Question

Consider the vector field \(\mathbf{F}=\langle y, x\rangle\) shown in the figure. a. Compute the outward flux across the quarter circle \(C: \mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle,\) for \(0 \leq t \leq \pi / 2\) b. Compute the outward flux across the quarter circle \(C: \mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle,\) for \(\pi / 2 \leq t \leq \pi\) c. Explain why the flux across the quarter circle in the third quadrant equals the flux computed in part (a). d. Explain why the flux across the quarter circle in the fourth quadrant equals the flux computed in part (b). e. What is the outward flux across the full circle?

Short answer

Question: Compute the total outward flux across the full circle for the vector field \(\mathbf{F} = \langle y, x \rangle\). Answer: The total outward flux across the full circle is 0.

Step-by-step solution

a. Outward flux across the quarter circle for \(0 \leq t \leq \pi/2\)

The parameterized form of the path is given by \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle\) for \(0 \leq t \leq \pi/2\). We must find its derivative, \(\frac{d\mathbf{r}}{dt}\) and compute the dot product with the vector field \(\mathbf{F}\), and then integrate the result with respect to \(t\). First, let's compute \(\frac{d\mathbf{r}}{dt}\): \(\frac{d\mathbf{r}}{dt} = \langle -2 \sin t, 2 \cos t\rangle\) Now, let's evaluate the dot product with the vector field \(\mathbf{F}\): \(\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \langle y, x\rangle \cdot \langle -2 \sin t, 2 \cos t\rangle\) To compute the flux, we will integrate the result with respect to \(t\): \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{\pi/2} (y(2\cos t) - x(2\sin t)) dt\) Now substituting \(\mathbf{r}(t)\) and integrating: \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{\pi/2} (2\sin t(2\cos t) - 2\cos t(2\sin t)) dt=\int_0^{\pi/2} (4\sin t\cos t - 4\sin t\cos t) dt = 0\) The outward flux across the quarter circle for \(0 \leq t \leq \pi/2\) is 0.

b. Outward flux across the quarter circle for \(\pi/2 \leq t \leq \pi\)

The parameterized form of the path is given by \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle\) for \(\pi/2 \leq t \leq \pi\). The same process as in part (a) must be applied for this case. We still have the derivative \(\frac{d\mathbf{r}}{dt} = \langle -2 \sin t, 2 \cos t\rangle\), so our evaluation proceeds: \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_{\pi/2}^{\pi} (2\sin t(2\cos t) - 2\cos t(2\sin t)) dt\) As the integrand has not changed from part (a), the result will be the same: 0. The outward flux across the quarter circle for \(\pi/2 \leq t \leq \pi\) is 0.

c. Why the flux in the third quadrant is equal to the flux in part (a)

In the third quadrant, both \(x\) and \(y\) values are negative. The vector field in the third quadrant can be represented as \(\mathbf{F} = \langle -y, -x\rangle\). For the third quadrant, the parameterized form of the path is \(\mathbf{r}(t) = \langle 2 \cos t, 2 \sin t\rangle\) for \(\pi \leq t \leq 3\pi/2\). The dot product of this vector field with the derivative of \(\mathbf{r}(t)\) turns out to be the same as in part (a), so their integration will result in the same flux value. Therefore, the flux across the quarter circle in the third quadrant equals the flux computed in part (a).

d. Why the flux in the fourth quadrant is equal to the flux in part (b)

In the fourth quadrant, \(x\) values are positive, and \(y\) values are negative. The vector field in the fourth quadrant can still be represented as \(\mathbf{F} = \langle y, x\rangle\). For the fourth quadrant, the parameterized form of the path is \(\mathbf{r}(t) = \langle 2 \cos t, 2 \sin t\rangle\) for \(3\pi/2 \leq t \leq 2\pi\). The dot product of this vector field with the derivative of \(\mathbf{r}(t)\) turns out to be the same as in part (b), so their integration will result in the same flux value. Therefore, the flux across the quarter circle in the fourth quadrant equals the flux computed in part (b).

e. Outward flux across the full circle

Since we have calculated the outward fluxes for all four quadrants, and their values are equal (0 in each case), the total outward flux across the full circle is the sum of the fluxes in each quadrant: Total outward flux = Flux in quadrant 1 + Flux in quadrant 2 + Flux in quadrant 3 + Flux in quadrant 4 = 0 + 0 + 0 + 0 = 0 Thus, the total outward flux across the full circle is 0.

Key concepts

Flux

In mathematics, particularly in vector calculus, the concept of flux refers to the measure of how much a vector field flows through a surface. Imagine a field of arrows representing wind velocity, and a surface like a net or a sail in this field. Flux is the amount of "wind" that flows through or escapes the surface.

To mathematically calculate flux, you need to perform an integration of the vector field over a surface. This requires considering both the direction and the magnitude of the field through the surface's normal vector. For example, if a vector field is given as \(\mathbf{F}=\langle y, x\rangle\), flux through a curve or surface can be computed using the integral:

• Determine the parameterized curve along which flux is measured.
• Compute the derivative of the parameterization.
• Take the dot product of the vector field with this derivative.
• Integrate this dot product over the specified interval.

In the exercise, the outward flux is computed across portions of a circle, with results indicating zero flux due to symmetry in each quadrant.

Quarter Circle

A quarter circle is a geometric figure representing one-fourth of a circle, specifically a 90-degree arc. In scenarios involving flux and vector fields, quarter circles are essential because they simplify calculations and help evaluate symmetrical properties.

To relate quarter circles to vector field problems, consider:

• The standard circle parametrically defined with radius \(r\), such as \(\langle r \cos t, r \sin t \rangle\).
• For a quarter circle in vector field exercises, the parameter \(t\) ranges over intervals like \([0, \pi/2]\) or \([\pi/2, \pi]\), representing specific quadrants.

Quarter circles are practical in problems involving symmetric calculations and help visualize flux computations because they divide the problem into manageable sections. In the steps above, quarter circles help determine flux to reveal that symmetry results in zero flux for each evaluated arc segment.

Parametric Equations

Parametric equations describe how a system evolves over time using a different set of variables than usual. For instance, in vector fields, they help to define paths or curves along which we want to calculate certain physical quantities like flux.

A parametric equation might take the form \(\mathbf{r}(t) = \langle f(t), g(t) \rangle\), where \(f(t)\) and \(g(t)\) describe the \(x\) and \(y\) components respectively. In the context of the problem, parametric equations describe a quarter circle's path. Here's how they are useful:

• Aids in visualizing geometry and motion along paths.
• Facilitates the integration process by providing expressions for calculating derivatives.

By understanding how to construct and manipulate parametric equations, we can effectively interpret complex geometries in two dimensions, reducing complexity when applied to vector fields and integrals.

Integration

Integration is a powerful mathematical tool for computing quantities that involve accumulation, such as area under curves or flux through surfaces. In vector field analysis, integration allows us to compute total effects over curves or surfaces by summing infinitesimal contributions.

In the case of evaluating flux across a prescribed path, the integration involves several key steps:

• Calculate the derivative of the parametric path, say \(\mathbf{r}(t)\).
• Perform a dot product with the vector field, \(\mathbf{F}\).
• Integrate this product over the interval specified by the parametric path.

The process of integration simplifies complex cumulative calculations, and in this exercise, illustrates underlying symmetries in both vector fields and geometrical paths, leading to the conclusion that the flux for every quarter circle evaluated is zero.