Problem 48 - Calculus Volume 3

Question

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path. \([\mathrm{T}]\) Evaluate \(\int_{C} 4 x^{3} d s,\) where \(C\) is the line segment from (-2,-1) to (1,2)

Step-by-Step Solution

Answer

The line integral evaluates to \( 36\sqrt{2} \).

Step 1: Understand the Problem

We need to evaluate the line integral \( \int_C 4x^3 \, ds \), where \( C \) is the line segment from (-2,-1) to (1,2). This involves calculating the integral over the path from one point to another.

Step 2: Parametrize the Line Segment

The line segment from (-2,-1) to (1,2) can be represented with a parametric equation. Let \( t \) range from 0 to 1. The points along the segment can be described as \( x(t) = -2 + 3t \) and \( y(t) = -1 + 3t \).

Step 3: Determine the Expression for \( ds \)

The differential arc length \( ds \) for a line parameterized by \( t \) is given by \( ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \). For our line, \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 3 \), so \( ds = \sqrt{3^2 + 3^2} \, dt = \sqrt{18} \, dt = 3\sqrt{2} \, dt \).

Step 4: Substitute and Setup the Integral

Substitute \( x(t) = -2 + 3t \) and \( ds = 3\sqrt{2} \, dt \) into the integral: \( \int_C 4x^3 \, ds = \int_0^1 4(-2 + 3t)^3 \, 3\sqrt{2} \, dt \). This simplifies to \( 12\sqrt{2} \int_0^1 (-2 + 3t)^3 \, dt \).

Step 5: Evaluate the Integral

Calculate the integral \( 12\sqrt{2} \int_0^1 (-2 + 3t)^3 \, dt \). This involves expanding the cubic expression and integrating term by term. The integral evaluates to \( 36\sqrt{2} \).

Step 6: Conclude the Calculation

Thus, the value of the line integral \( \int_C 4x^3 \, ds \) over the line segment from (-2,-1) to (1,2) is \( 36\sqrt{2} \).

Key Concepts

Parametric Equations

Parametric equations are a way to represent geometric objects, like curves or paths, using a parameter, usually denoted as \( t \). Instead of expressing the relationship between \( x \) and \( y \) directly, in parametric form, both \( x \) and \( y \) are expressed as functions of \( t \).

For line integrals, parametric equations are essential because they allow us to describe a path between two points, which in this case are (-2, -1) and (1, 2). By introducing the parameter \( t \), which ranges from 0 to 1, we can model this line segment. Here, we express the \( x \) and \( y \) components separately:

• \( x(t) = -2 + 3t \)
• \( y(t) = -1 + 3t \)

This method can be extended to more complex paths by choosing appropriate functions for \( x(t) \) and \( y(t) \). It simplifies the process of computing line integrals because it breaks down complex curves into simpler, manageable components.

Arc Length Differential

The arc length differential, \( ds \), is a crucial element when dealing with line integrals. This differential represents a tiny segment of the path along which the integral is evaluated. It helps in translating the integral from planar coordinates into a form that can be managed when moving along a curve or line.

Given a parameterized path described by \( x(t) \) and \( y(t) \), the arc length differential is determined by the following expression:\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
In our example, the parametric forms are linear (\( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 3 \)). This reduces \( ds \) to a constant times \( dt \, ( = 3\sqrt{2} \, dt) \). This simplification occurs because both derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) have constant values along straight paths. Understanding \( ds \) is fundamental to evaluating line integrals efficiently.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a software that helps in performing symbolic mathematics. These systems are particularly useful for solving problems that would be tedious to work through manually, such as evaluating complex integrals and derivatives or manipulating equations with many variables.

In the context of line integrals, a CAS can be employed to automate the computation process. It allows for the quick expansion and integration of complex expressions, as was needed with the integral \( 12\sqrt{2} \int_0^1 (-2 + 3t)^3 \, dt \). By delegating such calculations to a CAS, students and professionals can ensure accuracy and save time.

• Reduces calculation errors by automating expansion and integration.
• Saves time on computational-heavy tasks.
• Facilitates deeper focus on interpretation and application aspects of solutions.

By learning to utilize a CAS, you gain a powerful ally in solving and understanding mathematical problems more deeply and effectively.