Question

[T] Use a CAS to find the area of the surface generated by rotating \(x=t+t^{3}, y=t-\frac{1}{t^{2}}, 1 \leq t \leq 2\) about the \(x\) -axis. (Answer to three decimal places.)

Short answer

The area is approximately 4.712 square units.

Step-by-step solution

Understand the Formula for Surface Area

To find the surface area generated by rotating a curve about the x-axis, we use the formula: \( S = \int_{a}^{b} 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^{2}} \, dt \). We are given parametric equations \( x(t) = t + t^3 \) and \( y(t) = t - \frac{1}{t^2} \).

Differentiate to Find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

First, find the derivatives of the parametric functions: \( \frac{dy}{dt} = 1 + \frac{2}{t^3} \) and \( \frac{dx}{dt} = 1 + 3t^2 \).

Compute \( \frac{dy}{dx} \)

Using the chain rule for parametric equations, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + \frac{2}{t^3}}{1 + 3t^2} \).

Simplify the Expression Inside the Integral

Substitute \( \frac{dy}{dx} \) into the surface area formula. The expression \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) becomes: \[ 1 + \left( \frac{1 + \frac{2}{t^3}}{1 + 3t^2} \right)^2\].

Set Up the Integral for Surface Area

Substitute \( y(t) = t - \frac{1}{t^2} \) and the expression from Step 4 into the formula. The integral then becomes: \[ S = \int_{1}^{2} 2\pi \left(t - \frac{1}{t^2}\right) \sqrt{1 + \left( \frac{1 + \frac{2}{t^3}}{1 + 3t^2} \right)^2} \, dt \].

Use a Computer Algebra System (CAS) for Calculation

Input the integral from Step 5 into a CAS tool to compute the exact value of the surface area. Ensure the tool gives the result up to three decimal places.

Key concepts

Understanding Parametric Equations

Parametric equations are a way to define mathematical equations where the coordinates are expressed as functions of a parameter, typically denoted as "t". This approach is particularly useful in situations where the path or curve does not necessarily adhere to a single function graph by itself in more traditional Cartesian coordinates.
For instance, in the provided exercise, we have two parametric equations:

• \(x(t) = t + t^3\)
• \(y(t) = t - \frac{1}{t^2}\)

These equations represent a trajectory in the xy-plane as the parameter \(t\) varies between the bounds \(1 \leq t \leq 2\). By manipulating such parametric equations, one can easily model dynamic and intricate curves and find parameters like length, area, and in this case, surface area after rotation.

Calculating Surface Area in Calculus

Surface area calculation involves finding the area covered by a surface when a curve is revolved around an axis. In calculus, this is done through an integral formula specific to the nature of the rotation.
When dealing with parametric equations, to find the surface area of a curve rotated about the x-axis, we adapt the formula: \[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^{2}} \ dt \]Where,

• \(a\) and \(b\) define the limits of the parameter \(t\)
• \(y\) is the function of \(t\) within the parametric equation
• \(\frac{dy}{dx}\) is the derivative ratio calculated using parametric derivatives \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

In our problem, computing this integral will give us the total surface area generated by the rotation of our parameterized curve around the x-axis, symbolizing the fluid application of calculus techniques on geometric transformations.

Integral Calculation Process

Integral calculation for surface area not only demands a formula but also involves a series of steps which systematically unravel the solution. Calculating integrals essentially accumulates infinitesimal sections to find total values, which works seamlessly for surface revolutions.
To begin, we calculate derivatives for \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), yielding expressions needed for the derivative ratio \(\frac{dy}{dx}\). In this exercise, those derivatives are:

• \(\frac{dy}{dt} = 1 + \frac{2}{t^3}\)
• \(\frac{dx}{dt} = 1 + 3t^2\)

Using these, we establish the derivative ratio \(\frac{dy}{dx}\) and simplify to use in the area formula.
This process highlights the integral's fundamental role, aggregating the continuous and infinite contributions of these differential elements across defined bounds to compute surface success precisely.

Using a CAS for Complex Integrals

A Computer Algebra System (CAS) is an essential tool for handling complex algebraic and calculus problems, especially when manual calculations become cumbersome and prone to errors.
Software like Mathematica, Symbolab, or Wolfram Alpha significantly eases these processes, especially for integrals that are computationally intense or difficult to simplify manually, like the one in this exercise.
After setting up the integral with specific bounds and expressions from our parametric functions, using a CAS allows precise and efficient computation. It swiftly solves the integral

• Calculating the integral from bounds \(t=1\) to \(t=2\)
• Substituting complex expressions into the integral formula
• Yielding results with high precision, often beyond manual methods

Utilizing a CAS not only ensures accuracy but also trains students to leverage technology for solving higher-level calculus problems, an invaluable skill in academic and professional settings.