Problem 338 - Calculus Volume 3

Question

[T] Use a CAS and Stokes' theorem to approximate line integral \(\int_{C}[(1+y) z d x+(1+z) x d y+(1+x) y d z]\), where \(C\) is a triangle with vertices \((1,0,0), \quad(0,1,0)\), and (0,0,1) oriented counterclockwise.

Step-by-Step Solution

Answer

The integral approximates to \(\frac{1}{6}\).

Step 1: Identify Given Information

We need to evaluate the line integral \(\int_{C}[(1+y)z dx + (1+z)x dy + (1+x)y dz]\) along the curve \(C\), which is a triangle with vertices \((1,0,0), (0,1,0), (0,0,1)\) oriented counterclockwise.

Step 2: State Stokes' Theorem

Stokes' Theorem relates a line integral around a closed curve \(C\) to a surface integral over a surface \(S\) bounded by \(C\). It states: \(\int_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S}\). Our task is to find \(abla \times \mathbf{F}\) and compute the surface integral instead of the line integral.

Step 3: Calculate the Curl of the Vector Field

Given \(\mathbf{F} = [(1+y)z, (1+z)x, (1+x)y]\), calculate \(abla \times \mathbf{F}\).The curl \(abla \times \mathbf{F}\) is:\[abla \times \textbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ (1+y)z & (1+z)x & (1+x)y \end{vmatrix}\]\[= \mathbf{i}((1+x) - x) - \mathbf{j}((1+x) - (1+y)) + \mathbf{k}((1+z) - z)\]\[= \mathbf{i}y - \mathbf{j} + \mathbf{k}\]

Step 4: Define the Surface

Choose the surface \(S\) as the face of the triangle (a plane), satisfying the equation \(x + y + z = 1\) with parameterization: \(x = u, y = v, z = 1 - u - v\). Here \(u, v \geq 0\) and \(u + v \leq 1\).

Step 5: Compute the Surface Integral

The differential area element \(d\mathbf{S}\) is upward normal. For the plane \(x + y + z = 1\), the normal vector is the gradient \(abla (x + y + z) = \langle 1, 1, 1 \rangle\), so \(d\mathbf{S} = \langle 1, 1, 1 \rangle dudv\).Find \(\iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S}\):\[\iint_{S} (\mathbf{i}y - \mathbf{j} + \mathbf{k}) \cdot \langle 1, 1, 1 \rangle dudv = \iint_{S} (y - 1 + 1) dudv = \iint_{S} y dudv\]Parametrize: \(y = v\), thus:\[= \int_{0}^{1} \int_{0}^{1-u} v \, dv \, du = \int_{0}^{1} \left[ \frac{v^2}{2} \right]_{0}^{1-u} du = \int_{0}^{1} \frac{(1-u)^2}{2} \, du\]\[= \frac{1}{2} \left[ \frac{1}{3} - \frac{u^3}{3} \right]_{0}^{1} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\]

Step 6: Conclusion of the Integral

By using Stokes' theorem and evaluating the surface integral, the approximation for the given line integral is computed as \(\frac{1}{6}\).

Key Concepts

line integral

Line integrals play a significant role when dealing with vector fields, especially when the path to be integrated over is a curve or loop. In the case of a line integral, you are essentially summing up the influence of a vector field along a path. Imagine it like tracing your finger around the perimeter of a shape while considering how a wind might push you in various directions along that path.
A line integral represented by \( \int_{C} \textbf{F} \cdot d\textbf{r} \) computes the integral of a vector field \( \textbf{F} \) over a curve \( C \). Here, \( d\textbf{r} \) represents a differential element of the path, and the "dot" indicates that you’re taking the dot product between the vector field and this differential path element.

• The path \( C \) can be an actual geometric loop or can be represented more abstractly.
• The integral accounts for both the magnitude and direction of the vector field relative to the path.

In the problem stated, the integral \( \int_{C}[(1+y)z dx + (1+z)x dy + (1+x)y dz] \) is a line integral where \( C \) is a triangular path with known vertices and orientation.

curl of a vector field

The curl of a vector field, denoted by \( abla \times \textbf{F} \), provides insight into the rotational behavior of the field at a given point. Think of the curl as a measure of how much the field "twists" around a given point.
In mathematical terms, the curl is calculated using a determinant involving the unit vectors \( \textbf{i} \), \( \textbf{j} \), \( \textbf{k} \), and the vector field \( \textbf{F} = [(1+y)z, (1+z)x, (1+x)y] \). The formula for the curl is:

• \( abla \times \textbf{F} = \left|\begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ (1+y)z & (1+z)x & (1+x)y \end{array}\right| \)

Solving the determinant for this specific problem, you get the vector \( \textbf{i}y - \textbf{j} + \textbf{k} \), representing the curl of the field. This result shows how the vector field circulates at each point of the plane surface.

surface integral

Surface integrals extend the concept of line integrals to two dimensions, where instead of summing over a path, you sum over a surface. When using Stokes' Theorem, you're replacing a hard-to-calculate line integral with a potentially simpler surface integral.
For a vector field \( \textbf{F} \), the surface integral \( \iint_{S} (abla \times \textbf{F}) \cdot d\textbf{S} \) represents the accumulation of the curl over the surface \( S \) bounded by \( C \).

• The choice of \( S \) is crucial as it must be bounded by the curve \( C \).
• The differential area element \( d\textbf{S} \) holds the normal vector to the surface, indicating the direction of integration.

In the given problem, the chosen surface is the triangle defined by the vertices, which allows for the calculation of the surface integral replacing the original line integral using Stokes' theorem.

triangle parameterization

Triangle parameterization simplifies the process of integrating over a triangular surface. In situations involving Stokes' Theorem, effective parameterization of the triangle is vital for accurately calculating the surface integral.
This problem involves the plane \( x + y + z = 1 \), and the triangle is parameterized using variables \( u \) and \( v \). The parameterization is:\

• \( x = u \)
• \( y = v \)
• \( z = 1 - u - v \)

These parameters have constraints \( u, v \geq 0 \) and \( u + v \leq 1 \). This effectively transforms the problem from a three-variable system \( (x, y, z) \) into a more manageable two-variable system \( (u, v) \).
By using these parameters, the integration over the triangle becomes straightforward, leading to the computation of the surface integral using the defined boundaries and the parameterized equations.