Multivariable calculus and vector analysis

Circulation per unit area

When discussing the idea behind Green’s theorem, I claimed that the “microscopic” circulation of a two-dimensional vector field $\mathbf{F}$ was

$$\begin{array}{lll}\hfill \text{ “microscopic circulation”}=\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}.& & \hfill \end{array}$$

We can prove that the quantity

$$\begin{array}{lll}\hfill \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}.& & \hfill \end{array}$$

does represent “circulation per unit area,” which is probably a better term for it than “microscopic circulation.”

Remeber that the circulation of $\mathbf{F}$ around the closed curve $C$ is

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}.& & \hfill \end{array}$$

so the “circulation per unit area” is simply the above integral divided by the area inside $C$:

$$\begin{array}{lll}\hfill \frac{{\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}}{\text{ Area inside }C}.& & \hfill \end{array}$$

It turns out if you let $C$ shrink down to a point, this ratio becomes

$$\begin{array}{lll}\hfill \frac{{\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}}{\text{ Area inside }C}\to \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y},& & \hfill \end{array}$$

assuming $C$ is oriented counterclockwise.

We will sketch a proof of this for a rectangular curve $C$, oriented counterclockwise. Let the lower-left point of $C$ be $\left(a,b\right)$, its width be $\Delta x$, and its height be $\Delta y$. Label the edges of the rectangle by $C_1$, $C_2$, $C_3$, and $C_4$.

We assume box small enough to approximate $\mathbf{F}$ as constant along each edge.

Along the curve $C_1$, the value of $y$ is constantly $b$, but the value of $x$ changes from $a$ to $a+\Delta x$. We assume we can ignore that change in $x$ (since the box is small), and simply approximate $\mathbf{F}\left(x,y\right)$ as $\mathbf{F}\left(a,b\right)$ all along the segment $C_1$.

Along the curve $C_2$, the value of $x$ is constantly $a+\Delta x$, but $y$ changes from $b$ to $b+\Delta y$. Since we assume we can ignore the change in $y$, we approximate $\mathbf{F}\left(x,y\right)$ as $\mathbf{F}\left(a+\Delta x,b\right)$ all along segment $C_2$.

Similiarly, along $C_3$, $y=b+\Delta y$, and we approximate $x$ as $a$ (even though it ranges between $a$ and $a+\Delta x$). We approximate $\mathbf{F}\left(x,y\right)$ as $\mathbf{F}\left(a,b+\Delta y\right)$ all along segment $C_3$.

Along $C_4$, $x=a$, and we approximage $y$ as $b+\Delta y$ (even though it ranges between $b$ and $b+\Delta y$). We approximate $\mathbf{F}\left(x,y\right)$ as $\mathbf{F}\left(a,b\right)$ all along segment $C_4$.

We summarize these approximations in the following figure.

Next, to compute the integral ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$, we remember that it is the same thing as integrating the scalar-valued function $\mathbf{F}\cdot \mathbf{T}$, where $\mathbf{T}$ is the unit tangent vector along $C$:

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot \mathbf{T}ds.& & \hfill \end{array}$$

We need to compute $\mathbf{F}\cdot \mathbf{T}$ along each segment of $C$. The tangent vector $\mathbf{T}$ is constant along each segment, and we are approximating $\mathbf{F}$ as constant along each segment, so the dot product $\mathbf{F}\cdot \mathbf{T}$ will be constant along each segment of $C$.

Along $C_1$, the path is directed in the positive $x$ direction, so the unit tangent vector is $\mathbf{T}=\left(1,0\right)$. The dot product $\mathbf{F}\cdot \mathbf{T}$ is simply the first component of $\mathbf{F}\left(a,b\right)$:

$$\begin{array}{lll}\hfill \mathbf{F}\cdot \mathbf{T}=F_1\left(a,b\right).& & \hfill \end{array}$$

Along $C_2$, the path is directed in the positive $y$ direction, so $\mathbf{T}=\left(0,1\right)$, and $\mathbf{F}\cdot \mathbf{T}$ is simply the second component of $\mathbf{F}\left(a+\Delta x,b\right)$:

$$\begin{array}{lll}\hfill \mathbf{F}\cdot \mathbf{T}=F_2\left(a+\Delta x,b\right).& & \hfill \end{array}$$

Along $C_3$, the path is directed in the negative $x$ direction, so the unit tangent vector is $\mathbf{T}=\left(-1,0\right)$. The dot product $\mathbf{F}\cdot \mathbf{T}$ is minus the first component of $\mathbf{F}\left(a,b+\Delta y\right)$:

$$\begin{array}{lll}\hfill \mathbf{F}\cdot \mathbf{T}=-F_1\left(a,b+\Delta y\right).& & \hfill \end{array}$$

Along $C_4$, the path is directed in the negative $y$ direction, so $\mathbf{T}=\left(0,-1\right)$, and $\mathbf{F}\cdot \mathbf{T}$ is minus the second component of $\mathbf{F}\left(a,b\right)$:

$$\begin{array}{lll}\hfill \mathbf{F}\cdot \mathbf{T}=-F_2\left(a,b\right).& & \hfill \end{array}$$

We summarize these findings in the following figure.

