Problem 16 - Thomas Calculus

Question

The paths of integration for Exercises 15 and 16 Integrate \(f ( x , y , z ) = x + \sqrt { y } - z ^ { 2 }\) over the path from \(( 0,0,0 )\) to \(( 1,1,1 )\) (see accompanying figure) given by $$ \begin{array} { l l } { C _ { 1 } : } & { \mathbf { r } ( t ) = t \mathbf { k } , \quad 0 \leq t \leq 1 } \\ { C _ { 2 } : } & { \mathbf { r } ( t ) = t \mathbf { j } + \mathbf { k } , \quad 0 \leq t \leq 1 } \\ { C _ { 3 } : } & { \mathbf { r } ( t ) = t \mathbf { i } + \mathbf { j } + \mathbf { k } , \quad 0 \leq t \leq 1 } \end{array} $$

Step-by-Step Solution

Answer

The total integral over the path is \(-\frac{1}{6}\).

Step 1: Setup C1 Path Integration

Consider the first path, \( C_1 \), which is parametric \( \mathbf{r}(t) = t\mathbf{k} \). On this path, \( x = 0 \), \( y = 0 \), and \( z = t \). Substitute into \( f(x, y, z) = x + \sqrt{y} - z^2 \) to obtain \( f(t) = -t^2 \). The differential is expressed as \( d\mathbf{r} = \mathbf{k} dt \). The integral becomes:\[\int_{C_1} f(x,y,z) \cdot d\mathbf{r} = \int_0^1 -t^2 dt.\]

Step 2: Compute C1 Integral

Evaluate the integral from Step 1:\[\int_0^1 -t^2 dt = \left[ -\frac{1}{3}t^3 \right]_0^1 = -\frac{1}{3}.\]

Step 3: Setup C2 Path Integration

For path \( C_2 \), where \( \mathbf{r}(t) = t\mathbf{j} + \mathbf{k} \), we have \( x = 0 \), \( y = t \), and \( z = 1 \). Substitute values into \( f(x, y, z) \) to get \( f(t) = \sqrt{t} - 1 \). The differentials give: \( d\mathbf{r} = \mathbf{j} dt \). The integral is:\[\int_{C_2} f(x,y,z) \cdot d\mathbf{r} = \int_0^1 (\sqrt{t} - 1) dt.\]

Step 4: Compute C2 Integral

Solve the integral from Step 3:\[\int_0^1 (\sqrt{t} - 1) dt = \left[ \frac{2}{3}t^{3/2} - t \right]_0^1 = \frac{2}{3} - 1 = -\frac{1}{3}.\]

Step 5: Setup C3 Path Integration

For path \( C_3 \), described by \( \mathbf{r}(t) = t\mathbf{i} + \mathbf{j} + \mathbf{k} \), set \( x = t \), \( y = 1 \), \( z = 1 \). Upon substitution into \( f(x, y, z) \), obtain \( f(t) = t \). The differential is \( d\mathbf{r} = \mathbf{i} dt \). The integral is:\[\int_{C_3} f(x,y,z) \cdot d\mathbf{r} = \int_0^1 t dt.\]

Step 6: Compute C3 Integral

Evaluate the integral from Step 5:\[\int_0^1 t dt = \left[ \frac{1}{2}t^2 \right]_0^1 = \frac{1}{2}.\]

Step 7: Combine Integrals For Total Integration

Combine results from the path integrals: \( C_1: -\frac{1}{3}, C_2: -\frac{1}{3}, \) and \( C_3: \frac{1}{2} \). The total integral over the path is:\[-\frac{1}{3} - \frac{1}{3} + \frac{1}{2} = -\frac{1}{6}.\]

Key Concepts

Parametric Equations

Parametric equations play a key role in describing the paths for path integrals. They allow us to define points on a curve using a parameter, usually denoted as \( t \), which represents the progress along the curve.
By using parametric equations, each coordinate (\( x \), \( y \), \( z \)) is expressed in terms of \( t \). This is particularly useful in vector calculus where you have multivariable functions.

• For each section of a path, a unique parametric equation defines points on that path.
• The path \( C_1 \) is defined by \( \mathbf{r}(t) = t\mathbf{k} \), indicating the movement happens only along the \( z \)-axis.
• The use of parametric equations helps break down complex 3D movements into manageable calculations, step by step.

Parametric transformation of variables simplifies the process of line integration by translating a three-variable function into a single-variable function. This is done by evaluating how \( x, y, \) and \( z \) change over the parameter \( t \) along the path.

Line Integrals

Line integrals are a type of integral where we integrate a function along a certain path or curve. In the context of vector calculus, it helps to compute the work done by a force field along a path, or the flux through a curve.
In our exercise, the line integral sums up the contributions of the function \( f(x, y, z) = x + \sqrt{y} - z^2 \) along each path \( C_1, C_2, \) and \( C_3 \).

• Each path \( C \) is integrated separately, reflecting specific constraints on coordinates \( x \), \( y \), and \( z \) given by their parametric forms.
• The integral along \( C_1 \) assesses the path where the function depends only on \( z \).
• The definite integral is computed from \( t = 0 \) to \( t = 1 \), considering the limits for \( t \) within each path integral.

This process ultimately involves combining the results from each path's integral to determine the overall path integral. Line integrals can measure much more than just lengths; they can capture the essence of a field along the curve.

Vector Calculus

Vector calculus is a branch of mathematics focused on differentiation and integration of vector fields. It is crucial for understanding the physics modeled through vectors, such as paths, or fields in space.
For the given problem, vector calculus helps in analyzing the vector paths \( C_1, C_2, \) and \( C_3 \). A critical aspect of vector calculus includes understanding vectors in terms of direction and magnitude.

• Each path can be represented as a position vector \( \mathbf{r}(t) \), with components denoted by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
• The vector differential \( d\mathbf{r} \) provides the direction of movement along the infinitesimal segment with respect to \( t \).
• Vector calculus allows the conversion of complex multidimensional integration into simpler parametric forms.

Using vector calculus methodologies, you can bridge the gap between pure mathematical integration and practical applications, such as calculating forces, work, and energy in physical systems.