Problem 13 - Thomas Calculus

Question

Find the line integral of \(f(x, y, z)=x+y+z\) over the straight-line segment from \((1,2,3)\) to \((0,-1,1)\) .

Step-by-Step Solution

Answer

The line integral is \(-3\).

Step 1: Parametrize the Path

First, we need to find a parameterization for the line segment from point \((1, 2, 3)\) to \((0, -1, 1)\). A linear parameterization can be given by: \( \mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1 \), where \(\mathbf{r}_0 = (1, 2, 3)\) and \(\mathbf{r}_1 = (0, -1, 1)\). Thus, the parameterization is: \( \mathbf{r}(t) = (1-t)(1, 2, 3) + t(0, -1, 1) = (1-t, 2-3t, 3-2t) \), where \( t \) varies from 0 to 1.

Step 2: Express Function Along the Path

Substitute the parameterization \( \mathbf{r}(t) = (1-t, 2-3t, 3-2t) \) into \( f(x, y, z) = x + y + z \). Along this path: \[ f(x, y, z) = (1-t) + (2-3t) + (3-2t) = 6 - 6t. \]

Step 3: Find the Differential of the Path

The differential of the path is given by \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \). The derivative of the parameterization is: \[ \frac{d\mathbf{r}}{dt} = \left(-1, -3, -2\right). \]Thus, \( d\mathbf{r} = (-1, -3, -2) dt. \)

Step 4: Compute the Line Integral

The line integral of \( f(x, y, z) \) over the path is given by \[ \int_C f(x, y, z) \, ds = \int_{0}^{1} f(\mathbf{r}(t)) \, \lVert \frac{d\mathbf{r}}{dt} \rVert \, dt. \]However, for straight-line paths, it's common to use the form \[ \int_{0}^{1} f(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt \]Substituting \( f(\mathbf{r}(t)) = 6 - 6t \) and \( \frac{d\mathbf{r}}{dt} = (-1, -3, -2) \), the integral becomes\[ \int_0^1 (6 - 6t) \cdot (-1) \, dt. \]Evaluate the integral:\[ \int_0^1 -6 + 6t \, dt = \int_0^1 (-6 + 6t) \, dt. \] This integral can be split and solved:\[ \int_0^1 -6 \, dt + \int_0^1 6t \, dt = [-6t]_0^1 + [3t^2]_0^1 = -6 + 3 = -3. \]

Step 5: Provide the Final Answer

Add the evaluated components of the integral to find the line integral:\[-6 + 3 = -3.\]Thus, the value of the line integral over the given path is \(-3\).

Key Concepts

Parametrization of Paths

When calculating a line integral, the first essential step is to parametrize the path you're integrating over. Parametrization effectively means expressing the line or curve through which you are integrating in terms of a single variable, often denoted as \( t \). This process simplifies the complex path into a form that's easier to manage and integrate.In our original exercise, we needed to find a parameterization for the line segment connecting the points \((1, 2, 3)\) and \((0, -1, 1)\).

• We started with the formula: \( \mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1 \), which is a linear interpolation between two points.
• By substituting \( \mathbf{r}_0 \) and \( \mathbf{r}_1 \), we derived our parameterization: \( \mathbf{r}(t) = (1-t, 2-3t, 3-2t) \) with \( t \) ranging from 0 to 1.

This parameterization is crucial because it allows us to express any point along the straight-line segment with a single variable.

Vector Calculus

Vector calculus offers the language and methods required to solve problems involving vectors and spaces. It's an extension of calculus into multiple dimensions, where not only numbers but vectors and functions of several variables are considered.

• In vector calculus, the concept of a vector path is invaluable. Paths are represented as vector functions \( \mathbf{r}(t) \) that trace through space as \( t \) changes.
• Crucially, we calculate the derivative of our path vector, \( \frac{d\mathbf{r}}{dt} \). This derivative points in the direction of the path and has a magnitude representing the rate at which you traverse the path.

For our problem, we found \( \frac{d\mathbf{r}}{dt} = (-1, -3, -2) \). This provides vital information about how the path changes over the interval for \( t \), and is necessary for calculating the line integral.

Integral Calculus

Integral calculus is a fundamental part of calculus focused on accumulated quantities, such as areas under curves and total effects over paths. When considering the line integral from vector calculus, we calculate how much a function accumulates along a specified path.A line integral for a scalar field like \( f(x, y, z) \) involves integrating the function over a path, denoted by \( \int_C f(x, y, z) \, ds \). However, in vector form, especially over straight line paths, the integrals are often simplified.

• In our case study, thanks to the parameterization, \( f(x, y, z) \) becomes a function of \( t \), changing our problem into the integral \( \int_0^1 f(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt \).
• This simplification allows calculation along the path using ordinary calculus techniques, finally resulting in the definite integral that yielded \(-3\).

This process showcases how integral calculus, combined with parametrization and vector calculus, paves the way for solving complex problems elegantly. The integral's result \(-3\) effectively measures the cumulative value of the function along the path segment.