Problem 5 - Thomas Calculus

Question

In Exercises \(5-14,\) use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(F\) and curve \(C .\) $$ \begin{array}{l}{\mathbf{F}=(x-y \mathbf{i}+(y-x) \mathbf{j}} \\ {\text { C: The square bounded by } x=0, x=1, y=0, y=1}\end{array} $$

Step-by-Step Solution

Answer

Circulation: 0; Flux: 2.

Step 1: Understand Green's Theorem

Green's Theorem relates the line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). It states \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \, dA \), where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \).

Step 2: Identify the Components of \( \mathbf{F} \)

For the vector field \( \mathbf{F} = (x-y) \mathbf{i} + (y-x) \mathbf{j} \), identify \( M = x-y \) and \( N = y-x \).

Step 3: Calculate the Partial Derivatives

Calculate \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \).\[\frac{\partial N}{\partial x} = \frac{\partial (y-x)}{\partial x} = -1\]\[\frac{\partial M}{\partial y} = \frac{\partial (x-y)}{\partial y} = -1\]

Step 4: Set Up and Evaluate the Double Integral for Circulation

Substitute the derivatives into Green's Theorem:\[\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \, dA = \iint_D (-1 + 1) \, dA = \iint_D 0 \, dA = 0\]Since the integrand is 0, the circulation is 0.

Step 5: Set Up and Evaluate the Double Integral for Flux

Green's Theorem for flux is expressed as \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_D (abla \cdot \mathbf{F}) \, dA \), where \( abla \cdot \mathbf{F} = \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \).Calculate \( abla \cdot \mathbf{F} \):\[\frac{\partial M}{\partial x} = 1, \quad \frac{\partial N}{\partial y} = 1\]\[abla \cdot \mathbf{F} = 1 + 1 = 2\]Evaluate the double integral over the region \( D \):\[\iint_D 2 \, dA = 2 \times \text{Area}(D) = 2 \times 1 = 2\]The outward flux is 2.

Key Concepts

Vector Fields

A vector field is a way to assign a vector to every point in a certain space or region. Imagine it as assigning a little arrow, which represents the force or velocity at each point in a fluid or force field. In this exercise, our vector field is given as \( \mathbf{F} = (x-y)\mathbf{i} + (y-x)\mathbf{j} \).

Here's what you need to know about vector fields:

• **Components**: The vector field \( \mathbf{F} \) is made up of two functions, \( M \) and \( N \), where \( M = x-y \) and \( N = y-x \). These are referred to as the components of the vector field, aligning with the \( x \)- and \( y \)-axes, respectively.
• **Dimensions**: A vector field can extend over surfaces (2D) or through volumes (3D). In this exercise, our field is over a square region in the plane (2D).
• **Use**: Vector fields describe phenomena like wind flow, magnetic fields, or gravitational forces. Understanding how they behave helps in fields like physics and engineering.

Under the hood, vector fields like \( \mathbf{F} \) are often utilized to calculate other important values, like the line integral or double integral.

Line Integral

A line integral is a way to add up values along a path or curve in a vector field. In this context, it measures quantities like work done by a force or fluid flow along the path.

To understand the line integral, consider these steps:

• **Path Dependency**: The line integral of a vector field along a curve \( C \) sums up the dot products of the field with small line segments along \( C \). This curve is typically closed, forming a loop around region \( D \).
• **Green's Theorem Connection**: By Green's Theorem, the line integral related to \( \mathbf{F} \) can be converted into a double integral over the region \( D \) bounded by \( C \). In particular, it correlates with the circulation around the curve, which was found to be zero in this problem.

Evaluating a line integral directly often requires calculating significant transformations. However, using the theorem simplifies the process by relating it to an easier-to-calculate double integral.

Double Integral

A double integral extends the idea of integration to two dimensions, allowing calculation over an area rather than just a line or curve. In this problem, double integrals help us find both the circulation and flux of a vector field across region \( D \).

Here's how double integrals come into play:

• **Area Calculation**: The double integral runs over the region \( D \), integrating the function of interest with respect to both \( x \) and \( y \). It's like summing quantities over a plane region, providing the total value across it.
• **Green's Theorem Use**: By applying Green's Theorem, the problem converts calculating the more complex line integral into a simpler double integral problem over a square \( D \).
• **Flux Calculation**: The flux, representing the amount of field flowing out of or through a surface, is obtained using a particular double integral of the divergence \( abla \cdot \mathbf{F} \), which sums up the local field expansion or contraction.

Using a double integral makes it feasible to handle computations for complex areas, especially with well-defined geometric regions like squares.