Add the turning inside, find the circulation outside, integrals from easy to advanced lesson

Travel once around the unit circle counterclockwise through the field $\mathbf{F}(x, y) = (-y, x)$ , whose arrows turn around the origin. How much does the field push along the direction of travel, in total? That total is the closed line integral $\oint_C (-y\,dx + x\,dy)$ , and this page finds it in two ways that both give $2\pi$ , once directly along the circle and once with Green’s theorem, which adds up the local turning of the field over the disk instead. It also explains why the boundary total and the interior total must agree, and why the direction of travel decides the sign.

Key terms

Circulation

Definition Let $C$ be a closed curve, traced once in a chosen direction by a piecewise continuously differentiable parametrization, and let $\mathbf{F} = (P, Q)$ be continuous on $C$ . The circulation of $\mathbf{F}$ around $C$ is the line integral

\begin{aligned} &\oint_C \mathbf{F} \cdot d\mathbf{r} \\[0.4em] &\quad = \oint_C (P\,dx + Q\,dy). \end{aligned}

The circle on the integral sign marks a closed path. Where the parametrization has a nonzero derivative, the integrand can also be written $\mathbf{F} \cdot \mathbf{T}\,ds$ , with $\mathbf{T}$ the unit tangent in the direction of travel and $ds$ the element of arc length.

In plain words Circulation adds up the push of the field along the curve, all the way around. Arrows that point with the motion add, arrows that point against it subtract, and arrows that cross the curve at right angles add nothing.

Example Parametrize the unit circle counterclockwise by $\mathbf{r}(t) = (\cos t, \sin t)$ , $0 \le t \le 2\pi$ . Then $\mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t)$ is exactly $\mathbf{r}'(t)$ , so the integrand is $\sin^2 t + \cos^2 t = 1$ and

\begin{aligned} &\oint_C (-y\,dx + x\,dy) \\[0.4em] &\quad = \int_0^{2\pi} 1\,dt = 2\pi. \end{aligned}

In geometric terms, on the unit circle the field has length $1$ and points along the direction of travel, so each piece of the circle contributes its own length, and the total is the circumference $2\pi$ .

Counterclockwise orientation

Definition Let $D$ be a bounded region in the plane whose boundary $C$ is a simple closed curve, a closed curve that does not cross itself. The positive orientation of $C$ is the direction of travel that keeps $D$ on the left, and for such a curve it is the counterclockwise direction. Reversing the orientation of a curve reverses every displacement $d\mathbf{r}$ , so it changes the sign of every line integral along it. For a region with holes, the positive orientation runs counterclockwise around the outer boundary and clockwise around each inner boundary, again keeping the region on the left.

In plain words Walk along the boundary with the region on your left. Around a single closed curve that walk goes counterclockwise, and walking the other way flips the sign of the answer.

Example Write $-C$ for the unit circle traced clockwise, by $\mathbf{r}(t) = (\cos t, -\sin t)$ , $0 \le t \le 2\pi$ . The field $(-y, x)$ now points against the motion at every point, the integrand is $-\sin^2 t - \cos^2 t = -1$ , and

\begin{aligned} &\oint_{-C} (-y\,dx + x\,dy) \\[0.4em] &\quad = \int_0^{2\pi} (-1)\,dt = -2\pi. \end{aligned}

Scalar curl

Definition Let $\mathbf{F} = (P, Q)$ have continuous first partial derivatives on an open set $D$ . The scalar curl of $\mathbf{F}$ is the function

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y},

also written $Q_x - P_y$ . It is the third component of the curl $\nabla \times (P, Q, 0)$ of the same field viewed in space. At each point $\mathbf{p}$ of $D$ it equals the circulation per unit area around small circles centered at $\mathbf{p}$ ,

\begin{aligned} &(Q_x - P_y)(\mathbf{p}) \\[0.4em] &\quad = \lim_{\varepsilon \to 0^+} \frac{1}{\pi \varepsilon^2} \oint_{C_\varepsilon} \mathbf{F} \cdot d\mathbf{r}, \end{aligned}

where $C_\varepsilon$ is the circle of radius $\varepsilon$ centered at $\mathbf{p}$ , traced counterclockwise.

