Problem 10 - Thomas Calculus

Question

Find the line integrals of \(\mathbf{F}\) from \((0,0,0)\) to \((1,1,1)\) over each of the following paths in the accompanying figure. $$ \begin{array}{l}{\text { a. The straight-line path } C_{1} : \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { b. The curved path } C_{2} : \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{4} \mathbf{k}, \quad 0 \leq t \leq 1} \\ {\text { c. The path } C_{3} \cup C_{4} \text { consisting of the line segment from }(0,0,0)} \\ {\text { to }(1,1,0) \text { followed by the segment from }(1,1,0) \text { to }(1,1,1)}\end{array} $$ $$ \mathbf{F}=x y \mathbf{i}+y z \mathbf{j}+x z \mathbf{k} $$

Step-by-Step Solution

Answer

C1: 1, C2: 11/18, C3∪C4: 5/6

Step 1: Evaluate the line integral for path C1

Given the path \( C_1: \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \), find \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \).Start by computing the derivative of \( \mathbf{r}(t) \):\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]Compute \( \mathbf{F}(t) = x(t) y(t) \mathbf{i} + y(t) z(t) \mathbf{j} + x(t) z(t) \mathbf{k} \):- \( x = t, \ y = t, \ z = t \)- Thus, \( \mathbf{F}(t) = t^2 \mathbf{i} + t^2 \mathbf{j} + t^2 \mathbf{k} \)Now evaluate the dot product:\[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = t^2 + t^2 + t^2 = 3t^2 \]Integrate over \( t \) from 0 to 1:\[ \int_0^1 3t^2 \, dt = \left[ t^3 \right]_0^1 = 1 \]

Step 2: Evaluate the line integral for path C2

Given the path \( C_2: \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \), compute \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \).Start by computing the derivative of \( \mathbf{r}(t) \):\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} + 4t^3 \mathbf{k} \]Compute \( \mathbf{F}(t) = x(t) y(t) \mathbf{i} + y(t) z(t) \mathbf{j} + x(t) z(t) \mathbf{k} \):- \( x = t, y = t^2, z = t^4 \)- Thus, \( \mathbf{F}(t) = t^3 \mathbf{i} + t^6 \mathbf{j} + t^5 \mathbf{k} \)Evaluate the dot product:\[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = t^3 + 2t^7 + 4t^8 \]Integrate over \( t \) from 0 to 1:\[ \int_0^1 (t^3 + 2t^7 + 4t^8) \, dt = \left[ \frac{t^4}{4} + \frac{t^8}{4} + \frac{t^9}{9} \right]_0^1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{9} = \frac{11}{18} \]

Step 3: Evaluate the line integral for path C3 from (0,0,0) to (1,1,0)

The path for \( C_3 \) is a straight line segment from \((0,0,0)\) to \((1,1,0)\), defined as \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} \) for \( 0 \leq t \leq 1 \).First, compute the derivative of \( \mathbf{r}(t) \):\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + \mathbf{j} \]Calculate \( \mathbf{F}(t) \) where \( z = 0 \):- \( \mathbf{F}(t) = t^2 \mathbf{i} \) (other components are zero since \( y,z \) products include zero)Evaluate the dot product:\[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = t^2 \]Integrate over \( t \) from 0 to 1:\[ \int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3} \]

Step 4: Evaluate the line integral for path C4 from (1,1,0) to (1,1,1)

The path for \( C_4 \) is a vertical line from \((1,1,0)\) to \((1,1,1)\), defined as \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k} \) for \( 0 \leq t \leq 1 \).First, compute the derivative:\[ \frac{d\mathbf{r}}{dt} = \mathbf{k} \]For this path, \( \mathbf{F}(t) = 1 \cdot 1 \mathbf{i} + 1 \cdot t \mathbf{j} + 1 \cdot t \mathbf{k} = \mathbf{i} + t \mathbf{j} + t \mathbf{k} \)Evaluate the dot product:\[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = t \]Integrate over \( t \) from 0 to 1:\[ \int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2} \]

Step 5: Combine results for path C3 and C4

The total line integral for path \( C_3 \cup C_4 \) is the sum of the results from Step 3 and Step 4.Thus, the result is:\[\text{Line integral for } C_3 \cup C_4 = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\]

Key Concepts

Vector Fields

In mathematics, a vector field is a function that assigns a vector to each point in a subset of space. You can imagine a vector field as a map of arrows. Each arrow represents both a direction and a magnitude at each point.

Vector fields are essential in physics and engineering as they are used to model various phenomena like the velocity of a moving fluid or the force field around a magnet. In the context of line integrals, the vector field is often represented as \( \mathbf{F} \), a function of position, that describes how the vector quantity varies over space.

For example, in this exercise, the vector field \( \mathbf{F} = xy \mathbf{i} + yz \mathbf{j} + xz \mathbf{k} \) gives us the vectors according to the position \((x, y, z)\). Understanding how vector fields work with line integrals will help you understand physical concepts such as work done by a force.

Straight-Line Path

A straight-line path is one of the simplest ways to connect two points in space. It is defined by linearly interpolating between these points. Mathematically, a straight-line path can be represented using a parameterized vector function.

In the exercise, path \( C_1 \) is defined as a straight line from \((0,0,0)\) to \((1,1,1)\) given by \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \) where \( 0 \leq t \leq 1 \). This path indicates that as \( t \) varies from 0 to 1, the position moves linearly along each of the x, y, and z axes.

• \(t = 0\) gives the initial position \((0,0,0)\)
• \(t = 1\) gives the final position \((1,1,1)\)

Using this parameterization simplifies the computation of the line integral because the derivation of a vector field across a straight path reduces to simple algebraic operations.

Curved Path

A curved path is more complex because it deviates from the straight line. Such paths evoke changes not only in direction but potentially in speed as well. Describing curved paths usually involves a parameterization that captures this non-linearity.

In the exercise, path \( C_2 \) is a curved path given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \) with \( 0 \leq t \leq 1 \). This path involves quadratic and quartic components in the y and z directions, respectively. Here:

• The \(x\)-component, \(t\), increases linearly.
• The \(y\)-component, \(t^2\), increases quadratically, making the curve bend.
• The \(z\)-component, \(t^4\), rises rapidly due to the degree of 4.

Understanding the nature of a curved path is crucial for correct evaluation of line integrals, as it influences how vector fields apply over the path.

Parameterization

Parameterization is a fundamental technique in calculus that simplifies difficult paths into a set of manageable equations. It involves defining a path by introducing a new variable, often called the parameter, which allows you to describe the path in terms of simpler, unified functions.

Parametric equations map out the paths through separate equations for each coordinate. In line integrals, parameterization allows the decomposition of complex paths into simpler segments that can be individually integrated.

• For a straight line, use linear equations like \( x = a + (b-a)t \).
• For curves, consider non-linear terms like \( y = t^2 \) or \( z = t^4 \).

This technique significantly boosts one's ability to compute line integrals, as it breaks down complex geometries into simpler computations. Parameterization transforms integrating a path into integrating a simple function with respect to the parameter values.