Curl and line integrals

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Green’s Theorem is a fundamental theorem of calculus.

In this section we will learn the fundamental derivative for vector fields, as well as a new fundamental theorem of calculus.

The curl of a vector field

Calculus has taught us that knowing the derivative of a function $f:\R \to \R$ can tell us important information about the function. In a similar way we have that seen that if we wish to understand a function of several variables $F:\R ^n\to \R$ , then the gradient, $\grad F$ , contains similar useful information. If you have a vector field $\vec {F}:\R ^n\to \R ^n,$ we now ask: “what is the natural analogue of a derivative in this setting?” When the vector field is two or three-dimensional, the curl is the analogue of the derivative that we are looking for:

In two dimensions, given a vector field $\vec {F}:\R ^2\to \R ^2$ , where $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$ the (scalar) curl is given by $\curl \vec F = \pp [N]{x}-\pp [M]{y}.$ In three dimensions, given a vector field $\vec {F}:\R ^3\to \R ^3$ , where $\vec {F}(x,y,z) = \vector {U(x,y,z)),V(x,y,z)),W(x,y,z)}$ the curl is given by $\begin{align*} \curl \vec F &= \det \begin{bmatrix} \veci & \vecj & \veck \\ \pp{x} & \pp{y} & \pp{z}\\ U & V & W \end{bmatrix}\\ &= \veci\left(\pp[W]{y}-\pp[V]{z}\right)- \vecj\left(\pp[W]{x}-\pp[U]{z}\right)+ \veck\left(\pp[V]{x}-\pp[U]{y}\right). \end{align*}$ Other authors sometimes use the notation $\mathop {\mathrm {rot}}\vec {F}$ for the scalar curl of a two-dimensional vector field $\vec {F}:\R ^2\to \R ^2$ , and $\mathop {\mathrm {curl}}\vec {F}$ for the curl of a three-dimensional vector field $\vec {F}:\R ^3\to \R ^3$ .

In two dimensions, is $\curl \vec {F}$ a scalar or a number?

number. vector.

In three dimensions, is $\curl \vec {F}$ a scalar or a number?

number. vector.

Consider the vector field $\vec {F}(x,y) = \vector {-y,x}$ . Compute: $\curl \vec {F}(x,y) \begin {prompt}= \answer {2}\end {prompt}$

Consider the vector field $\vec {F}(x,y,z) = \vector {-z,x,y}$ . Compute: $\curl \vec {F}(x,y,z) \begin {prompt} = \vector {\answer {1},\answer {-1},\answer {1}} \end {prompt}$

Now for something you’ve seen before, but in a different form.

Let $F:\R ^2\to \R$ . Compute: $\curl \grad F(x,y) \begin {prompt}= \answer {0}\end {prompt}$

Let $F:\R ^3\to \R$ . Compute: $\curl \grad F(x,y,z) \begin {prompt}= \vector {\answer {0},\answer {0},\answer {0}}\end {prompt}$

When $\curl \vec {F} = \vec {0}$ , then you know:

$\vec {F}$ is a gradient field. $\vec {F}$ is a conservative field. $\vec {F}:\R ^3\to \R ^3$ .

You can be assured that $\vec {F}:\R ^3\to \R ^3$ , since $\vec {0}$ is a vector. We only know a definition for curl in two and three dimensions; however, the two-dimensional definition is a scalar, not a vector. So if $\curl \vec {F} = \vec {0}$ , then $\vec {F}:\R ^3\to \R ^3$ .

What does the curl measure?

The curl of a vector field measures the rate that the direction of field vectors “twist” as $x$ and $y$ change. Imagine the vectors in a vector field as representing the current of a river. A positive curl at a point tells you that a “beach-ball” floating at the point would be rotating in a counterclockwise direction. A negative curl at a point tells you that a “beach-ball” floating at the point would be rotating in a clockwise direction. Zero curl means that the “beach-ball” would not be rotating. Below we see our “beach-ball” with two field vectors. If $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$

we see that the right field vector is larger than the left, thus giving the “beach-ball” a counterclockwise rotation. In an entirely similar way, if we have

we see that the bottom field vector is larger than the top, thus giving the “beach-ball” a counterclockwise rotation. Thus the curl combines $\pp [N]{x}$ and $-\pp [M]{y}$ $\curl \vec {F} = \pp [N]{x} - \pp [M]{y}$ to obtain the infinitesimal rotation of the field. The most obvious example of a vector field with nonzero curl is $\vec {F}(x,y) = \vector {-y,x}$ .

Unfortunately, while we can sometimes identify nonzero curl from a graph, it can be difficult. In our next example, we see a field that does not have global rotation but does have local rotation.

