Chapter 5: Problem 30
Compute the circulation and net flux for the given flow and the indicated closed contour \(C\). \(\mathbf{F}(x, y)=(x+2 y) \mathbf{i}+(x-y) \mathbf{j}\), where \(C\) is the square with vertices \(z=0\) \(z=1+i, z=2 i, z=-1+i\)
Short Answer
Expert verified
The circulation is 7.5 and the net flux is -4.
Step by step solution
01
Parameterize the Square Contour
The contour \( C \) is a square with vertices at \((0, 0)\), \((1, 1)\), \((0, 2)\), and \((-1, 1)\). We'll break it into four segments for parameterization:1. From \((0, 0)\) to \((1, 1)\): \( \mathbf{r}_1(t) = (t, t) \), \(0 \leq t \leq 1\).2. From \((1, 1)\) to \((0, 2)\): \( \mathbf{r}_2(t) = (1-t, 1+t) \), \(0 \leq t \leq 1\).3. From \((0, 2)\) to \((-1, 1)\): \( \mathbf{r}_3(t) = (-t, 2-t) \), \(0 \leq t \leq 1\).4. From \((-1, 1)\) to \((0, 0)\): \( \mathbf{r}_4(t) = (-1+t, 1-t) \), \(0 \leq t \leq 1\).This parameterization allows us to evaluate line integrals over each segment.
02
Compute the Circulation using Line Integrals
To find the circulation \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), compute the line integral of \( \mathbf{F} \) along each segment. For each segment, parameterize \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \) and compute \( \mathbf{F} \cdot d\mathbf{r} \):1. Segment 1: \( \mathbf{F}(t,t) = (t+2t)\mathbf{i} + (t- t)\mathbf{j} = 3t \mathbf{i} \). \(d\mathbf{r} = (1,1) dt \). \( \int_0^1 3t \cdot (1 + 1) dt = 6 \int_0^1 t dt = 3 \).2. Segment 2: Similarly compute this line integral: \( \int_0^1 3(1 - t) dt = 3 \times \left[1 - \frac{1}{2}\right] = 3 \times 0.5 = 1.5 \).3. Segment 3: \( \mathbf{F}(-t, 2-t) = -3t \mathbf{i} \). Compute \( \int_0^1 -3t \cdot (-1 - 1) dt = 6 \int_0^1 t dt = 3 \).4. Segment 4: \( \mathbf{F}(-1+t,1-t) = (-1 + t) + 2(1 - t) = 1-3t \mathbf{i} \). Compute \( \int_0^1 (1-3t) \cdot (1 -1) dt = 0 \).Add the contributions: \( 3 + 1.5 + 3 + 0 = 7.5 \).
03
Compute the Net Flux using Green's Theorem
The net flux is given by \( \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). For \( M = x+2y \) and \( N = x-y \), we find:\[ \frac{\partial N}{\partial x} = 1, \quad \frac{\partial M}{\partial y} = 2 \]\[ \iint_S \left(1 - 2\right) \, dA = -\iint_S 1 \, dA \]The area of the square \((R)\) is \(2 \times 2 = 4\), so the flux is \[ \iint_R (-1) \, dA = -4 \].
04
Conclusion of Calculations
The calculated circulation around the contour \( C \) is \(7.5\) and the net flux is \(-4\). These values represent the flow of \( \mathbf{F} \) around and through the square, respectively.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals
Line integrals are a fundamental concept in vector calculus, allowing us to integrate a vector field along a curve. They are used to measure the total effect, such as work done, of a vector field along a defined path. In simpler terms, they help us understand how a vector field behaves as it travels around a path.
To compute a line integral, we need to parameterize the path or curve along which we're integrating. In the context of our problem, the contour of a square is parameterized into four line segments, which makes it easier to handle and evaluate the line integral piece by piece.
With the function given as \( \mathbf{F}(x, y)=(x+2 y) \mathbf{i}+(x-y) \mathbf{j} \), the task is to integrate this vector field over the contour. For each segment of the path, we calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} \) and then integrate with respect to the parameter.
To compute a line integral, we need to parameterize the path or curve along which we're integrating. In the context of our problem, the contour of a square is parameterized into four line segments, which makes it easier to handle and evaluate the line integral piece by piece.
With the function given as \( \mathbf{F}(x, y)=(x+2 y) \mathbf{i}+(x-y) \mathbf{j} \), the task is to integrate this vector field over the contour. For each segment of the path, we calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} \) and then integrate with respect to the parameter.
- Parameterization simplifies dealing with complex paths by breaking them into manageable segments.
- Line integrals help determine the total work done by or against the field around a path.
- For closed paths like our square contour, line integrals can also indicate circulation—the total spin or rotation effect around the path.
Green's Theorem
Green's Theorem forms a bridge between a line integral around a simple closed curve and a double integral over the plane region bounded by the curve. It is one of the essential theorems in vector calculus and relates these integrals in a plane.
The theorem states that the line integral around a simple closed curve \( C \) of the vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \) is equivalent to a double integral over the region \( R \) that \( C \) encloses. Mathematically, it can be expressed as:
\[ \oint_C (M\, dx + N\, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
In our problem, we're given \( M = x+2y \) and \( N = x-y \). By applying Green's Theorem, we transform the task of calculating circulation from a line integral around \( C \) to a region integral over the square.
The theorem states that the line integral around a simple closed curve \( C \) of the vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \) is equivalent to a double integral over the region \( R \) that \( C \) encloses. Mathematically, it can be expressed as:
\[ \oint_C (M\, dx + N\, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
In our problem, we're given \( M = x+2y \) and \( N = x-y \). By applying Green's Theorem, we transform the task of calculating circulation from a line integral around \( C \) to a region integral over the square.
- Green's Theorem simplifies complex circulation calculations by converting them into area integrals.
- The theorem helps in visualizing how flow behaves over a region rather than just around the boundary.
- This approach effectively computes the net circulation and flux through the region.
Flux
Flux is a concept in vector calculus that measures how much of a vector field passes through a surface. The net flux through a closed region helps quantify the 'flow' exiting or entering the enclosed area.
For a vector field \( \mathbf{F} \), the net flux through a closed region \( R \) with boundary \( C \) is computed using the surface integrated form of divergence:
\[ \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
In our exercise, the vector field is \( \mathbf{F}(x, y)=(x+2 y) \mathbf{i}+(x-y) \mathbf{j} \), and the net flux through the square region was calculated using the divergence form \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \). The result, \(-4\), indicates a negative flux, meaning more of the vector field is entering the region than leaving which aligns with the difference in these partial derivatives.
For a vector field \( \mathbf{F} \), the net flux through a closed region \( R \) with boundary \( C \) is computed using the surface integrated form of divergence:
\[ \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
In our exercise, the vector field is \( \mathbf{F}(x, y)=(x+2 y) \mathbf{i}+(x-y) \mathbf{j} \), and the net flux through the square region was calculated using the divergence form \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \). The result, \(-4\), indicates a negative flux, meaning more of the vector field is entering the region than leaving which aligns with the difference in these partial derivatives.
- Flux quantifies the rate at which 'stuff' passes through a boundary.
- Negative flux suggests internal sinks, where flow enters more than exits.
- By evaluating divergence, flux can reveal the behavior of a field over a specific region.