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Magazine Chapter 6: Problem 48
For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path. [T] Evaluate \(\int_{C} 4 x^{3} d s,\) where \(C\) is the line segment from \((-2,-1)\) to \((1,2)\)
Short Answer
Expert verified
The integral evaluates to \(-24\sqrt{2}\).
Step by step solution
01
Understanding the Problem
We need to evaluate the line integral \( \int_{C} 4x^3 \, ds \) along the path defined by the line segment from \((-2, -1)\) to \((1, 2)\). This involves parameterizing the path and expressing the integral in terms of a single variable.
02
Parameterize the Path
The path \( C \) is a line segment from \((-2, -1)\) to \((1, 2)\). A linear parameterization can be given by \( \mathbf{r}(t) = (-2, -1) + t((1, 2) - (-2, -1)) \). This simplifies to \( \mathbf{r}(t) = (-2 + 3t, -1 + 3t) \), where \( t \) ranges from 0 to 1.
03
Express the Integrand
Since the integrand is \( 4x^3 \), we substitute the parameterized \( x(t) = -2 + 3t \) into it. The expression becomes \( 4(-2 + 3t)^3 \).
04
Find the Expression for \( ds \)
The differential arc length \( ds \) is given by \( \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt \). Here, \( dx/dt = 3 \) and \( dy/dt = 3 \), which leads to \( ds = \sqrt{3^2 + 3^2} \, dt = 3\sqrt{2} \, dt \).
05
Set Up the Integral
Substitute the expressions for the integrand and \( ds \) into the line integral, resulting in \[ \int_{0}^{1} 4(-2 + 3t)^3 \, 3\sqrt{2} \, dt. \]
06
Evaluate the Integral
We compute \[ \int_{0}^{1} 3\sqrt{2} \times 4(-2 + 3t)^3 \, dt = 12\sqrt{2} \int_{0}^{1} (-2 + 3t)^3 \, dt. \] Use a CAS to evaluate this integral, which gives \(-24\sqrt{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parameterization
Parameterization is the process of expressing a path, curve, or surface using parameters, typically represented by the variable \( t \). For a line integral, this helps us convert a multidimensional problem into a single-variable problem, making it easier to compute.
In our example, we parameterize the line segment between points \((-2, -1)\) and \((1, 2)\). This is done by finding a function \( \mathbf{r}(t) \) that smoothly moves from the starting point to the endpoint as \( t \) changes from 0 to 1.
Our parameterization is given by:
In our example, we parameterize the line segment between points \((-2, -1)\) and \((1, 2)\). This is done by finding a function \( \mathbf{r}(t) \) that smoothly moves from the starting point to the endpoint as \( t \) changes from 0 to 1.
Our parameterization is given by:
- Start with initial point: \((-2, -1)\)
- Find direction vector to endpoint: \((1, 2) - (-2, -1) = (3, 3)\)
- Parameterize path: \( \mathbf{r}(t) = (-2, -1) + t(3, 3) = (-2+3t, -1+3t) \)
Arc Length Differential
The arc length differential \( ds \) is essential in line integrals because it represents a small segment of the path. Calculating it correctly ensures we take into account the length of the path as we integrate along it.
For our parameterized path, \( \mathbf{r}(t) = (-2 + 3t, -1 + 3t) \), we find the derivatives:
For our parameterized path, \( \mathbf{r}(t) = (-2 + 3t, -1 + 3t) \), we find the derivatives:
- \( \frac{dx}{dt} = 3 \)
- \( \frac{dy}{dt} = 3 \)
- \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} \cdot dt = \sqrt{3^2 + 3^2} \cdot dt = 3\sqrt{2} \cdot dt \)
Computer Algebra System
A Computer Algebra System (CAS) is a software tool that can perform symbolic mathematics. It is particularly useful for evaluating integrals that are complex or cumbersome to compute by hand.
In our exercise, we have an integral \[ \int_{0}^{1} 12\sqrt{2}(-2+3t)^3 \, dt \]This is a polynomial integral that can be expanded and solved symbolically by a CAS, such as Mathematica, Maple, or Wolfram Alpha. These systems are capable of:
In our exercise, we have an integral \[ \int_{0}^{1} 12\sqrt{2}(-2+3t)^3 \, dt \]This is a polynomial integral that can be expanded and solved symbolically by a CAS, such as Mathematica, Maple, or Wolfram Alpha. These systems are capable of:
- Symbolically expanding and simplifying complex expressions
- Performing definite and indefinite integrals
- Offering insight into the function behavior by plotting
Line Segment Path
A line segment path is the simplest path one can define between two points. It is a straight line that connects starting and ending points in a vector space.
Utilizing a line segment path, as in our exercise, simplifies the parameterization and the computation of the line integral greatly.
For a line segment from \((-2, -1)\) to \((1, 2)\), the parameterization is linear and straightforward since:
Utilizing a line segment path, as in our exercise, simplifies the parameterization and the computation of the line integral greatly.
For a line segment from \((-2, -1)\) to \((1, 2)\), the parameterization is linear and straightforward since:
- A line segment has constant direction: given by the vector direction \((3, 3)\)
- The endpoints clearly define the start and end of the path
- Parameter \( t \) takes values from 0 to 1 providing a normalized transition along the path
