Question

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

Short answer

In conclusion, we have shown that for any given \(a, b, c\), and \(d\), provided the condition \(b=c\) holds, the vector field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\) has zero circulation on any oriented circle centered at the origin. This result was demonstrated by parameterizing the circle, evaluating the vector field and its derivative at the points on the circle, computing the line integral, and finally showing that the integral equaled zero under the given condition.

Step-by-step solution

Parameterize the circle

Let's consider an oriented circle centered at the origin with radius \(r\). We can parameterize the circle with the following parameterization: $$ \mathbf{r}(t) = \langle r\cos t, r\sin t\rangle, \quad 0 \leq t \leq 2\pi $$ Note that the parameterization traces the circle counterclockwise as \(t\) increases.

Compute the derivative of the parameterization

To evaluate the line integral, we need the derivative of the parameterization with respect to \(t\): $$ \frac{d\mathbf{r}}{dt} = \langle -r\sin t, r\cos t\rangle $$

Evaluate the vector field at the parameterization points

For the given vector field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\), we need to evaluate it at the points on the circle parameterized by \(\mathbf{r}(t)\): $$ \mathbf{F}(\mathbf{r}(t)) = \langle a(r\cos t) + b(r\sin t), c(r\cos t) + d(r\sin t)\rangle $$

Compute the dot product of the evaluated vector field and derivative of the parameterization

Next, we compute the dot product of the evaluated vector field, \(\mathbf{F}(\mathbf{r}(t))\), and the derivative of the parameterization, \(\frac{d\mathbf{r}}{dt}\): $$ \begin{aligned} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} &= \left\langle a(r\cos t) + b(r\sin t), c(r\cos t) + d(r\sin t) \right\rangle \cdot \left\langle -r\sin t, r\cos t \right\rangle \\ &= [a(r\cos t) + b(r\sin t)](-r\sin t) + [c(r\cos t) + d(r\sin t)](r\cos t) \end{aligned} $$

Evaluate the line integral around the circle

Now, we need to compute the line integral of the vector field around the circle: $$ \begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt \\ &= \int_0^{2\pi} \left[ [a(r\cos t) + b(r\sin t)](-r\sin t) + [c(r\cos t) + d(r\sin t)](r\cos t) \right] dt \end{aligned} $$

Use the condition \(b=c\) to simplify the integral

Given the condition \(b=c\), we can simplify the line integral: $$ \begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= \int_0^{2\pi} \left[ [a(r\cos t) + b(r\sin t)](-r\sin t) + [b(r\cos t) + d(r\sin t)](r\cos t) \right] dt \\ &= \int_0^{2\pi} \left[ -r^2(a\cos t \sin t + b\sin^2 t) + r^2(b\cos^2 t + d\sin t \cos t) \right] dt \\ &= r^2 \int_0^{2\pi} \left[ -(a\cos t \sin t + b\sin^2 t) + b\cos^2 t + d\sin t \cos t \right] dt. \end{aligned} $$

Show that the integral equals zero

Now we need to show that the integral is equal to zero: $$ \begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= r^2 \int_0^{2\pi} \left[-(a\cos t \sin t + b\sin^2 t) + b\cos^2 t + d\sin t \cos t \right] dt \\ &= r^2 \left[ -a \int_0^{2\pi} (\cos t \sin t) \, dt - b\int_0^{2\pi} (\sin^2 t) \, dt + b\int_0^{2\pi} (\cos^2 t) \, dt + d\int_0^{2\pi} (\sin t \cos t) \, dt \right] \end{aligned} $$ Since $$\int_0^{2\pi} (\cos t \sin t) \, dt = \int_0^{2\pi} (\sin t \cos t) \, dt = 0$$ and $$\int_0^{2\pi} (\sin^2 t) \, dt = \int_0^{2\pi} (\cos^2 t) \, dt = \frac{\pi}{2},$$ then $$ \begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= r^2 \left[ -a \cdot 0 - b\cdot \frac{\pi}{2} + b\cdot \frac{\pi}{2} + d \cdot 0\right] \\ &= 0 \end{aligned} $$ Hence, the vector field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\) has zero circulation on any oriented circle centered at the origin for any \(a, b, c\), and \(d\), provided \(b=c\).

Key concepts

Line Integrals

Line integrals allow us to calculate the integral of a vector field along a curve. Suppose we have a vector field \( \mathbf{F} \) defined in a region and a smooth curve \( C \) within this region. The line integral of \( \mathbf{F} \) along \( C \) is essentially the accumulation of dot products of \( \mathbf{F} \) with differential line elements \( d\mathbf{r} \) of the curve.

The expression for evaluating a line integral is: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt \] where \( \mathbf{r}(t) \) is a parameterization of the curve \( C \), and \( t \) varies over some interval \([a, b]\).

Line integrals of vector fields have several applications, including calculating work done by a force along a path. They are particularly useful in physics and engineering for studying fields like electrostatics and fluid dynamics.

Zero Circulation

A vector field has zero circulation around a closed path if the line integral around the path evaluates to zero. In our problem statement, the condition \(b = c\) causes the field \( \mathbf{F} \) to have this characteristic.

Circulation measures the total 'twisting' effect produced by a vector field around a closed curve. When we say zero circulation, it means there is no net rotating influence along the curve.

Mathematically, for a vector field \(\mathbf{F} = \langle a x + b y, b x + d y \rangle\) and an oriented circle \(C\), the integral simplifies as shown in the solution:

• The contributions of \(a\) and \(d\) drop out due to symmetry,
• The terms involving \(b\), due to \(b = c\), perfectly counteract over the circle,

which leads to the integral being zero.

This property of zero circulation is an essential concept in fields such as fluid mechanics, where it can correspond to irrotational flow.

Parameterization

Parameterizing a curve involves representing it using a continuous function of a single variable, usually written as a vector function \( \mathbf{r}(t) \). For a circle of radius \( r \) centered at the origin, we use:

\[ \mathbf{r}(t) = \langle r\cos t, r\sin t \rangle, \quad 0 \leq t \leq 2\pi \]

Parameterization is crucial when working with integrals over curves, as it allows us to translate a curve into a form that makes computation manageable.

With this simplification, we can determine the derivative \( \frac{d\mathbf{r}}{dt} \) and substitute it along with \( \mathbf{r}(t) \) into the integrand. Thus, parameterization turns a geometric curve into algebraic expressions that are much easier to handle with calculus.

Good parameterization choices often simplify complex integrals, revealing underlying properties like symmetry.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric terms such as \(\sin t\) and \(\cos t\). In our problem, these integrals arise because the circle parameterization uses these functions.

To evaluate line integrals with trigonometric components, we often rely on identities and standard integral results. For instance,

• \( \int_0^{2\pi} \sin^2 t \, dt = \int_0^{2\pi} \cos^2 t \, dt = \pi \)
• \( \int_0^{2\pi} \sin t \cos t \, dt = 0 \)

These identities help simplify the calculations significantly.

Understanding these integrals is vital as they are prevalent in many fields, especially those involving wave-like or periodic behavior. Problems like the one presented here often highlight the utility of converting complex expressions into a form where these standard integrals can be used.