Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the surface generated by revolving the curve \(x=t^{2} / 2+a t, y=t+a\), for \(-\sqrt{a} \leq t \leq \sqrt{a}\) about the \(x\) -axis.

Step-by-Step Solution

Answer

Calculate the integral \(A = \int_{0}^{2a} 2\pi u \sqrt{u^2 + 1} \, du\) to find the surface area.

Step 1: Understand the Problem Statement

We are given a parametric curve defined by \(x = \frac{t^2}{2} + at\) and \(y = t + a\). We need to find the surface area of the solid formed by rotating this curve over the interval \(-\sqrt{a} \leq t \leq \sqrt{a}\) around the \(x\)-axis.

Step 2: Recognize the Formula for Surface Area of Revolution

When a curve \(y = f(t)\) is revolved around the \(x\)-axis, the formula for the surface area \(A\) is \[ A = \int_a^b 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] Here, \(y = t + a\), \(x = \frac{t^2}{2} + at\), and the limits of integration are \(-\sqrt{a}\) to \(\sqrt{a}\).

Step 3: Compute Derivatives

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). - For \(x = \frac{t^2}{2} + at\), \(\frac{dx}{dt} = t + a\). - For \(y = t + a\), \(\frac{dy}{dt} = 1\).

Step 4: Calculate the Integrand

Substitute the derivatives into the integrand of the surface area formula: \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(t + a)^2 + 1} \).

Step 5: Set Up the Integral

The integral for the surface area is given by: \[ A = \int_{-\sqrt{a}}^{\sqrt{a}} 2\pi (t + a) \sqrt{(t + a)^2 + 1} \, dt \]

Step 6: Evaluate the Integral

This integral can be evaluated using a substitution method. Let \(u = t + a\), then \(du = dt\), and change the limits accordingly. Integrate:\[ A = 2\pi \int_{0}^{2a} u \sqrt{u^2 + 1} \, du \] Evaluate this integral to find the surface area.

Key Concepts

Parametric Equations

Parametric equations provide a powerful way to describe complex curves and lines, allowing us to express each coordinate as a separate function of a parameter, typically denoted by \( t \). In our exercise, we have two equations, \( x = \frac{t^2}{2} + at \) and \( y = t + a \), each defining the \( x \) and \( y \) coordinates, respectively. These equations give us a flexible representation of the curve, making it possible to consider the parameter \( t \) over its defined interval, from \(-\sqrt{a}\) to \(\sqrt{a}\).

This approach offers numerous advantages, especially when dealing with calculus problems that involve rotations, lengths, or other geometrical properties of curves. Key benefits include:

• Separating complex dependencies: Each coordinate is a function of \( t \), simplifying individual analyses.
• Facilitating integration and differentiation: Calculations can focus on individual parameter functions.
• Providing greater geometric flexibility: Easier manipulation and transformation of complex shapes.

Integration Techniques

Integration plays a crucial role when determining areas, lengths, or volumes in calculus, especially when linked with parametric equations. In this problem, we need to find the surface area by integrating over a specified interval. The formula for the surface area of revolution requires integrating:

\[ A = \int_a^b 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

This expression can appear complex, but using strategic substitutions can simplify the process. In our solution, the substitution \( u = t + a \) reduces the complexity by transforming the integral into terms that are easier to handle. This method highlights:

• The utility of substitution: Simplifies integrands and maintains proper variable relationships.
• Adjusting integral limits: Correctly changing limits is critical when converting variables.
• Using symmetry properties: When possible, leverage problem qualities to simplify calculations further, such as equal intervals around symmetry axes.

Calculus Problem Solving

Solving calculus problems involves applying a systematic approach to break down complex questions into manageable steps, as seen in our area of revolution problem. The process typically starts by understanding the problem, then identifying relevant formulas and derivative calculations, followed by setting up and evaluating the integral.

Key aspects of this method include:

• Understanding the context: Grasp the problem statement and know what is being asked.
• Identifying the correct formulas: Select appropriate calculus tools for the problem at hand.
• Derivative and integrand analysis: Use derivatives to construct integrands needed for integration.
• Evaluation and interpretation: Compute accurately and relate results back to the problem.

This structured strategy not only ensures thorough problem-solving but also builds essential calculus skills that are reusable in various mathematical contexts.

Derivatives in Calculus

Derivatives are fundamental in calculus for analyzing how functions change. In our parametric equations, derivatives help us understand the rate of change for each coordinate, assisting in building the integrand for the surface area of revolution. Specifically, we find:

- \( \frac{dx}{dt} = t + a \)
- \( \frac{dy}{dt} = 1 \)

These derivatives are essential for expressing the curve's orientation and fluctuation as \( t \) varies, ensuring the surface area calculation reflects accurate geometry. Here’s why derivatives are important in this context:

• Provide slope information: Determine how each parameter influences the curve's direction.
• Critical for integrand construction: Needed to calculate the differential arc length, tangential effects in rotation integrals.
• Facilitate checking calculations for errors: Verify steps through derivative consistency checks.

Understanding how derivatives work in this framework ensures students can confidently tackle similar calculus challenges.