Question

Zero flux fields. Consider the vector field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle .\) Show that \(\mathbf{F}\) has zero flux across any oriented circle centered at the origin, for any \(a, b, c,\) and \(d,\) provided \(a=-d\)

Short answer

Question: Show that the vector field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\) has zero flux across any oriented circle centered at the origin for any values of \(a, b, c\), and \(d\), given that the condition \(a=-d\). Answer: The flux is given by \(\Phi = r^2\int_0^{2\pi} \left[(-d \cos(\theta) + b \sin(\theta))(-\sin(\theta)) + (c \cos(\theta) - a \sin(\theta))(\cos(\theta))\right]d\theta\), and using the condition \(a=-d\), the flux \(\Phi\) equals zero. Therefore, the vector field \(\mathbf{F}\) has zero flux across any oriented circle centered at the origin for any values of \(a, b, c\), and \(d\), provided the condition \(a=-d\) holds.

Step-by-step solution

Parametrization of the curve

We will consider an oriented circle of radius \(r\) (\(r > 0\)) centered at the origin described by the equation $$ x^2+y^2=r^2. $$ Parametrize the circle using polar coordinates: $$ x=r\cos(\theta), \quad y=r\sin(\theta), $$ where \(0 \le \theta \le 2\pi\).

Compute the vector field in polar coordinates

Now, express the vector field in terms of \(\theta\) by substituting the parametrization into \(\mathbf{F}\): $$ \mathbf{F}(\theta)=\langle a r\cos(\theta)+b r\sin(\theta), c r\cos(\theta)+d r\sin(\theta)\rangle $$

Compute the derivative of the parametrization with respect to \(\theta\)

Compute the derivative of the parametrization with respect to \(\theta\): $$ \frac{d}{d\theta}\left(\langle r\cos(\theta),r\sin(\theta)\rangle\right)=\langle-r\sin(\theta), r\cos(\theta)\rangle $$

Compute the flux of the vector field across the circle

Now, calculate the line integral of the vector field \(\mathbf{F}\) dot product with the derivative of the parametrization with respect to \(\theta\) to find the flux around the circle. $$ \Phi = \oint_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\theta) \cdot \frac{d}{d\theta}\left(\langle r\cos(\theta),r\sin(\theta)\rangle\right) d\theta $$ Substitute the expressions of \(\mathbf{F}(\theta)\) and \(\frac{d}{d\theta}\left(\langle r\cos(\theta),r\sin(\theta)\rangle\right)\) into the formula: $$ \Phi = r^2\int_0^{2\pi} \left[(a \cos(\theta) + b \sin(\theta))(-\sin(\theta)) + (c \cos(\theta) + d \sin(\theta))(\cos(\theta))\right]d\theta $$

Apply the given condition \(a = -d\)

Now, substitute the given condition \(a=-d\) into the flux equation and then simplify: $$ \Phi = r^2\int_0^{2\pi} \left[(-d \cos(\theta) + b \sin(\theta))(-\sin(\theta)) + (c \cos(\theta) - a \sin(\theta))(\cos(\theta))\right]d\theta $$

Evaluate the integral

Since we have a product of sines and cosines, we know that integrating over \(2\pi\) will result in zero because the two functions are orthogonal in the range \([0, 2\pi]\). Thus, we have: $$ \Phi = r^2\int_0^{2\pi} \left[(-d \cos(\theta) + b \sin(\theta))(-\sin(\theta)) + (c \cos(\theta) - a \sin(\theta))(\cos(\theta))\right]d\theta = 0 $$ Therefore, the vector field \(\mathbf{F}\) has zero flux across any oriented circle centered at the origin for any values of \(a, b, c\), and \(d\), provided the condition \(a=-d\) holds.

Key concepts

Vector Field

A vector field is a mathematical construct that assigns a vector to every point in a subset of space. In two dimensions, a vector field can be represented as \( \mathbf{F} = \langle P(x, y), Q(x, y) \rangle \), where \( P \) and \( Q \) are functions that describe the horizontal and vertical components of the vector at the point \( (x, y) \).

When we talk about flux, we are referring to the amount of the vector field that 'flows' through a given surface or curve. In the exercise provided, the vector field is given in Cartesian coordinates, but as we will see later, can also be expressed in polar coordinates. This expression of the vector field is crucial for understanding how it interacts with geometrical shapes such as circles, especially when computing flux, which involves integration along a path or over a surface.

Line Integral

The line integral of a vector field represents the net effect of the field along a path. For a curve \( \mathcal{C} \) parametrized by \( \mathbf{r}(t) \) with \( t \) in the interval \( [a, b] \) and a vector field \( \mathbf{F} \) defined on \( \mathcal{C} \) the line integral is given by \( \int_\mathcal{C} \mathbf{F} \cdot d\mathbf{r} \).

In simple terms, this calculates how much of the field is aligned with the path — it's a measure of 'flow' along the path. In the context of the exercise, we are looking to find the line integral of the vector field across a circle, noting that if the integral evaluates to zero, it implies that the net flow of the vector field across the path is null, thus the term 'zero flux'.

Flux Computation

The computation of flux through a curve encompasses how much a vector field 'flows' through that curve per unit time. Mathematically, flux is calculated as the line integral of the vector field across the curve. In the context of the exercise, we want to show that the flux of the vector field \( \mathbf{F} \) through any circle centered at the origin is zero, under a specific condition on the coefficients of the field.

To achieve this, one evaluates the dot product of the vector field with the differential element of the path expressed via the derivative of the parametrization – in this case, the circle – and integrates over the entire path. The condition \( a = -d \) simplifies the integral and leads to the result that the flux is indeed zero.

Polar Coordinates

The polar coordinates system is an alternative to the Cartesian coordinates system for representing points in a plane. A point \( (x, y) \) in Cartesian coordinates can be represented as \( (r, \theta) \) in polar coordinates, where \( r \) is the distance from the origin (radial coordinate) and \( \theta \) is the angle from the positive x-axis (angular coordinate).

In our exercise, using polar coordinates allows us to describe the circle's points efficiently by keeping \( r \) constant and varying \( \theta \) from \( 0 \) to \( 2\pi \). The assignment of vectors to points in polar coordinates is crucial for the computation of integrals over circular paths, as it greatly simplifies the process, particularly when dealing with symmetrical fields and boundaries like circles centered at the origin.