary.

Step 3: Set Up the Volume Integral

Write an expression for the volume of the solid of revolution by revolving the area about the \( x \)-axis. Using the disk method, the function for the outer radius at a point \( x \) is \( e^x \) and the inner radius is \( \frac{e}{x} \). The volume \( V \) is given by \[ V = \pi \int_{0}^{3} \left[(e^x)^2 - \left(\frac{e}{x}\right)^2\right] \, dx. \]

Step 4: Integrate to Find the Volume

Evaluate the integral you've set up. Calculate \[ \int_{0}^{3} (e^{2x} - \left(\frac{e}{x}\right)^2) \, dx = \int_{0}^{3} e^{2x} \, dx - \int_{0}^{3} \frac{e^2}{x^2} \, dx. \] Compute each of these integrals separately and subtract the second from the first.

Step 5: Compute Each Integral

The first integral \( \int_{0}^{3} e^{2x} \, dx = \frac{1}{2} e^{2x} \bigg|_{0}^{3} \) evaluates to \( \frac{1}{2}(e^6 - 1) \). The second integral \( \int_{0}^{3} \frac{e^2}{x^2} \, dx \) evaluates to \( \left[ - \frac{e^2}{x} \right]_{0}^{3} = e^2 \left( -\frac{1}{3} + 1 \right) = \frac{2e^2}{3} \).

Step 6: Final Volume Calculation

Combine and simplify the results of the integrals in the formula for volume:\[ V = \pi \left( \frac{1}{2}(e^6 - 1) - \frac{2e^2}{3} \right). \] This expression can be simplified to obtain the final volume.

Key Concepts

Disk Method

The Disk Method is a technique used in calculus to find the volume of a solid of revolution. Imagine you have a shape that you want to rotate around an axis to create a three-dimensional object. The Disk Method helps you calculate the volume of this object. When using the Disk Method:

• Think of stacking thin disks (like coins) along the axis of rotation.


• The radius of each disk represents the distance from the axis to the function's value at a certain x.


• By calculating the volume of each disk and summing them, you get the total volume.

In our example, we revolve a region bounded by the curves around the x-axis. The volume expression uses the outer radius formula, which is defined by the equation of the curve furthest from the axis and the inner radius close to the axis. Briefly, the formula involves definite integrals of squared radii.

Definite Integrals

Definite Integrals are used in calculus to find the area under a curve or between curves over a specific interval. When finding the volume of solids of revolution, definite integrals are crucial to determine how much space the solid occupies. In the problem we are discussing, definite integrals help to:

• Systematically add up the volumes of the infinitesimally thin disks generated by revolving the region.


• Calculate from one limit, x = 0, to another, x = 3, ensuring the entire region's volume is accounted for.


• Provide an exact volume calculation when coordinates and exact functions are known.

Without definite integrals, we'd rely on less systematic, less accurate methods for volume calculation.

Calculus Integration Techniques

To solve definite integrals like the ones in the disk method, specific calculus integration techniques are employed. These techniques help solve complex integrals that can't be easily calculated:

• Substitution Method: Useful for transforming integrals into simpler forms.


• Integration by Parts: Helps in breaking down integrals of products of functions.

For our volume problem, separate integrals need evaluating: integrating powers of the exponential function and rational functions. Understanding each integration's technique helps calculate accurately and reach a solution.

Function Graphing

Function Graphing is fundamental in visualizing regions involved when discussing volumes of solids of revolution. By sketching curves such as \(y = e^x\), \(y = \frac{e}{x}\), and \(y = 0\), you identify where these functions intersect and bound the region to be considered:

• Helps in understanding the shapes and boundaries involved.


• Provides a clearer picture of the area being revolved.


• Assists in determining which function serves as the outer or inner radius in the disk method setup.

Without graphing, it becomes challenging to precisely set limits for integration or determine the computational plan for volume calculation.