The Theorem of Stokes in the Plane

Section 11.3 The Theorem of Stokes in the Plane

Given a vector field \(\vect f=(f_1,f_2)\text{,}\) Green’s theorem asserts that

\begin{equation*} \iint_D\Bigl(\frac{\partial}{\partial x_1}f_2 -\frac{\partial}{\partial x_2}f_1\Bigr)\,dx_1\,dx_2 =\int_{\partial D}f_1\,dx_1+f_2\,dx_2\text{.} \end{equation*}

As we saw in Section 7.4 the line integral on the right hand side represents the circulation of the vector field \(\vect f\) along the curve \(\partial D\) in the positive direction. In Remark 8.17 we called the integrand on the left hand side the curl of a plane vector field, so we set

\begin{equation*} \curl\vect f:=\frac{\partial}{\partial x_1}f_2 - \frac{\partial}{\partial x_2}f_1 \end{equation*}

in this section. With this we can reformulate Green’s theorem as follows.

Theorem 11.7. Theorem of Stokes.

If \(\vect f\) is a smooth vector field defined on a piecewise smooth domain \(D\) then

\begin{equation} \int_D\curl\vect f(\vect x)\,d\vect x =\int_{\partial D}\vect f\cdot\vect\tau\,ds\text{,}\tag{11.3} \end{equation}

where \(\vect\tau\) is the positive unit tangent to \(\partial D\text{.}\)

We next want to obtain a physical interpretation of \(\curl\vect f\text{.}\)

Observation 11.8. Physical interpretation of curl.

Let \(\vect f\) be the velocity of a fluid or a gas in a region \(D\text{.}\) Fix a point \(\vect a=(a_1,a_2)\) in \(D\text{,}\) and denote by \(B_\delta\) a disc centred at \(\vect a\) with radius \(\delta\text{.}\) Then

\begin{equation*} \int_{\partial B_\delta}\vect f\cdot\vect\tau\,ds \end{equation*}

is the net flow of \(\vect f\) along \(\partial B_\delta\text{.}\) In other words it is the circulation of the field along \(\partial B_\delta\text{.}\) By the theorem of Stokes

\begin{equation*} \frac{1}{\area(B_\delta)} \int_{B_\delta}\curl\vect f\,d\vect x =\frac{1}{\area(B_\delta)} \int_{\partial B_\delta}\vect f\cdot\vect\tau\,ds\text{.} \end{equation*}

The above is a density of the circulation of \(\vect f\) in \(B_\delta\text{.}\) If we shrink \(B_\delta\) to \(\vect a\) we get (see Lemma 11.6)

\begin{equation*} \lim_{\delta\to 0}\frac{1}{\area(B_\delta)} \int_{B_\delta}\curl\vect f\,d\vect x =\curl\vect f(\vect a)\text{.} \end{equation*}

Hence \(\curl\vect f\) is a measure for the circulation of the vector field \(\vect f\) at \(\vect a\text{.}\) If we fix a little paddle wheel at \(\vect a\) then \(\curl\vect f(\vect a)\) tells us how it turns. If \(\curl\vect f(\vect a)>0\) it turns in the counterclockwise direction, if \(\curl\vect f(\vect a)<0\) it turns in the clockwise direction, and if \(\curl\vect f(\vect a)=0\) it does not turn at all.

Example 11.9.

Let \(\vect f=(y,0)\) for \((x,y)\in\mathbb R^2\text{.}\) Then the vector field looks as shown in Figure 11.10. A simple computation shows that \(\curl\vect f=-1\text{,}\) so a little paddle wheel would turn in the clockwise direction. This is clear as the current is different on the top and the bottom.

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Figure 11.10. The vector field \(\vect f(x,y):=(y,0)\) in \(\mathbb R^2\text{.}\)

Example 11.11.

Let \(\vect f=(-y,x)\) for \((x,y)\in\mathbb R^2\text{.}\) Then the vector field looks as shown in Figure 11.12. A simple computation shows that \(\curl\vect f=2\text{,}\) so a little paddle wheel would turn in the counterclockwise direction.

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Figure 11.12. The vector field \(\vect f(x,y):=(-y,x)\) in \(\mathbb R^2\text{.}\)

As the vector field circles around the origin it seems `obvious’ that the curl is positive. Note however, that the curl is a local property of the vector field, and that a global picture can be quite deceptive as the following example shows!

Example 11.13.

Let \(\vect f(x,y):=\Bigl(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\Bigr)\) be the vector field studied in Example 8.12. It looks as shown in Figure 8.13. It follows from the computations there that \(\curl\vect f=0\) except at the origin, where \(\vect f\) is not defined. On the other hand the field looks at first quite similar to the one from the previous example, where the curl was positive.

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