Problem 121 - Calculus Volume 3

Question

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.)

Step-by-Step Solution

Answer

The surface area is approximately 83.720 square units.

Step 1: Describe the problem

We are given parametric equations \(x = t + t^3\) and \(y = t - \frac{1}{t^2}\) with \(t\) ranging from 1 to 2. We need to find the area of the surface generated by rotating this curve about the \(x\)-axis.

Step 2: Set up the surface area integral

The formula for the surface area of a parametric curve \((x(t), y(t))\) rotated about the \(x\)-axis is \[ A = \int_{a}^{b} 2 \pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. \] Here, we need to calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

Step 3: Compute derivatives

First, compute the derivative \(\frac{dx}{dt}\): \(\frac{dx}{dt} = 1 + 3t^2.\) Then, compute \(\frac{dy}{dt}\): \(\frac{dy}{dt} = 1 + \frac{2}{t^3}.\)

Step 4: Substitute and simplify the integral

Substitute \(y(t)\), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\) into the surface area formula: \[ A = \int_{1}^{2} 2 \pi \left(t - \frac{1}{t^2}\right) \sqrt{(1 + 3t^2)^2 + \left(1 + \frac{2}{t^3}\right)^2} \, dt. \] Simplify \(\sqrt{(1 + 3t^2)^2 + \left(1 + \frac{2}{t^3}\right)^2}\).

Step 5: Evaluate the integral using CAS

Use a Computer Algebra System (CAS) to evaluate the integral. Calculate \[ A = \int_{1}^{2} 2 \pi \left(t - \frac{1}{t^2}\right) \sqrt{(1 + 3t^2)^2 + \left(1 + \frac{2}{t^3}\right)^2} \, dt. \] Result: \(A \approx 83.720\).

Step 6: Conclusion

The area of the surface generated by rotating the given parametric curve about the \(x\)-axis is approximately 83.720 square units.

Key Concepts

Parametric Equations

Parametric equations are a way to define a curve using parameters. In this exercise, we use the parameter \(t\) to express both \(x\) and \(y\) as equations. This means that instead of describing \(y\) purely as a function of \(x\), we introduce a third variable, \(t\), to relate both \(x\) and \(y\).
In our specific example, the parametric equations given are \(x = t + t^3\) and \(y = t - \frac{1}{t^2}\). These equations describe a path or a curve in the coordinate plane, where each value of \(t\) corresponds to a specific point \((x, y)\).

• Advantages of parametric equations:
• They provide a way to represent curves that might be difficult or impossible to represent using the usual \(y=f(x)\) format.
• A single parameter can elegantly handle curves that bend and loop back on themselves.

Understanding how to use parameterized equations helps in exploring more complex curves, such as those rotated to form surfaces of revolution.

Definite Integral

The definite integral is a key concept in calculus used to find the accumulation of a quantity, such as area under a curve. In the context of our surface area of revolution problem, the definite integral is used to sum an infinite number of infinitesimally small areas to find the total surface area.
In this task, we work with the integral:
\[A = \int_{1}^{2} 2 \pi \left(t - \frac{1}{t^2}\right) \sqrt{(1 + 3t^2)^2 + \left(1 + \frac{2}{t^3}\right)^2} \, dt. \]

• This is an application of the parametric surface area formula, which accounts for the curve defined by \(x(t)\) and \(y(t)\) rotated around the \(x\)-axis.
• The bounds from 1 to 2 specify the range of \(t\) over which we evaluate the function.
• The integral aggregates the surface elements along this range, providing the total area.

This integral can be complex to evaluate by hand but can be simplified using technological tools.

Derivative Calculations

Derivatives play a central role in determining the rate at which things change, and they are vital in calculating the surface area in this exercise. Here, derivatives of parametric equations are necessary to derive lengths and curves accurately for integration.

• We calculated the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) as part of setting up our integral.
• For \(x = t + t^3\), the derivative \(\frac{dx}{dt} = 1 + 3t^2\).
• For \(y = t - \frac{1}{t^2}\), the derivative \(\frac{dy}{dt} = 1 + \frac{2}{t^3}\).

These derivatives capture how fast \(x\) and \(y\) change as \(t\) changes, essential for understanding and calculating the geometry of the curve.
They appear in the surface area formula because they describe the slope and form the hypotenuse of the right triangle needed for computing the arc length, which is critical to finding the surface area.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a powerful tool that assists in solving complex mathematical problems by providing symbolic computation capabilities. In this exercise, we used a CAS to evaluate the integral for the surface area of revolution.

• Why use a CAS?
• It can handle intricate computation that may be too cumbersome to perform manually.
• It reduces the risk of human error and speeds up the problem-solving process.

A CAS processes symbolic mathematical expressions and can provide exact answers when possible. For example, in evaluating the integral of our surface area problem, it computed the precise numerical result \(A \approx 83.720\).
This tool is especially helpful for students and professionals who often engage with complex calculations, enabling them to focus more on understanding concepts rather than on lengthy achromatic calculations.