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Magazine Chapter 14: Problem 23
Consider the following regions \(R\) and vector fields \(\mathbf{F}\). a. Compute the two-dimensional divergence of the vector field. b. Evaluate both integrals in Green's Theorem and check for consistency. c. State whether the vector field is source free. $$\mathbf{F}=\langle x, y\rangle ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}$$
Short Answer
Expert verified
Answer: No, the vector field is not source-free.
Step by step solution
01
Calculate the Divergence of the Vector Field
To find the two-dimensional divergence of the given vector field \(\mathbf{F} = \langle x, y\rangle\), we will use the formula:
$$\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y}$$
Therefore, the divergence is:
$$\nabla \cdot \mathbf{F} = 1 + 1 = 2$$
02
Set up the Green's Theorem Tensor and Region R
Green's Theorem states:
$$\oint_{C} (P dx + Q dy) = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$$
For our given vector field \(\mathbf{F} = \langle x, y\rangle\), we have \(P = x\) and \(Q = y\). Thus, the tensor for Green's theorem can be given as:
$$\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y} = 0 - 0 = 0$$
The region \(R\) is defined as:
$$R = \{(x, y): x^2 + y^2 \leq 4\}$$which is a circle with radius 2 centered at the origin.
03
Evaluate the Line Integral
Now, let's evaluate the line integral part of Green's theorem. To do this, we will parameterize the boundary \(C\) of the region \(R\). The parameterization of the circle can be given as:
$$x(t) = 2\cos t$$
$$y(t) = 2\sin t$$with \(0 \leq t \leq 2\pi\).
Now, we will find the derivatives of the parameterization with respect to \(t\):
$$x'(t) = -2\sin t$$
$$y'(t) = 2\cos t$$
Now, substitute the parameterization and their derivatives into the line integral:
$$\oint_{C} (P dx + Q dy) = \int_{0}^{2\pi} (x(-2\sin t) + y(2\cos t)) dt$$
Substitute the parameterization to get:
$$\oint_{C} (P dx + Q dy) = \int_{0}^{2\pi} (2\cos t(-2\sin t) + 2\sin t(2\cos t)) dt$$
04
Calculate the Line Integral
Now we need to compute the integral:
$$\oint_{C} (P dx + Q dy) = \int_{0}^{2\pi} (4\sin t\cos t) dt = 2\int_{0}^{2\pi} (2\sin t\cos t) dt$$
Apply the double-angle formula for sine, \(\sin 2t = 2\sin t \cos t\):
$$\oint_{C} (P dx + Q dy) = 2\int_{0}^{2\pi} \sin 2t dt = 4\left[-\frac{1}{2}\cos 2t\right]_{0}^{2\pi} = 0$$
05
Evaluate the Double Integral
Now, we evaluate the double part of Green's Theorem:
$$\iint_{D} (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) dA = \iint_{D} 0\ dA = 0$$
06
Check for Consistency
Since both the line integral and double integral are equal to zero, we can conclude that the results are consistent.
07
Determine if the Vector Field is Source-Free
A vector field is source-free if its divergence is zero. In this case, the divergence of the vector field \(\mathbf{F} = \langle x, y\rangle\) is 2, which is not equal to zero. Therefore, the vector field is not source-free.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergence of a Vector Field
The divergence of a vector field is a scalar function that describes the net rate of outflow of a fluid from a given point. It is a measure of how much the vector field 'spreads out'. Mathematically, for a vector field \( \mathbf{F} = \langle P, Q \rangle \), the divergence is given by:
\[ abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \]
In our exercise, with the vector field \( \mathbf{F} = \langle x, y \rangle \), the divergence is \( 2 \) which suggests a uniform outflow at every point in the field. To visualize the concept, imagine a balloon inflating, where every point on the surface is moving away from the center, producing divergence.
\[ abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \]
In our exercise, with the vector field \( \mathbf{F} = \langle x, y \rangle \), the divergence is \( 2 \) which suggests a uniform outflow at every point in the field. To visualize the concept, imagine a balloon inflating, where every point on the surface is moving away from the center, producing divergence.
Line Integral
The line integral is a type of integral where a function is integrated along a curve. For vector fields, it is used to measure the work done by a force field along a path. In the context of Green's Theorem, which relates a line integral around a simple, closed curve \( C \) to a double integral over the plane region \( D \), it's given by:
\[ \oint_{C} (P dx + Q dy) \]
Parameterizing the boundary of the region simplifies the evaluation of this integral. It's crucial to consider the orientation of the path; it should be counter-clockwise for Green's Theorem to apply. In our problem, the line integral of the vector field over the circle with radius 2 is computed and found to be zero, indicating no 'net work' done along the path.
\[ \oint_{C} (P dx + Q dy) \]
Parameterizing the boundary of the region simplifies the evaluation of this integral. It's crucial to consider the orientation of the path; it should be counter-clockwise for Green's Theorem to apply. In our problem, the line integral of the vector field over the circle with radius 2 is computed and found to be zero, indicating no 'net work' done along the path.
Double Integral
A double integral allows for the calculation of the volume under a surface over a planar region. It is a powerful tool used in multiple areas such as physics, engineering, and economics. Within Green's Theorem, the double integral is related to the rotation of the vector field and is represented by:
\[ \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA \]
This component of Green's Theorem often represents things like fluid circulation or the 'curl' of the field. In the given exercise, when applying Green's Theorem, the double integral over the region \( R \) was zero, which was expected due to the specific form of the vector field and the nature of the path.
\[ \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA \]
This component of Green's Theorem often represents things like fluid circulation or the 'curl' of the field. In the given exercise, when applying Green's Theorem, the double integral over the region \( R \) was zero, which was expected due to the specific form of the vector field and the nature of the path.
Source-Free Vector Field
A vector field is considered source-free if its divergence everywhere in the field is zero. This means that there is no creation or destruction of flux within the field, similar to an incompressible fluid. If the divergence is not equal to zero, as in our case where it's equal to \( 2 \), then the vector field is not source-free. This concept is fundamental in many areas of physics and engineering, such as electromagnetism and fluid dynamics, as it can be related to conservation laws. In our exercise, the non-zero divergence indicates that the vector field \( \mathbf{F} = \langle x, y \rangle \) has sources or sinks at points within the region.
