Question

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. \(\mathbf{F}=\langle y,-x\rangle ; R\) is the square with vertices (0,0),(1,0),(1,1), and (0,1).

Short answer

No, the given vector field is not source free. This is because the computed two-dimensional divergence of the vector field is 1, which is not zero.

Step-by-step solution

Compute the two-dimensional divergence of the vector field.

The given vector field is \(\mathbf{F} = \langle y,-x \rangle\). To calculate the two-dimensional divergence, we will compute the partial derivative of the first component with respect to \(x\) and the second component with respect to \(y\) and then sum the results: \(\text{div} \ \mathbf{F} = \frac{\partial}{\partial x} y - \frac{\partial}{\partial y} x = 0 - (-1) = 1\)

Evaluate both integrals in Green's Theorem.

Green's Theorem states that for a positively oriented, simple, closed curve \(C\) and a continuously differentiable vector field \(\mathbf{F}(x,y) = \langle M, N \rangle\) on region \(R\) enclosed by \(C\), we have: \(\oint_C (\mathbf{F} \cdot \mathbf{ds}) = \iint_R (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \ dA\) In this problem, \(M = y\) and \(N = -x\). Thus, we have: \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} y = 1\) and \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (-x) = -1\) Now substituting these values into Green's theorem: \( \oint_C (\mathbf{F} \cdot \mathbf{ds}) = \iint_R (-1-1) \ dA = -\int_0^1 \int_0^1 2 \ dy \ dx\) To evaluate this double integral, we first integrate with respect to \(y\) and then with respect to \(x\): \(-\int_0^1 \int_0^1 2 \ dy \ dx = -2\int_0^1 \left[y\right]_0^1 \ dx = -2\int_0^1 (1-0) \ dx = -2\int_0^1 1 \ dx\) Now, integrate with respect to \(x\): \(-2\int_0^1 1 \ dx = -2\left[x\right]_0^1 = -2(1-0) = -2\) Now let's evaluate the circulation integral (line integral): \( \oint_C (\mathbf{F} \cdot \mathbf{ds}) = \oint_C (y \ dx - x \ dy)\) We break the integral into four parts, one for each side of the square: \( \oint_C (y \ dx - x \ dy) = \int_{C_1} (y \ dx - x \ dy) + \int_{C_2} (y \ dx - x \ dy) + \int_{C_3} (y \ dx - x \ dy) + \int_{C_4} (y \ dx - x \ dy)\) After evaluating the line integrals for each side of the square using the respective parametric equations, we find that their sum is indeed -2, which is consistent with the value of the flux integral computed earlier.

Determine if the vector field is source free.

A vector field is source free if its divergence is zero everywhere. In this case, we computed the divergence of the vector field to be 1: \(\text{div} \ \mathbf{F} = 1\) Since the divergence is not zero, the vector field \(\mathbf{F}\) is not source free.

Key concepts

Two-Dimensional Divergence

The concept of two-dimensional divergence is pivotal in understanding the behavior of vector fields in the plane. Imagine we're examining the air flow in a room - divergence tells us if there's a point where air seems to be 'oozing out' (a source) or 'sucked in' (a sink). Mathematically, for a vector field \(\mathbf{F} = \langle P, Q \rangle\), the two-dimensional divergence is defined as \(\text{div} \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\).

In our exercise, with \(\mathbf{F} = \langle y, -x \rangle\), this amounts to \(\text{div} \mathbf{F} = 0 - (-1) = 1\). A nonzero divergence implies that the vector field has sources or sinks, thus affecting the fluid flow within the region, analogous to having a fan or drain in our room example.

Line Integrals

Line integrals are like a stroll down a path where we add up contributions along the way - instead of counting steps, we're summing up the vector field's influence. In formal terms, a line integral of a vector field \(\mathbf{F}\) along a curve \(C\) is denoted \(\oint_C (\mathbf{F} \cdot \mathbf{ds})\). Here, \(\mathbf{ds}\) represents a tiny segment of the path, and we're essentially looking at how much the field 'pushes' along each segment.

When we computed the line integrals for our square's sides, this 'stroll' around the edges summed up to -2, indicating the total effect of the field around the region. The consistency of this value with our area-based calculation via Green's Theorem is a magical check on the theorem's reliability in connecting local field behavior with global accumulation.

Vector Fields

Let's think of a vector field as a way to assign a vector to every point in space. For example, imagine that each point on a map has an arrow showing wind direction and strength; this is a vector field over two dimensions. Mathematically, a two-dimensional vector field is a function that assigns each point \((x, y)\) a two-dimensional vector \(\mathbf{F}(x, y)\).

Vector fields are not limited to representing physical quantities like wind or water currents; they can represent any two-dimensional flow, including abstract concepts in mathematics or economics. Our focus is on how these fields interact within a certain region, such as \(R\), and how their characteristics lead to various integral expressions and theorems.

Source Free Vector Field

The notion of a source free vector field might conjure up images of a serene pond, where no water is bubbling up from a spring, nor is it draining away. Mathematically, a vector field is source free if its divergence is zero at all points. Returning to our two-dimensional example \(\mathbf{F} = \langle y, -x \rangle\), we found that \(\text{div} \mathbf{F} = 1\).

Because we're not getting a divergence of zero, this particular field isn't source free, much like our hypothetical pond with an incoming stream or a whirlpool at its center. In physical terms, a source free field implies conservation, such as constant mass for a fluid, since there are no 'leaks' or 'additions' within the flow domain.