Consider the following vector field $\vec {F}$ :

Setting $\d x = 1$ and $\d y = 1$ , estimate: $\curl \vec {F}(\vec {a})$

First note that one should imagine a vector at every point. We’ll assume that the magnitudes of the vectors are constant along vertical lines. Set $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$ . To estimate $\pp [N]{x}$ , we examine the change in $N(x,1)$ between $x=1$ and $x=2$ : $\frac {N(2,1) - N(1,1)}{2-1} = \answer [given]{1}$ and we should also check the change of $N(x,1)$ between $x=2$ and $x=3$ : $\frac {N(3,1) - N(2,1)}{3-2} = \answer [given]{1}$ Averaging these values together we find $\pp [N]{x} \approx \answer [given]{1}$ To estimate $\pp [M]{y}$ , we examine the change in $M(2,y)$ between $y=0$ and $y=1$ : $\frac {M(2,1) - M(2,0)}{1-0} = \answer [given]{0}$ and we should also check the change of $M(2,y)$ between $y=1$ and $y=2$ : $\frac {M(2,2) - M(2,1)}{2-1} = \answer [given]{0}$ Averaging these values together we find $\pp [M]{y} \approx \answer [given]{0}$ So we approximate $\curl \vec {F}(\vec {a})\approx \answer [given]{1}.$ So field above example does not have global rotation, but it does have local rotation.

Now we’ll show you a field that has global rotation but no local rotation!

Consider the vector field $\vec {F}(x,y) = \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}$

Compute: $\curl \vec {F}(x,y)$

In this case, $\curl \vec {F}(x,y) =\answer [given]{0}$ when $\vector {x,y}\ne \vec {0}$ . At $\vec {0}$ , the curl of the vector field is zeroinfinityundefined , and hence this field is not a gradient field. This field has global rotation, but it does not have local rotation.

Finally, we’ll show you a field that has global rotation with local rotation in the opposite direction!

Consider the vector field $\vec {F}(x,y) = \vector {\frac {-y}{(x^2+y^2)^{3/2}},\frac {x}{(x^2+y^2)^{3/2}}}$

Compute: $\curl \vec {F}(x,y)$

In this case, $\curl \vec {F}(x,y) =\answer [given]{\frac {-1}{(x^2+y^2)^{3/2}}}$ when $\vector {x,y}\ne \vec {0}$ . Note, this is positivenegative for all values of $x$ and $y$ . This corresponds to local rotation in the clockwisecounterclockwise direction. This field has global rotation in the clockwisecounterclockwise direction, but has local rotation in the clockwisecounterclockwise direction.

At this point we only know how to take the derivative (via the curl) of a vector field of two or three dimensions. You can take another course to learn more about derivatives of $n$ -dimensional vector fields.

A new fundamental theorem of calculus

Recall that a fundamental theorem of calculus says something like:

To compute a certain sort of integral over a region, we may do a computation on the boundary of the region that involves one fewer integrations.

In the single variable case we have:

With line integrals we have:

We now introduce a new fundamental theorem of calculus involving the curl. It’s called Green’s Theorem:

Green’s Theorem If the components of $\vec {F}:\R ^2\to \R ^2$ have continuous partial derivatives on a closed region $R$ where $C$ is a boundary of $R$ and $\vec {p}(t)$ parameterizes $C$ in a counterclockwise direction with the interior on the left, then $\iint _R \curl \vec {F}\d A = \oint _C \vec {F}\dotp \d \vec {p}$

Let $C$ be the rectangle with corners $(2,-1)$ , $(2,4)$ , $(3,4)$ , and $(3,-1)$ . Compute: $\oint _C \frac {\sin (x)}{x} \d x + x^2 \d y$

We’ll use Green’s Theorem to squash this scary integral with ease. First note that if we imagine $\oint _C \frac {\sin (x)}{x} \d x + x^2 \d y = \oint _C\vec {F}\dotp \d \vec {p}$ we set: $\vec {F}(x,y) = \vector {\answer [given]{\frac {\sin (x)}{x}},\answer [given]{x^2}}$ Further note that our field is continuous on the interior of the rectangle. Thus we may apply Green’s Theorem! Write with me now, $\curl \vec {F}(x,y) = \answer [given]{2x}$ So by Green’s Theorem $\oint _C \frac {\sin (x)}{x} \d x + x^2 \d y = \int _{\answer [given]{-1}}^{\answer [given]{4}}\int _{\answer [given]{2}}^{\answer [given]{3}} \answer [given]{2x}\d x\d y$ Now, keep writing with me, $\begin{align*} \int_{\answer[given]{-1}}^{\answer[given]{4}}\int_{\answer[given]{2}}^{\answer[given]{3}} \answer[given]{2x}\d x\d y &= \int_{\answer[given]{-1}}^{\answer[given]{4}} \answer[given]{5} \d y\\ &= \answer[given]{25}. \end{align*}$ The upshot is that we were able to use Green’s Theorem to transform a tedious integral into a trivial one.

Suppose that the curl of a vector field $\vec {F}:\R ^2\to \R ^2$ is constant, $\curl \vec {F} = 3$ .

If $\int _{C_1} \vec {F} \dotp \d \vec {p} = 20$ estimate $\int _{C_2} \vec {F}\dotp \d \vec {p} \begin {prompt} = \answer {46} \end {prompt}$

Use Green’s Theorem.

How is Green’s Theorem a fundamental theorem of calculus? Well consider this:

Are there more fundamental theorems of calculus? Absolutely, read on young mathematician!