Problem 10 - Calculus Early Transcendentals

Question

Use Stokes' Theorem to evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k} ; C \text { is the triangle in the plane }} \\ {x+y+z=1 \text { with vertices }(1,0,0),(0,1,0), \text { and }(0,0,1)} \\ {\text { with a counterclockwise orientation looking from the first }} \\ {\text { octant toward the origin. }}\end{array} $$

Step-by-Step Solution

Answer

The integral \( \oint_C \mathbf{F} \cdot d \mathbf{r} = -\frac{1}{2} \).

Step 1: Convert the line integral to a surface integral

Stokes' Theorem states that \( \oint_C \mathbf{F} \cdot d \mathbf{r} = \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} \), where \( S \) is any surface bounded by the curve \( C \). We first need to find \( abla \times \mathbf{F} \).

Step 2: Compute the curl of \( \mathbf{F} \)

For the vector field \( \mathbf{F}(x, y, z) = xy \mathbf{i} + yz \mathbf{j} + zx \mathbf{k} \), the curl \( abla \times \mathbf{F} \) is calculated using the determinant formula:\[abla \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \xy & yz & zx\end{vmatrix}= (z-y) \mathbf{i} - (z-x) \mathbf{j} + (y-x) \mathbf{k}.\]

Step 3: Set up the surface integral

The triangle defined by \((1,0,0), (0,1,0), (0,0,1)\) lies in the plane \(x + y + z = 1\). We parameterize this surface with \( z = 1 - x - y \). The normal vector area element is given by using the gradient of the plane equation: \( d\mathbf{S} = abla g \, dx \, dy = \langle 1, 1, 1 \rangle \, dx \, dy \).

Step 4: Evaluate the surface integral

Using the parameterization, the curl becomes \( (1-x-y - (1-y)) \mathbf{i} - (1-(x+(1-x-y))) \mathbf{j} + (y-x) \mathbf{k} \), which simplifies to \( -y \mathbf{i} + (x+y-1) \mathbf{j} + (y-x) \mathbf{k} \). Dotting this with the normal vector \( \langle 1, 1, 1 \rangle \), we have:\[-y + (x+y-1) + (y-x) = -1.\]Integrating over the region bounded by the triangle:\[\int_0^1 \int_{0}^{1-x} (-1) \, dy \, dx = -\int_0^1 (1-x) \, dx = -\left[ x - \frac{x^2}{2} \right]_0^1 = -\left( 1 - \frac{1}{2} \right) = -\frac{1}{2}.\]

Step 5: Conclusion from Stokes' Theorem

Thus, by Stokes' Theorem, the original line integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) evaluates to the surface integral \( -\frac{1}{2} \).

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics focused on vector fields, which are essentially functions that assign a vector to every point in a space. These fields are typically represented in two or three dimensions and are used to model physical quantities such as force, velocity, or acceleration that have both magnitude and direction. In vector calculus, we utilize operations like divergence, gradient, and curl to study these vector fields.

Some key operations in vector calculus include:

• Gradient: Determines the rate and direction of change in scalar fields, resulting in a vector field.
• Divergence: Measures the magnitude of a source or sink at a given point, giving a scalar quantity.
• Curl: Represents the rotation of a vector field, providing pivotal information about fluid flow or electromagnetic fields.

Understanding these operations allows us to analyze complex systems succinctly, and their mathematical computations often lay the groundwork for applying theorems like Stokes’ Theorem in physics and engineering disciplines.

Line Integrals

Line integrals are a fundamental concept within vector calculus, useful for calculating quantities along curves. Unlike regular integrals that evaluate functions over intervals, line integrals work over a path in a field, making them invaluable for analyzing physical phenomena like electromagnetism or fluid dynamics where force or flow along paths is crucial.

For vector fields and curves, a line integral can determine 'work done' by a force field on a particle moving along a curve. Mathematically, this is expressed as: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} \]where \( \mathbf{F} \) is the vector field, and \( C \) is the curve. In our example, this translates the questions of movement through a field into solvable math problems.

Line integrals can be connected to surface integrals through the application of theorems like Stokes’ Theorem by converting a line integral over a closed curve into a surface integral over the surface bounded by the curve, which simplifies computations significantly.

Surface Integrals

Surface integrals extend the concept of integrals to functions over surfaces. They play a key role in multivariable calculus, especially in calculating flux—the quantity of a field passing through a surface—and in understanding several physical phenomena like electromagnetic fields and gravitational fields.

The surface integral of a vector field \( \mathbf{F} \) over a surface \( S \) is calculated using:\[ \iint_S \mathbf{F} \cdot d\mathbf{S} \] where \( d\mathbf{S} \) represents a differential vector area of the surface. In our exercise, using Stokes' Theorem, we convert a complicated line integral into a more manageable surface integral, evaluating the flux through the given triangular plane.

Surface integrals help translate complex physical questions of how a field interacts through a surface into comprehensible mathematical logic. This powerful tool helps engineers and physicists model real-world systems efficiently and accurately.