In plain words The scalar curl measures how strongly the field turns around a point, as the circulation around a tiny loop divided by the area of the loop. It is not the length of the arrows. Parallel arrows can still have a nonzero curl when their length changes from side to side, across the direction in which they point, because a small loop then gets a stronger push along one side than along the opposite side.

Example For $\mathbf{F} = (-y, x)$ , the partial derivatives are $P_y = -1$ and $Q_x = 1$ , so the scalar curl is $1 - (-1) = 2$ at every point. The arrows get longer away from the origin, but the curl stays $2$ . Writing $P = y$ by mistake would give the false value $0$ . The conservative field $(y, x)$ of the previous lesson has scalar curl $1 - 1 = 0$ . The field $(0, x)$ has parallel vertical arrows, yet its scalar curl is $1 - 0 = 1$ . Its counterclockwise circulation around the square $[0, 1] \times [0, 1]$ is $1$ , all of it from the right side $x = 1$ , since the left side has $x = 0$ and the other two sides are perpendicular to the field.

Cancellation along shared edges

Definition Suppose a region $D$ is cut into finitely many subregions $D_1, \dots, D_n$ with piecewise smooth boundaries, each traced in its positive orientation, and let $\mathbf{F}$ be continuous on $D$ together with its boundary. Each cut between two neighboring subregions is traveled once in each direction, so its two contributions cancel, and

\begin{aligned} &\sum_{k=1}^{n} \oint_{\partial D_k} \mathbf{F} \cdot d\mathbf{r} \\[0.4em] &\quad = \oint_{\partial D} \mathbf{F} \cdot d\mathbf{r}. \end{aligned}

For a small rectangle with sides $\Delta x$ and $\Delta y$ , the right and left sides together contribute about $Q_x\,\Delta x\,\Delta y$ , and the bottom and top together about $-P_y\,\Delta x\,\Delta y$ . So the circulation around a small cell is approximately $(Q_x - P_y)\,\Delta x\,\Delta y$ , and adding over a grid of cells gives a Riemann sum for the double integral of the scalar curl.

In plain words Inside the region, every edge belongs to two cells that travel it in opposite directions, so it adds nothing to the total. Only the outer edges are traveled once, and together they make up the boundary of the whole region.

Example Cut the square $[0, 1] \times [0, 1]$ along the line $x = \tfrac{1}{2}$ and use $\mathbf{F} = (-y, x)$ . The left half has circulation $1$ , with $\tfrac{1}{2}$ from its right side and $\tfrac{1}{2}$ from its top. The right half also has circulation $1$ , with $1$ from its right side, $\tfrac{1}{2}$ from its top and $-\tfrac{1}{2}$ from its left side. The shared edge contributes $\tfrac{1}{2}$ going up for the left half and $-\tfrac{1}{2}$ going down for the right half, so it cancels. The total $1 + 1 = 2$ is the circulation around the whole square, which equals the scalar curl $2$ times the area $1$ .

Green’s theorem

Definition Let $D$ be a bounded region in the plane whose boundary $C$ is a piecewise smooth simple closed curve, traced in the positive, counterclockwise direction. If $P$ and $Q$ have continuous first partial derivatives on an open set containing $D$ and $C$ , then

\begin{aligned} &\oint_C (P\,dx + Q\,dy) \\[0.4em] &\quad = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \end{aligned}

The same formula holds for a region with finitely many holes when its whole boundary is traced in the positive orientation. Both conditions matter. A clockwise boundary gives the negative of the double integral, and a singularity inside $D$ can break the formula. The field

\left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right)

has scalar curl $0$ at every point except the origin, where it is undefined, yet its counterclockwise circulation around the unit circle is $2\pi$ , not $0$ .

