Problem 60 - Calculus Early Transcendentals

Question

Find the exact area of the surface obtained by rotating the given curve about the \(x\) -axis. $$x=3 t-t^{3}, \quad y=3 t^{2}, \quad 0 \leq t \leqslant 1$$

Step-by-Step Solution

Answer

The exact area involves evaluating \(\int_0^1 18\pi t^2 \sqrt{1 + 2t^2 + t^4} \, dt\), which requires advanced integration techniques or numeric evaluation.

Step 1: Calculate the Derivative of x and y with respect to t

To find the surface area, first calculate the derivatives of the parametric equations. The derivative of \(x\) with respect to \(t\) is: \[\frac{dx}{dt} = 3 - 3t^2\]The derivative of \(y\) with respect to \(t\) is: \[\frac{dy}{dt} = 6t\].

Step 2: Compute the Differential Arc Length Element

The differential arc length \(ds\) is given by \[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt.\]Substitute the derivatives from Step 1: \[\begin{align*}\frac{dx}{dt} &= 3 - 3t^2, \\frac{dy}{dt} &= 6t.\end{align*}\]Now compute:\[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (3 - 3t^2)^2 + (6t)^2 = 9 - 18t^2 + 9t^4 + 36t^2 = 9 + 18t^2 + 9t^4.\]

Step 3: Set Up and Simplify the Surface Area Integral

The surface area \(A\) of a curve rotated about the \(x\)-axis is given by \[A = \int_{a}^{b} 2\pi y \cdot ds,\]where \(y = 3t^2\) and \(ds = \sqrt{9 + 18t^2 + 9t^4} \, dt\). Set up the integral from 0 to 1:\[\begin{align*}A &= \int_{0}^{1} 2\pi (3t^2) \cdot \sqrt{9 + 18t^2 + 9t^4} \, dt.\end{align*}\]

Step 4: Evaluate the Integral

To evaluate the integral, substitute and simplify:\[A = \int_{0}^{1} 6\pi t^2 \sqrt{9(1 + 2t^2 + t^4)} \, dt = \int_{0}^{1} 6\pi t^2 \sqrt{9} \sqrt{1 + 2t^2 + t^4} \, dt = \int_{0}^{1} 18\pi t^2 \sqrt{1+ 2t^2 + t^4} \, dt.\]Using a trigonometric substitution or numerical integration for full evaluation could simplify this further, potentially by factoring, but the exact form needs to be determined computationally or by advanced calculus techniques.

Step 5: Final Evaluation of the Integral (Optional Information)

Depending on techniques available, the integral can be solved using substitution or numeric methods. For exact solutions, further simplification required involves advanced techniques that might include recognizing forms, substitutions, and/or partial fractions.

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in calculus, allowing us to describe curves by defining both the x and y coordinates in terms of a third parameter, often denoted as t. This approach is particularly useful for describing curves that cannot be easily expressed as a single function of x or y. For example, in our problem, the parametric equations are given by:

• x = 3t - t^3
• y = 3t^2

Here, the parameter t runs from 0 to 1, smoothly tracing the path of the curve in the xy-plane. This representation is not only elegant but also simplifies the process of finding derivatives and integrals related to the curve, as we can work directly with these parametric forms.

Arc Length

The concept of arc length is vital in calculus when dealing with curves. It represents the length of a curve between two points. When curves are expressed using parametric equations, we calculate this length using the differential arc length element, denoted as ds.

To find ds, we utilize the following expression:\[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]This formula integrates the contributions of both x and y changes by incorporating the derivatives of x and y with respect to t. For our specific case:

• \(\frac{dx}{dt} = 3 - 3t^2\)
• \(\frac{dy}{dt} = 6t\)

By substituting these derivatives, we obtain the expression for the arc length element ds, which is essential for calculating curve properties such as the surface area.

Derivative Calculation

Derivatives play a crucial role in finding the surface area of a rotation. By determining how fast each coordinate of a curve changes with respect to t, we can derive accurate representations of the curve's geometry. For parametric equations, we differentiate each one separately:

• For x, the derivative is \(\frac{dx}{dt} = 3 - 3t^2\)
• For y, the derivative is \(\frac{dy}{dt} = 6t\)

Understanding these derivatives helps us determine the slope and direction of the curve at any point. These derivatives are also integral to calculating the arc length and the surface area when the curve is revolved around an axis.

Integral Evaluation

Integral evaluation is the process of calculating the value of an integral, which, in this context, helps determine the surface area of a surface of revolution. The surface area, when a curve is rotated around the x-axis, is calculated by:\[A = \int_{a}^{b} 2\pi y \cdot ds\]For our curve, with y given as \(3t^2\) and the differential arc length \(ds = \sqrt{9 + 18t^2 + 9t^4} \, dt\), we form the integral:\[A = \int_{0}^{1} 6\pi t^2 \sqrt{9 + 18t^2 + 9t^4} \, dt\]Evaluating this integral can be challenging, often requiring substitutions or computational tools, especially when the inner expression is complex. By simplifying or using advanced calculus methods, the integral can be accurately evaluated, providing the exact surface area for the given rotated curve.