Formal definition of curl in two dimensions

Background

• Curl in two dimensions
• Line integrals in a vector field

What we're building to

Formalizing fluid rotation

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  Positive

  A

  Positive
• (Choice B)  

  Negative

  B

  Negative
• (Choice C)  

  Zero

  C

  Zero

Check

Answer

Positive.

As you walk around the circle, the arrows of the vector field, $\mathbf{\text{F}}$‍, always point in the same direction as your motion, $d\mathbf{\text{r}}$‍. Therefore the dot product $\mathbf{\text{F}}\cdot d\mathbf{\text{r}}$‍ will always be positive, and hence the integral as a whole must be positive.

Letting the size of the region change

$\begin{array}{r}{\oint }_{C_R}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}=\end{array}$‍

Check

Answer

First we need to parameterize this circle. We can do this with the function

$\begin{array}{r}\mathbf{\text{r}}(t)=\left[\begin{array}{c}R\cos (t)\\ R\sin (t)\end{array}\right]\end{array}$‍

For this parameterization to cover the circle exactly once, let $t$‍ range from $0$‍ to $2\pi$‍. With this, we expand the line integral as follows:

$\begin{array}{rl}{\oint }_{C_R}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}& ={\int }_0^{2\pi }\mathbf{\text{F}}(\mathbf{\text{r}}(t))\cdot {\mathbf{\text{r}}}^{\prime }(t)dt\end{array}$‍

Now plug the components of $\mathbf{\text{r}}(t)$‍ into the definition of $\mathbf{\text{F}}(x,y)$‍

$\begin{array}{rl}\mathbf{\text{F}}(x,y)& =\left[\begin{array}{}-y\\ x\end{array}\right]\\ & \Downarrow \\ \mathbf{\text{F}}(\mathbf{\text{r}}(t))& =\mathbf{\text{F}}(R\cos (t),R\sin (t))\\ & =\left[\begin{array}{}-R\sin (t)\\ R\cos (t)\end{array}\right]\end{array}$‍

Also, take the derivative of $\mathbf{\text{r}}(t)$‍

$\begin{array}{r}{\mathbf{\text{r}}}^{\prime }(t)=\left[\begin{array}{}{\displaystyle \frac{d}{dt}}R\cos (t)\\ {\displaystyle \frac{d}{dt}}R\sin (t)\end{array}\right]=\left[\begin{array}{}-R\sin (t)\\ R\cos (t)\end{array}\right]\end{array}$‍

The fact that $\mathbf{\text{F}}(\mathbf{\text{r}}(t))={\mathbf{\text{r}}}^{\prime }(t)$‍ in this case is a coincidence, peculiar to this example. It makes the completion of the line integral pretty smooth, though.

$\begin{array}{rl}{\oint }_{C_R}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}& ={\int }_0^{2\pi }\mathbf{\text{F}}(\mathbf{\text{r}}(t))\cdot {\mathbf{\text{r}}}^{\prime }(t)dt\\ & ={\int }_0^{2\pi }\left[\begin{array}{}-R\sin (t)\\ R\cos (t)\end{array}\right]\cdot \left[\begin{array}{}-R\sin (t)\\ R\cos (t)\end{array}\right]dt\\ & ={\int }_0^{2\pi }(R^2{\sin }^2(t)+R^2{\cos }^2(t))dt\\ & ={\int }_0^{2\pi }{R}^2(\underset{\text{equals 1}}{\underbrace{{\sin }^2(t)+{\cos }^2(t)}})dt\\ & =R^2{\int }_0^{2\pi }dt\\ & =2\pi R^2\end{array}$‍

Average rotation per unit area

Key idea: Maybe if you take ${\displaystyle {\oint }_C\mathbf{\text{F}}\cdot d\mathbf{\text{r}}}$‍, which measures the fluid flow around a region, and divide it by the area of that region, it can give you some notion of the average rotation per unit area.

Check

Answer

When I say "rotation-per-unit-area", what I mean by "rotation" is the value of the integral

$\begin{array}{r}{\oint }_C\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$‍

This is because this integral indicates how much the fluid is flowing counterclockwise around the curve $C$‍.

The "unit area" in question is the area enclosed by $C$‍.

We found in the previous question that, for our example, the integral equals $2\pi R^2$‍. Since the area of the circle is $\pi R^2$‍, the ratio is always $2$‍.

$\begin{array}{r}\text{2d-curl}\mathbf{\text{F}}=\end{array}$‍

Check

Answer

Applying the formula, plug in the following values:

$F_1=-y$‍

$F_2=x$‍,

Here's what we get:

$\begin{array}{rl}\text{2d-curl}\mathbf{\text{F}}& ={\displaystyle \frac{\partial }{\partial x}}(x)-{\displaystyle \frac{\partial }{\partial y}}(-y)\\ & =1-(-1)\\ & =2\end{array}$‍

Defining two-dimensional curl

• $\mathbf{\text{F}}$‍ is a two-dimensional vector field.
• $(x,y)$‍ is some specific point in the plane.
• $A_{(x,y)}$‍ represents some region around the point $(x,y)$‍. For instance, a circle centered at $(x,y)$‍.
• $|A_{(x,y)}|$‍ indicates the area of $A_{(x,y)}$‍.
• ${\displaystyle \underset{|A_{(x,y)}|\to 0}{lim}}$‍ indicates we are considering the limit as the area of $A_{(x,y)}$‍ goes to $0$‍, meaning this region is shrinking around $(x,y)$‍.
• $C$‍ is the boundary of $A_{(x,y)}$‍, oriented counterclockwise.
• ${\displaystyle {\oint }_C}$‍ is the line integral around $C$‍, written as $\oint$‍ instead of $\int$‍ to emphasize that $C$‍ is a closed curve.

One more feature of conservative vector fields

Summary

• If a vector field represents fluid flow, you can quantify "fluid rotation in a region" by taking the line integral of that vector field along the border of that region.
• To go from the idea of fluid rotation in a region to fluid flow around a point (which is what curl measures), we introduce the idea of "average rotation per unit area" in a region. Then consider what this value approaches as your region shrinks around a point.
• In formulas, this gives us the definition of two-dimensional curl as follows:

  ${\displaystyle \text{2d-curl}\mathbf{\text{F}}(x,y)=\underset{A_{(x,y)}\to 0}{lim}\underset{\text{Average rotation per unit area}}{\underbrace{({\displaystyle \frac{1}{|A_{(x,y)}|}}{\oint }_C\mathbf{\text{F}}\cdot d\mathbf{\text{r}})}}}$‍

• This relationship between curl and closed-loop line integrals implies that irrotational fields and conservative fields are one and the same.