Question

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: Prove that the vector field \(\mathbf{F} = \langle ax+by, cx+dy \rangle\) has zero flux across any circle centered at the origin if \(a = -d\). Solution: We have shown that the flux integral of the vector field \(\mathbf{F}\) is given by the expression \(R^2\int_0^{2\pi} (-2a\cos\theta\sin\theta + (c-b)(\cos^2\theta - \sin^2\theta)) \, d\theta\). After integrating and evaluating, we found that the flux integral equals zero. Therefore, the given vector field has zero flux across any circle centered at the origin under the condition \(a = -d\).

Step-by-step solution

Define Flux Integral and Parameterizing the Circle

The flux of \(\mathbf{F}\) across a curve \(C\) is given by the line integral: \(\text{Flux}(\mathbf{F}) = \oint_C \mathbf{F} \cdot d\mathbf{r}\) Consider a circle of radius \(R\) centered at the origin. We can parameterize the circle \(C\) using angle \(\theta\) as follows: \(\mathbf{r}(\theta) = \langle R\cos\theta, R\sin\theta \rangle\), where \(0 \leq \theta \leq 2\pi\) Now, we need to find the derivative of \(\mathbf{r}\) with respect to \(\theta\): \(\frac{d\mathbf{r}}{d\theta} = \langle -R\sin\theta, R\cos\theta \rangle\)

Calculate the Flux Integral

We can now calculate the flux integral of the vector field \(\mathbf{F}\): \(\text{Flux}(\mathbf{F}) = \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(\theta)) \cdot \frac{d\mathbf{r}}{d\theta} \, d\theta\) Substitute the parameterization of the circle and its derivative: \(\text{Flux}(\mathbf{F}) = \int_0^{2\pi} \langle aR\cos\theta + bR\sin\theta, cR\cos\theta - aR\sin\theta\rangle \cdot \langle -R\sin\theta, R\cos\theta \rangle \, d\theta\) Calculate the dot product: \(\text{Flux}(\mathbf{F}) = \int_0^{2\pi} (-aR^2\cos\theta\sin\theta - bR^2\sin^2\theta + cR^2\cos^2\theta - aR^2\sin\theta\cos\theta) \, d\theta\) Combine terms and factor out \(R^2\): \(\text{Flux}(\mathbf{F}) = R^2 \int_0^{2\pi} (-2a\cos\theta\sin\theta + (c-b)\cos^2\theta - (c+b)\sin^2\theta) \, d\theta\)

Evaluate the Integral and Show It Equals Zero

Using the trigonometric identity \(\cos^2\theta + \sin^2\theta = 1\), we can rewrite the integral as: \(\text{Flux}(\mathbf{F}) = R^2 \int_0^{2\pi} (-2a\cos\theta\sin\theta + (c-b)(\cos^2\theta -\sin^2\theta)) \, d\theta\) Now, we integrate each term: \(\int_0^{2\pi} -2a\cos\theta\sin\theta \, d\theta = -a\int_0^{2\pi} \sin (2\theta) \, d\theta = 0\) \(\int_0^{2\pi} (c-b)\cos^2\theta \, d\theta = \int_0^{2\pi} \frac{(c-b)(1+\cos(2\theta))}{2} \, d\theta = \frac{c-b}{2} \cdot 2\pi\) \(\int_0^{2\pi} (c-b)\sin^2\theta \, d\theta = \int_0^{2\pi} \frac{(c-b)(1-\cos(2\theta))}{2} \, d\theta = -\frac{c-b}{2} \cdot 2\pi\) Finally, summing all three integrals: \(\text{Flux}(\mathbf{F}) = R^2(0 + \frac{c-b}{2} \cdot 2\pi - \frac{c-b}{2} \cdot 2\pi) = 0\) Thus, the flux integral of the vector field \(\mathbf{F}\) across any circle centered at the origin is zero under the condition \(a = -d\).

Key concepts

Vector Field

In the realm of multivariable calculus, a vector field is a map that assigns a vector to each point in a subset of space. For example, in a two-dimensional space, a vector field might represent the flow of a fluid across an area, with each vector indicating the fluid's velocity at that point.

With respect to our exercise, the vector field given is \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\). This field can be visualized as sets of arrows on the plane, with the direction and magnitude of each arrow varying based on their position in relation to the origin. While this vector field is quite general, it conforms to a certain condition when \(a=-d\), which reveals interesting properties, such as the result described in the exercise regarding the flux across a circle centered at the origin.

Line Integral

A line integral is a means to measure the total influence of a vector field along a certain curve. When you think about a line integral, picture walking along a path and feeling a force, such as wind, pushing against you in various directions. The line integral calculates how much this force contributes to your movement along the path.

In our case, calculating the flux is akin to taking the line integral of the vector field \(\mathbf{F}\) along a closed curve \(C\), which represents the cumulative effect of the field as it permeates the curve. The line integral involves dotting the vector field with the differential element of the curve, \(d\mathbf{r}\), to focus on the component of the field that's normal (or perpendicular) to the curve.

Parameterization

To compute the line integral along a curve, such as a circle, we need to describe the curve mathematically. This process is known as parameterization. Essentially, we need to define a function that traces out the curve as we vary a parameter. This allows us to replace the vague concept of traveling around a curve with a more precise description involving only one variable.

In the context of our exercise, the circle is parameterized using the trigonometric functions sine and cosine, connecting every angle \(\theta\) to a point on the circle's circumference. The parameter \(\theta\) ranges from 0 to \(2\pi\) as it completes a full revolution around the circle. Once parameterized, the curve's properties can be plugged back into the expression for the line integral, marinating the calculation in concrete terms.

Trigonometric Identity

Trigonometry plays a crucial role in manipulating and simplifying the integral of functions related to circles and periodic phenomena. A fundamental trigonometric identity that is particularly useful is \(\cos^2\theta + \sin^2\theta = 1\). This identity allows us to simplify expressions that arise in calculations involving circles or oscillations.

In our exercise, after we've set up the line integral, we utilize this identity to transform and ultimately simplify the integral. It assists us in reducing a complex-looking integral into a sum of simpler terms, which can then be integrated over the interval from 0 to \(2\pi\). The result of these integrations can often reveal properties about the vector field and the curve in question, such as the zero flux result explained in the solution.