Chapter 15: Problem 5
Use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(\mathbf{F}\) and curve \(C\). \(\mathbf{F}=(x-y) \mathbf{i}+(y-x) \mathbf{j}\) \(C:\) The square bounded by \(x=0, x=1, y=0, y=1\)
Short Answer
Expert verified
Circulation: 0; Outward flux: 2.
Step by step solution
01
Understand Green's Theorem
Green's Theorem relates a line integral around a closed curve \(C\) to a double integral over the plane region \(D\) it encloses. It states that \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \).
02
Identify M and N in \( \mathbf{F} \)
Given \( \mathbf{F} = (x-y) \mathbf{i} + (y-x) \mathbf{j} \), we identify \( M = x-y \) and \( N = y-x \).
03
Compute Partial Derivative \( \frac{\partial N}{\partial x} \)
The partial derivative of \( N = y-x \) with respect to \( x \) is \( \frac{\partial N}{\partial x} = -1 \).
04
Compute Partial Derivative \( \frac{\partial M}{\partial y} \)
The partial derivative of \( M = x-y \) with respect to \( y \) is \( \frac{\partial M}{\partial y} = -1 \).
05
Set Up Double Integral
The double integral becomes \( \iint_D \left( -1 - (-1) \right) \; dA = \iint_D 0 \; dA \).
06
Evaluate the Double Integral
Since the integrand is zero, the double integral evaluates to zero: \( \iint_D 0 \; dA = 0 \).
07
Outward Flux Calculation
Using Green's Theorem for flux, we need to calculate \( \iint_D \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \). Compute \( \frac{\partial M}{\partial x} = 1 \) and \( \frac{\partial N}{\partial y} = 1 \). Thus, the double integral becomes \( \iint_D 2 \, dA \).
08
Integrate for Outward Flux
For the square region from \( x=0 \) to \( x=1 \) and \( y=0 \) to \( y=1 \), the area \( dA = dx \, dy \). The integral becomes \( \int_0^1 \int_0^1 2 \, dy \, dx = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals
A line integral is a concept from calculus where you integrate a function around a curve. Imagine it like taking a walk around a path, but in mathematical terms, you are measuring something along that path.
In the context of Green's Theorem, if you have a vector field \( \mathbf{F} \), a line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) sums up the effect of \( \mathbf{F} \) along the curve \( C \). Here, \( d\mathbf{r} \) represents a tiny segment of the path that you're measuring along.
In the context of Green's Theorem, if you have a vector field \( \mathbf{F} \), a line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) sums up the effect of \( \mathbf{F} \) along the curve \( C \). Here, \( d\mathbf{r} \) represents a tiny segment of the path that you're measuring along.
- The path in question is a closed curve, meaning it loops back on itself without any breaks.
- Green's Theorem helps convert this line integral into a simpler form, by converting it into a double integral over the area encircled by the path, which often makes calculations easier.
Vector Fields
Vector fields are a crucial part of understanding both line integrals and Green's Theorem. A vector field can be visualized as a collection of arrows, where each arrow represents a vector at a point in space.
In this exercise, the vector field is \( \mathbf{F} = (x-y) \mathbf{i} + (y-x) \mathbf{j} \). This means:
By examining how these vectors interact with a closed curve, Green's Theorem provides a way to relate this local behavior of vectors to a global summary in terms of circulation or flux.
In this exercise, the vector field is \( \mathbf{F} = (x-y) \mathbf{i} + (y-x) \mathbf{j} \). This means:
- At any point \((x, y)\), the vector is determined by these expressions: the \( x \)-component is \( x-y \), and the \( y \)-component is \( y-x \).
- The direction and magnitude of these vectors change as you move across different points in the plane.
By examining how these vectors interact with a closed curve, Green's Theorem provides a way to relate this local behavior of vectors to a global summary in terms of circulation or flux.
Partial Derivatives
Partial derivatives play a pivotal role in Green's Theorem, as they allow us to explore how the components of a vector field change.
For a vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \), Green's Theorem involves the expressions \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \).
In our example:
For a vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \), Green's Theorem involves the expressions \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \).
- \( \frac{\partial N}{\partial x} \) measures how the \( y \)-component \( N \) changes in the \( x \) direction.
- \( \frac{\partial M}{\partial y} \) shows how the \( x \)-component \( M \) changes in the \( y \) direction.
In our example:
- \( \frac{\partial N}{\partial x} = -1 \) and \( \frac{\partial M}{\partial y} = -1 \) for circulation.
- \( \frac{\partial M}{\partial x} = 1 \) and \( \frac{\partial N}{\partial y} = 1 \) for flux.