Problem 48 - Calculus of a Single Variable

Question

Finding the Area of a Surface of Revolution In Exercises 47 and \(48,\) use the integration capabilitities of a graphing utility to approximate the surface area of the solid of revolution. $$ \begin{array}{ll}{\text { Function }} & {\text { Interval}} & {\text {Axis of Revolution }} \\ {y=\ln x} & {[1, e]} & {y \text { -axis }}\end{array} $$

Step-by-Step Solution

Answer

In order to solve this problem, one would need to use a graphing utility to approximate the integral. The computation of this integral \(2 \pi \int_{1}^{e} x \sqrt{1 + (1/x)^2} dx\) using the utility will give the approximate surface area.

Step 1: Formula recall and setup

The formula for the surface area of a solid of revolution, when rotated around the y-axis, is \[A = 2\pi \int_{a}^{b} x \sqrt{1 + (f'(x))^2} dx \]. Here, \(f(x) = \ln x\), a is the lower limit 1, and b is the upper limit \(e\). Next, compute the derivative of \( f(x) = \ln x\). This gives \(f'(x) = 1/x \). Now set up the integral for the surface area.

Step 2: Substitute into the formula

Replace \(f(x)\) and \(f'(x)\) into the formula for the surface area and evaluate the integral: \[ A = 2 \pi \int_{1}^{e} x \sqrt{1 + (1/x)^2} dx \]. This integral could be complicated to solve by hand but we've been asked to use a graphing utility to approximate the surface area. This integral represents the area under the curve \(x \sqrt{1 + (1/x)^2}\) from 1 to \(e\).

Step 3: Approximate the integral

Calculate the integral using a graphing utility to approximate the value. Note that you might need to select a suitable method for the numeric computation and provide the accuracy required. The surface area will be approximately the numerical value of this integral.

Key Concepts

Integration

Integration is a fundamental concept in calculus that deals with finding the total accumulation of quantities. Imagine you have little slices of areas under a curve on a graph; integration is like adding up all those little slices to get the total area. This process is crucial when determining properties such as the area under a curve, the displacement of an object given its velocity, and, in our case, the surface area of a solid of revolution.

When we revolve a curve around a given axis, we essentially create a 3-dimensional object. To find the surface area of this object, we must integrate the function that represents the shape of its surface. The general formula for the surface area of a solid of revolution is an integral that involves the radius of revolution and the derivative of the function being revolved, a notion that is both elegant and powerful for such calculations.

Graphing Utility

A graphing utility is a powerful tool that can plot graphs, solve equations, and perform numerous calculations including integrals. When approached with an integral that is complex or difficult to solve by hand, a graphing utility becomes extremely useful. It can provide a visual representation of the function as well as an accurate approximation of the integral.

In the context of finding the surface area of a solid of revolution, after setting up the integral, one can enter the integral function into a graphing utility to visualize the curve whose area we want to calculate. This helps not only in understanding the problem better but also in computing the integral numerically, which can be a real timesaver in complicated scenarios.

Definite Integral

A definite integral has limits of integration, which indicates that the integration is to be performed over a specific interval on the x-axis. It provides a precise answer instead of a general function, encapsulating the exact accumulation of area between the set bounds. The outcome of a definite integral is a number, which often represents a physical quantity, depending on the context of the problem.

In the example exercise, the definite integral \( 2\pi \int_{1}^{e} x \sqrt{1 + (1/x)^2} dx \) is used to find the exact surface area of the solid of revolution created by revolving the function \( y = \ln x \) around the y-axis, from \( x = 1 \) to \( x = e \) (the natural logarithm base). This numerical result represents the total surface area across that interval.

Calculus of Single Variable

Calculus of a single variable deals with functions that involve only one independent variable. It encompasses both differential and integral calculus. Differential calculus focuses on the rate of change of functions, while integral calculus concentrates on the accumulation of quantities.

Our current exercise is an application of single-variable calculus as we are only dealing with functions of \( x \) and their behavior as we revolve them around an axis. Calculus of a single variable is perfectly suited for solving problems involving rates of change and areas, such as our surface area of a solid of revolution–an elegant example of how a one-dimensional function can reveal insights into three-dimensional geometry.