Problem 3 - Thomas Calculus

Question

In Exercises \(1-8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface, too. c. Use your grapher's or computer's integral evaluator to find the surface's area numerically. \(x y=1, \quad 1 \leq y \leq 2 ; \quad y\) -axis

Step-by-Step Solution

Answer

The surface area is approximately 7.458 square units.

Step 1: Understand the Problem

We need to find the surface area of the curve defined by the equation \(xy = 1\), which is a hyperbola, as it revolves around the \(y\)-axis from \(y = 1\) to \(y = 2\).

Step 2: Express x in terms of y

Since we have the equation \(xy = 1\), we can express \(x\) as a function of \(y\): \[ x = \frac{1}{y} \]

Step 3: Set up the Formula for Surface Area

The formula for the surface area of a curve \(x = f(y)\) rotated about the \(y\)-axis is: \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \] Here, \(f(y) = \frac{1}{y}\), \(c = 1\), and \(d = 2\).

Step 4: Compute the Derivative

First, find the derivative of \(x\) with respect to \(y\):\[ \frac{dx}{dy} = \frac{d}{dy} \left( \frac{1}{y} \right) = -\frac{1}{y^2} \]

Step 5: Plug into the Surface Area Formula

Substitute \(x = \frac{1}{y}\) and \(\frac{dx}{dy} = -\frac{1}{y^2}\) into the formula: \[ S = 2\pi \int_{1}^{2} \frac{1}{y} \sqrt{1 + \left( -\frac{1}{y^2} \right)^2} \, dy \] This simplifies to: \[ S = 2\pi \int_{1}^{2} \frac{1}{y} \sqrt{1 + \frac{1}{y^4}} \, dy \]

Step 6: Calculate the Integral

Use a graphing calculator or computer to evaluate the integral: \[ S = 2\pi \int_{1}^{2} \frac{1}{y} \sqrt{1 + \frac{1}{y^4}} \, dy \approx 7.458 \]

Step 7: Graph the Curve

Use graphing software to plot the hyperbola \(xy = 1\), and visualize the surface generated by revolving it around the \(y\)-axis from \(y = 1\) to \(y = 2\).

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand the behavior of functions and their changes. It involves concepts like differentiation and integration, which are crucial for solving problems involving rates of change and areas under curves.
In this exercise, we are dealing with the concept of calculus in the context of surface area. By using calculus, specifically integration, we can find the surface area of a curve when it is revolved around an axis. Calculus allows us to understand how small changes in one variable affect the other, providing us with a way to predict and calculate things like surface areas efficiently.
When we revolve a line or curve around an axis, calculus provides us with the tools to calculate exact values, forming the foundation for methods used in this exercise.

Surface Area Integral

The surface area integral is an application of calculus used to find the area of a surface generated by rotating a curve around an axis. In our exercise, the curve described by the equation \( xy = 1 \) is revolved around the \( y \)-axis, and the surface area integral is used to calculate the area of the resulting surface.
This involves:

• Expressing the curve with a function (in this case, \( x = \frac{1}{y} \)).
• Using a specific formula for the surface area when rotating about the \( y \)-axis: \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]
• Evaluating the integral to get a numerical value for the area.

Utilizing the derivative \( \frac{dx}{dy} \), the integral accounts for the curvature's effect on the final surface area. The precise calculations ensure that we capture all the intricacies of the surface formed by the rotation.

Hyperbola

A hyperbola is a type of smooth curve lying in a plane, defined by its relationship in a mathematical equation. In the exercise, the equation \( xy = 1 \) defines a hyperbola. This means that as one variable increases, the other decreases, maintaining a consistent product.
The graph of this specific hyperbola consists of two disconnected curves, but in our focus range from \( y = 1 \) to \( y = 2 \), we're looking at a single branch that lies in the first quadrant of the Cartesian plane.
Hyperbolas are characterized by their scalable symmetry and provide fascinating geometrical and mathematical properties, necessary for understanding concepts in calculus, such as the revolving surfaces in this problem. The reliable features of a hyperbola allow us to confidently apply the principles of calculus and integration.

Graphing

Graphing is a fundamental skill in mathematics that allows us to visualize equations and functions. In this exercise, graphing the hyperbola \( xy = 1 \) can be particularly beneficial.
By graphing, we can:

• Visualize the range and behavior of the curve between \( y = 1 \) and \( y = 2 \).
• Understand how the rotation around the \( y \)-axis will generate a surface.
• Identify symmetry and other characteristics of the graph that might simplify our problem-solving.

Graphing tools or software are highly effective at revealing the potential complexities in the curves and making mathematical concepts more accessible. In this particular problem, they play a crucial role in not only plotting the hyperbola but also in visualizing the surface area generated by its revolution, ultimately enhancing comprehension.