Problem 7 - 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} $$\mathbf{F}=3 \mathrm{yi}+2 x \mathbf{j}+4 z \mathbf{k}$$

Step-by-Step Solution

Answer

\( C_1: \frac{9}{2}, C_2: \frac{7}{2}, C_3 \cup C_4: \frac{9}{2} \).

Step 1: Understand the problem

You are given a vector field \( \mathbf{F} = 3y\mathbf{i} + 2x\mathbf{j} + 4z\mathbf{k} \) and asked to evaluate the line integral of \( \mathbf{F} \) over three different paths from \((0,0,0)\) to \((1,1,1)\).

Step 2: Path \(C_1\) - Straight line path

For the straight-line path \( C_1 \) given by \( \mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k} \), compute the line integral \( \int_{C_1} \mathbf{F} \cdot d\mathbf{r} \). Start by computing \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), then substitute \( \mathbf{r}(t) \) into \( \mathbf{F} \) to get \( \mathbf{F}(t) = 3t\mathbf{i} + 2t\mathbf{j} + 4t\mathbf{k} \). The dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 3t + 2t + 4t = 9t \). Integrate \( 9t \) with respect to \( t \) from 0 to 1.\[ \int_0^1 9t \, dt = \left[ \frac{9}{2}t^2 \right]_0^1 = \frac{9}{2} \].

Step 3: Path \(C_2\) - Curved path

For the curved path \( C_2 \) given by \( \mathbf{r}(t) = \mathbf{i} + t^2\mathbf{j} + t^4\mathbf{k} \), compute the line integral. Calculate \( \frac{d\mathbf{r}}{dt} = 2t\mathbf{j} + 4t^3\mathbf{k} \), then substitute \( \mathbf{r}(t) \) in \( \mathbf{F} \) to get \( \mathbf{F}(t) = 3t^2\mathbf{i} + 2\mathbf{j} + 4t^4\mathbf{k} \). The dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 6t^3 + 16t^7 \). Integrate \( 6t^3 + 16t^7 \) with respect to \( t \) from 0 to 1.\[ \int_0^1 (6t^3 + 16t^7) \, dt = \left[ \frac{3}{2}t^4 + 2t^8 \right]_0^1 = \frac{3}{2} + 2 = \frac{7}{2} \].

Step 4: Paths \(C_3\) and \(C_4\) - Segmented path

For \(C_3\) and \( C_4 \), consider them as two separate paths. \( C_3 \) is from \( (0,0,0) \) to \( (1,1,0) \) with \( \mathbf{r}_1(t) = t\mathbf{i} + t\mathbf{j} \), and \( C_4 \) is from \( (1,1,0) \) to \( (1,1,1) \) with \( \mathbf{r}_2(t) = 1\mathbf{i} + 1\mathbf{j} + t\mathbf{k} \). For \( C_3 \), \( \mathbf{F}(t) = 3t\mathbf{i} + 2t\mathbf{j} \) and \( \frac{d\mathbf{r}_1}{dt} = \mathbf{i} + \mathbf{j} \). Thus, \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}_1}{dt} = 5t \). Integrate from 0 to 1, giving \( \frac{5}{2} \). For \( C_4 \), \( \mathbf{F} = 2\mathbf{j} + 4z\mathbf{k} \), \( \frac{d\mathbf{r}_2}{dt} = \mathbf{k} \), leading to \(4t\). Integrate \( 4t \) from 0 to 1, giving \( 2 \). Summing the integrals for \( C_3 \) and \( C_4 \) results in \( \frac{5}{2} + 2 = \frac{9}{2} \).

Step 5: Conclusion

The line integrals for the paths are: \( C_1 = \frac{9}{2} \), \( C_2 = \frac{7}{2} \), and \( C_3 \cup C_4 = \frac{9}{2} \).

Key Concepts

Vector Field

A vector field is a function that assigns a vector to each point in space. In the context of line integrals, we generally deal with vector fields that represent forces, velocities, or other quantities that have both magnitude and direction at every point. For instance, in this exercise, we are given the vector field \( \mathbf{F} = 3y\mathbf{i} + 2x\mathbf{j} + 4z\mathbf{k} \). This vector field describes a force acting at each point \((x, y, z)\). The components of the vector define how the force changes with the spatial coordinates.
Understanding the vector field is crucial because the line integral of a vector field along a path depends on the values of this field along that path. This is why we substitute the path function \( \mathbf{r}(t) \) into the vector field to tailor it to the path in question.

Curve Paths

Curve paths are essential when dealing with line integrals. These paths define the trajectory through which we evaluate the vector field. Curve paths are represented by vector-valued functions. Each component of these functions corresponds to a coordinate direction, such as \( \,\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k} \.\) for a straight line.
Different curve paths (e.g., straight or curved) will interact differently with the vector field. In this exercise, you're working with multiple paths: a straight line, a curved path, and a segmented path combining two line segments. Each type of path brings its own complexity and requires understanding how the vector field behaves along these paths to compute the line integral.

Dot Product

The dot product is a mathematical operation that takes two vectors and results in a scalar. It essentially measures how much one vector goes in the direction of another. When evaluating line integrals, the dot product is used to combine the vector field \( \mathbf{F} \) with the differential path vector \( d\mathbf{r} \). The formula \( \mathbf{F} \cdot d\mathbf{r} \) simplifies the computation by capturing the component of \( \mathbf{F} \) in the direction of the path.
For example, in the integration over path \( C_1 \), the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 9t \) is a crucial step. This result is vital because it sets up the next step of integration by providing a scalar function ready for integration over the specified interval.

Integration

Integration is the process we use to find the line integral of a vector field along a path. Once we have computed the dot product of the vector field and the differential path vector, integration allows us to sum this product over the entire path. This operation gives us the total effect of the field along the path.
In this exercise, the integration involves straightforward definite integrals like \( \int_0^1 9t \, dt \) for the straight-line path \( C_1 \). For curved paths like \( C_2 \), the integration might involve more complex polynomials, but the principle remains the same: integrating over the path gives a single scalar value representing the line integral. Techniques like substitution and using integral properties help simplify these calculations.