Problem 78 - Calculus Volume 3

Question

For the following exercises, use a CAS to evaluate the given line integrals. \([\mathrm{T}] \quad\) Evaluate \(\quad \mathbf{F}(x, y, z)=x^{2} z \mathbf{i}+6 y \mathbf{j}+y z^{2} \mathbf{k},\) where \(\quad C \quad\) is \(\quad\) represented \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\ln t \mathbf{k}, 1 \leq t \leq 3\)

Step-by-Step Solution

Answer

The line integral evaluates to approximately 330.365.

Step 1: Understand the Problem

We need to evaluate the line integral of the vector field \( \mathbf{F}(x, y, z) = x^2 z \mathbf{i} + 6y \mathbf{j} + yz^2 \mathbf{k} \) over a curve \( C \), represented by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \ln t \mathbf{k} \) with \( 1 \leq t \leq 3 \). This means we're finding \( \int_C \mathbf{F} \cdot d\mathbf{r} \).

Step 2: Find \( \mathbf{F}(\mathbf{r}(t)) \)

Substitute \( x = t \), \( y = t^2 \), and \( z = \ln t \) into \( \mathbf{F}(x, y, z) \):\[ \mathbf{F}(\mathbf{r}(t)) = (t^2 \ln t) \mathbf{i} + 6(t^2) \mathbf{j} + (t^2)(\ln t)^2 \mathbf{k}. \]

Step 3: Compute \( d\mathbf{r} \)

Calculate the derivative of the vector function \( \mathbf{r}(t) \) with respect to \( t \):\[ d\mathbf{r} = \left( \frac{d}{dt}(t) \right) \mathbf{i} + \left( \frac{d}{dt}(t^2) \right) \mathbf{j} + \left( \frac{d}{dt}(\ln t) \right) \mathbf{k} = \mathbf{i} + 2t \mathbf{j} + \frac{1}{t} \mathbf{k}. \]

Step 4: Compute \( \mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r} \)

Compute the dot product of \( \mathbf{F}(\mathbf{r}(t)) \) and \( d\mathbf{r} \):\[ (t^2 \ln t) \cdot 1 + 6t^2 \cdot 2t + (t^2(\ln t)^2) \cdot \frac{1}{t} = t^2 \ln t + 12t^3 + t(\ln t)^2. \]

Step 5: Set Up the Integral

Now, set up the integral over \( t \) from 1 to 3:\[ \int_{1}^{3} (t^2 \ln t + 12t^3 + t(\ln t)^2) \, dt. \]

Step 6: Evaluate the Integral Using CAS

Use a computer algebra system (CAS) to evaluate the integral:- Input the integral \( \int_{1}^{3} (t^2 \ln t + 12t^3 + t(\ln t)^2) \, dt \).- Calculate the result, which the CAS will provide as a numeric value.

Key Concepts

Vector Field

In mathematics and physics, a vector field is an assignment of a vector to each point in a subset of space. Imagine a vector field like a gentle breeze where, at any location, the wind has a certain direction and magnitude. In the exercise provided, the vector field is \( \mathbf{F}(x, y, z) = x^2 z \mathbf{i} + 6y \mathbf{j} + yz^2 \mathbf{k} \). Each component \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) represents the direction and size of the vector in the 3D space defined by \( x, y, \) and \( z \).

To work with a vector field in a line integral, you evaluate it along a specified path or curve. This is akin to taking a snapshot of the vector field only along the curve, so you can analyze the "force" along the path. Understanding vector fields is vital for many applications in physics and engineering, especially in electromagnetic fields and fluid flows.

Parameterization of a Curve

Parameterization involves expressing a curve as a function of a parameter, typically \( t \). Instead of describing a curve using \( x \), \( y \), and \( z \) all together, you break it down in terms of a simpler, more maneuverable variable.

In our exercise, the curve \( C \) is given by the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \ln t \mathbf{k} \), where \( t \) ranges from 1 to 3.

• This function defines the path from start to finish as \( t \) continuously varies.
• Each expression of \( t \mathbf{i} \), \( t^2 \mathbf{j} \), and \( \ln t \mathbf{k} \) gives the position in the 3D space for any value of \( t \).

Parameterization is powerful because it breaks down complex curves into easily calculable sections that can be dealt with analytically or computationally.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. Geometrically, it reflects the product of the magnitudes of the two vectors and the cosine of the angle between them.

In the context of line integrals, the dot product helps us determine how much of one vector field flows through another vector along a curve. The exercise shows this by:

• Computing the dot product of \( \mathbf{F}(\mathbf{r}(t)) \) and the derivative \( d\mathbf{r} \).
• This operation gives us a scalar function \( t^2 \ln t + 12t^3 + t(\ln t)^2 \), which is important for setting up the line integral.

It illuminates how the vector field affects or interacts with the parameterized curve by telling you how the curve "aligns" with the vector field at each point.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is software that facilitates symbolic mathematics. It helps in performing algebraic operations on expressions and solving equations that may be too tedious or complex for manual computation.

In the exercise, a CAS is crucial for evaluating the line integral of a complex expression \( \int_{1}^{3} (t^2 \ln t + 12t^3 + t(\ln t)^2) \, dt \). This step is about inputting the integral into the CAS and letting it output the result quickly and accurately. Some advantages of using a CAS include:

• Handling complicated integrals that involve numerous steps.
• Checking the correctness of manually computed results.
• Visualizing changes in the parameters or expressions.

This shows how technology augments mathematical analysis, ensuring precision and saving significant time in calculations.