The integrals are now easy to compute. Along $C_1$, $\mathbf{F}\cdot \mathbf{T}$ is constant, so the integral is simply its value $F_1\left(a,b\right)$ times the length of the segment $\Delta x$:

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot \mathbf{T}ds={\int \; <!--nolimits\; -->}_{C_1}{F}_1\left(a,b\right)ds=F_1\left(a,b\right)\Delta x.& & \hfill \end{array}$$

Along $C_2$, the integral is $F_2\left(a+\Delta x,b\right)$ times its length $\Delta y$:

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}=F_2\left(a+\Delta x,b\right)\Delta y.& & \hfill \end{array}$$

Similarly,

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_3}\mathbf{F}\cdot d\mathbf{s}=-F_1\left(a,b+\Delta y\right)\Delta x,& & \hfill \end{array}$$

and

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_4}\mathbf{F}\cdot d\mathbf{s}=-F_2\left(a,b\right)\Delta y.& & \hfill \end{array}$$

The integral around all of $C$ is just the sum along the four segments

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& ={\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_3}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_4}\mathbf{F}\cdot d\mathbf{s}& \hfill & \\ \hfill & =F_1\left(a,b\right)\Delta x+F_2\left(a+\Delta x,b\right)\Delta y-F_1\left(a,b+\Delta y\right)\Delta x-F_2\left(a,b\right)\Delta y.& \hfill & \end{array}$$

The circulation per unit area is the integral divided by the area of the rectangle, which is $\Delta x\Delta y$

$$\begin{array}{lll}\hfill \frac{{\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}}{\Delta x\Delta y}=\frac{F_2\left(a+\Delta x,b\right)\Delta y-F_2\left(a,b\right)\Delta y-\left(F_1\left(a,b+\Delta y\right)\Delta x-F_1\left(a,b\right)\Delta x\right)}{\Delta x\Delta y},& & \hfill \end{array}$$

where I simply rearranged the terms in the numerator.

Half of the numerator is multiplied by $\Delta y$ and half is multiplied by$\Delta x$. If we separate these into two fractions, we can cancel the $\Delta y$ in the first fraction with the $\Delta y$ in the demoninator

$$\begin{array}{lll}\hfill \frac{F_2\left(a+\Delta x,b\right)\Delta y-F_2\left(a,b\right)\Delta y}{\Delta x\Delta y}=\frac{F_2\left(a+\Delta x,b\right)-F_2\left(a,b\right)}{\Delta x}.& & \hfill \end{array}$$

In the second fraction, we can cancel the $\Delta x$,

$$\begin{array}{lll}\hfill \frac{F_1\left(a,b+\Delta y\right)\Delta x-F_1\left(a,b\right)\Delta x}{\Delta x\Delta y}=\frac{F_1\left(a,b+\Delta y\right)-F_1\left(a,b\right)}{\Delta y}.& & \hfill \end{array}$$

Putting these back together, we have

$$\begin{array}{lll}\hfill \frac{{\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}}{\Delta x\Delta y}=\frac{F_2\left(a+\Delta x,b\right)-F_2\left(a,b\right)}{\Delta x}-\frac{F_1\left(a,b+\Delta y\right)-F_1\left(a,b\right)}{\Delta y}.& & \hfill \end{array}$$

Now, we let the curve $C$ shrink down to a point. This means that $\Delta x\to 0$ and $\Delta y\to 0$. In this limit, the two fractions become something familar: partial derivatives of $\mathbf{F}$.

$$\begin{array}{llll}\hfill \underset{\Delta x,\Delta y\to 0}{lim}\frac{{\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}}{\Delta x\Delta y}& =\underset{\Delta x\to 0}{lim}\frac{F_2\left(a+\Delta x,b\right)-F_2\left(a,b\right)}{\Delta x}-\underset{\Delta y\to 0}{lim}\frac{F_1\left(a,b+\Delta y\right)-F_1\left(a,b\right)}{\Delta y}& \hfill & \\ \hfill & =\frac{\partial F}{\partial x}\left(a,b\right)-\frac{\partial F}{\partial y}\left(a,b\right).& \hfill & \end{array}$$

We have shown the the circulation per unit area around the point $\left(x,y\right)=\left(a,b\right)$ is

$$\begin{array}{lll}\hfill \frac{\partial F}{\partial x}\left(a,b\right)-\frac{\partial F}{\partial y}\left(a,b\right).& & \hfill \end{array}$$

This is exactly what we integrate over a region $D$ to obtain the total circulation around the border of $D$, according to Green’s theorem

$$\begin{array}{lll}\hfill \int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_D\left(\frac{\partial F}{\partial x}\left(x,y\right)-\frac{\partial F}{\partial y}\left(x,y\right)\right)dA={\int \; <!--nolimits\; -->}_{\partial D}\mathbf{F}\cdot d\mathbf{s}.& & \hfill \end{array}$$

where $C=\partial D$ is the path going counterclockwise around $D$.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

5a. Green's theorem

• The idea behind Green's theorem
• Other ways of writing Green's theorem
• Green's theorem with multiple boundary components
• Using Green's theorem to find area
• Sketch of proof for circulation per unit area
• Green's theorem examples

5b. The theorem for conservative vector fields

5c. Stokes' theorem

5d. The divergence theorem

6. Review material