In plain words The circulation around the edge of a region equals the total turning inside it, as long as the edge is closed, traveled with the region on the left, and the field is continuously differentiable on the whole region. Then a line integral around the boundary can be traded for a double integral over the region.

Example For $\oint_C (-y\,dx + x\,dy)$ around the unit circle counterclockwise, the functions $P = -y$ and $Q = x$ are polynomials, so their partial derivatives are continuous on the whole plane, and the circle is a smooth simple closed curve traced positively. The scalar curl is $Q_x - P_y = 2$ and the unit disk $D$ has area $\pi$ , so

\begin{aligned} &\oint_C (-y\,dx + x\,dy) \\[0.4em] &\quad = \iint_D 2\,dA = 2\pi, \end{aligned}

in agreement with the direct parametrization.

Check yourself

Let $C$ be the circle $x^2 + y^2 = 4$ , traced counterclockwise. Use Green’s theorem to evaluate $\oint_C (P\,dx + Q\,dy)$ for $P = x - 2y$ and $Q = 3x + y$ .

Show the answer

The functions $P$ and $Q$ are polynomials, so their partial derivatives are continuous everywhere, and $C$ is a positively oriented simple closed curve around the disk $D$ of radius $2$ . The scalar curl is $Q_x - P_y = 3 - (-2) = 5$ , and $D$ has area $4\pi$ , so the circulation is

\iint_D 5\,dA = 5 \cdot 4\pi = 20\pi.

Let $C$ be the boundary of the square $[0, 1] \times [0, 1]$ , traced clockwise, starting up the $y$ -axis from the origin. Evaluate $\oint_C (y^2\,dx + 3xy\,dy)$ .

Show the answer

Green’s theorem gives the counterclockwise value. The scalar curl is $Q_x - P_y = 3y - 2y = y$ , so the counterclockwise circulation is

\int_0^1 \int_0^1 y\,dx\,dy = \frac{1}{2}.

The curve is traced clockwise, so the answer is $-\tfrac{1}{2}$ . Leaving out the sign change is the usual mistake here.

As a direct check, the left side has $x = 0$ and $dx = 0$ and contributes $0$ . The top, from $(0, 1)$ to $(1, 1)$ , has $y = 1$ and contributes $\int_0^1 1\,dx = 1$ . The right side, from $(1, 1)$ down to $(1, 0)$ , contributes $\int_1^0 3y\,dy = -\tfrac{3}{2}$ , and the bottom has $y = 0$ and contributes $0$ . The total is $1 - \tfrac{3}{2} = -\tfrac{1}{2}$ .

Use Green’s theorem to find $\int_C (-y\,dx + x\,dy)$ , where $C$ is the upper half of the unit circle from $(1, 0)$ to $(-1, 0)$ .

Show the answer

The curve is not closed, so Green’s theorem does not apply to it directly. Close it with the segment $L$ from $(-1, 0)$ to $(1, 0)$ along the $x$ -axis. Together, $C$ and $L$ trace the boundary of the upper half disk $H$ counterclockwise, with $H$ on the left. Since the scalar curl is $2$ and $H$ has area $\tfrac{\pi}{2}$ ,

\begin{aligned} &\int_C \mathbf{F} \cdot d\mathbf{r} + \int_L \mathbf{F} \cdot d\mathbf{r} \\[0.4em] &\quad = \iint_H 2\,dA = \pi, \end{aligned}

where $\mathbf{F} = (-y, x)$ . On $L$ both $y = 0$ and $dy = 0$ , so the integrand is $0$ and the integral along $L$ is $0$ . Therefore the integral along the semicircle is $\pi$ . The parametrization $(\cos t, \sin t)$ , $0 \le t \le \pi$ , confirms this, since the integrand is $1$ on an interval of length $\pi$ .

Add the turning inside, find the circulation outside

Key terms

Circulation

Counterclockwise orientation

Scalar curl

Cancellation along shared edges

Green’s theorem

Check yourself

How this lesson connects

Builds on

Used later in