Problem 11 - Thomas Calculus

Question

In Exercises \(7-12\) , 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{equation} \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)=\mathbf{i} \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} \end{equation} $$\mathbf{F}=\left(3 x^{2}-3 x\right) \mathbf{i}+3 z \mathbf{j}+\mathbf{k}$$

Step-by-Step Solution

Answer

The line integrals are: C1 = 2.5, C2 = 2, C3 & C4 = 1.

Step 1: Understand the Line Integral

The line integral of a vector field \( \mathbf{F} \) over a path \( C \) is given by \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) is the differential displacement vector along the path \( C \).

Step 2: Calculate for Path C1

For the straight-line path \( C_1 \), parametrize with \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \) where \( 0 \leq t \leq 1 \). Compute \( d\mathbf{r} = (1, 1, 1) dt \). Substitute \( x = t \), \( z = t \) in \( \mathbf{F}(x, y, z) = (3t^2 - 3t, 3t, 1) \). Evaluate the integral \( \int_{0}^{1} [(3t^2 - 3t)\(1 +(3t)\)1 + 1$1] \, dt = \int_{0}^{1} (3t^2 + t + 1) \, dt \).

Step 3: Solve the Integral for Path C1

Simplify \( \int_{0}^{1} (3t^2 + t + 1) \, dt \). This integral expands to \( \left[ t^3 + \frac{1}{2}t^2 + t \right]_{0}^{1} = \left( 1^3 + \frac{1}{2}\cdot1^2 + 1 \right) - \left(0 \right) \). Result is \(1 + 0.5 + 1 = 2.5 \).

Step 4: Calculate for Path C2

For the curved path \( C_2 \), use \( \mathbf{r}(t) = \mathbf{i} + t^2 \mathbf{j} + t^4 \mathbf{k} \), \( 0 \leq t \leq 1 \). Compute \( d\mathbf{r} = (0, 2t, 4t^3) dt \). Substitute \( x = 1 \), \( z = t^4 \) in \( \mathbf{F}(x, y, z) = (0, 3t^4, 1) \). Integrate \( \int_{0}^{1} [0 \times 0 + (3t^4) \, (2t) + 1 \, (4t^3)] \, dt = \int_{0}^{1} (6t^5 + 4t^3) \, dt \).

Step 5: Solve the Integral for Path C2

Calculate \( \int_{0}^{1} (6t^5 + 4t^3) \, dt \), which simplifies to \( \left[ t^6 + t^4 \right]_{0}^{1} = (1^6 + 1^4) - (0) = 1 + 1 = 2 \).

Step 6: Calculate for Path C3 and C4

For the path \( C_3 \cup C_4 \), divide the problem into segments: first for \( C_3 \) with \( \mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} \), \( 0 \leq t \leq 1 \), \( d\mathbf{r} = (1, 1, 0) dt \). Substitute \( x = t \), \( z = 0 \) into \( \mathbf{F}(x, y, z) = (3t^2 - 3t, 0, 1) \), evaluate the integral. For \( C_4 \), from \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k} \) for \( 0 \leq t \leq 1 \), compute \( d\mathbf{r} = (0, 0, 1) dt \). Substitute \( x = 1 \), \( z = t \) into \( \mathbf{F}(x, y, z) = (0, 3, 1) \). Evaluate the integral.

Step 7: Combine Results for C3 and C4

For \( C_3 \): \( \int_{0}^{1} (3t^2 - 3t) \cdot 1 \, dt = \int_{0}^{1} (3t^2 - 3t) \, dt = 0 \). For \( C_4 \): \( \int_{0}^{1} 1 \cdot 1 \, dt = [t]_{0}^{1} = 1 \). Add results: \( 0 + 1 = 1 \).

Key Concepts

Vector Field

In mathematics, a vector field is a function that assigns a vector to each point in a subset of space. Think of it as a map that shows a direction and magnitude at every point in space. Vector fields are used to represent a variety of physical phenomena, such as fluid flow, gravitational fields, and electromagnetic fields.

For example, in the exercise given, the vector field is defined as:

• \( \mathbf{F} = (3x^2 - 3x) \mathbf{i} + 3z \mathbf{j} + \mathbf{k} \)

This means at every point \( (x, y, z) \) in space, the vector at that point has its components determined by these functions of \( x \), \( y \), or \( z \).

Understanding vector fields is essential because they provide a way to model real-world situations where quantities are distributed through space, and can influence motion or behavior of objects.

Parametrization

Parametrization is the process of defining a path or curve in terms of a single parameter, usually denoted as \( t \). It is a crucial concept in calculus for simplifying complex curves into a form that is easier to analyze and compute.

In our exercise, each path \( C_1, C_2, C_3, \) and \( C_4 \) had their own parametrization. For example, for the straight-line path \( C_1 \), the curve was defined as:

• \( \mathbf{r}(t) = t \mathbf{i} + t \mathbf{j} + t \mathbf{k} \)
• where \( 0 \leq t \leq 1 \)

This means as \( t \) changes from 0 to 1, the path traces a straight line between the points \( (0,0,0) \) and \( (1,1,1) \).

Parametrization simplifies calculations and allows for evaluating integrals along curves, which would otherwise be difficult to compute.

Integral Calculus

Integral calculus studies accumulation of quantities and the spaces under curves, pivotal for analyzing total quantities over continuous regions. Line integrals extend this concept to vector fields, measuring the cumulative effect of a field along a path.

In the exercise, we computed line integrals of vector field \( \mathbf{F} \) along paths:

• \( \int_C \mathbf{F} \cdot d\mathbf{r} \)

This formula represents the work done by the vector field along the curve \( C \).

Through integral calculus, we solve integrals like \( \int_{0}^{1} (3t^2 + t + 1) \, dt \), enabling the computation of total quantities such as work, energy, or flow through lines and surfaces.

Differential Displacement

Differential displacement, expressed as \( d\mathbf{r} \), refers to a tiny change or increment in position along a path or curve. It's the elemental vector that points in the direction of the path and has a magnitude representing an infinitesimally small distance.

For example, on the straight line path \( C_1 \), the differential displacement is:

• \( d\mathbf{r} = (1, 1, 1) \, dt \)

This small vector can be thought of as the 'tangent' to the path at each point, crucial for computing line integrals since it represents how the path changes at tiny intervals.

Differential displacement helps in evaluating line integrals by determining directions and lengths over which the vector field \( \mathbf{F} \) exerts its influence, thereby playing a key role in the integration process over curves.