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A circulation round a closed curve rewritten as a double integral of $Q_x - P_y$ over the region inside, with the orientation, the smoothness and the holes that the statement depends on.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Green's theorem with its hypotheses, use it to replace a circulation round a polygon by a double integral, compute an area from a line integral round its outline, handle a region with a hole in it, and say exactly why the theorem fails when the field has a missing point inside the curve.
Line integrals round closed curves, double integrals over general regions, and the fact from two lessons back that the third component of a curl measures circulation per unit area. Green's theorem is that last sentence integrated: a local measurement at every point, added up, equals a global one round the edge.
A curve is simple when it does not cross itself and closed when it ends where it began. Its orientation is positive when the region it encloses stays on the left as you walk — counterclockwise for an ordinary loop. A region is multiply connected when it has holes; the inner boundary of such a region is traversed clockwise when the whole boundary is taken positively. $\oint_C$ denotes the integral round a closed curve.
The theorem. Let $C$ be a simple closed piecewise smooth curve, traversed counterclockwise, enclosing a region $D$; let $\mathbf{F} = \langle P, Q \rangle$ have continuous partial derivatives on an open set containing $D$. Then
$$\oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA.$$
The left side is a line integral round the edge; the right side is a double integral over the inside. Either may be computed to get the other, and which is easier depends entirely on the problem.
Every hypothesis is load-bearing. Simple rules out a curve that crosses itself, which would enclose two regions with opposite orientations. Closed is what makes an inside exist. Counterclockwise fixes the sign; the other way round gives the negative. And the field must be smooth on $D$, not merely on $C$: a single missing point inside is enough to break it, which is exactly what the vortex of the last lesson does.
Area from a line integral. Choosing $P = 0$, $Q = x$ makes the integrand $1$, so
$$\text{area}(D) = \oint_C x\,dy = -\oint_C y\,dx = \tfrac12 \oint_C (x\,dy - y\,dx).$$
A planimeter is a mechanical device that computes exactly this, and the shoelace formula for the area of a polygon is the last of these three applied to a closed chain of segments.
Regions with holes. If $D$ lies between an outer curve $C_1$ and an inner curve $C_2$, then with both taken counterclockwise
$$\iint_D (Q_x - P_y)\,dA = \oint_{C_1} \mathbf{F} \cdot d\mathbf{r} - \oint_{C_2} \mathbf{F} \cdot d\mathbf{r}.$$
Another way: picture
Tile the region with tiny squares and walk round each one counterclockwise. Every internal edge is walked twice, once in each direction, and cancels; only the outer edges survive. So the sum of all the little circulations is the circulation round the boundary — and each little circulation is $(Q_x - P_y)$ times the area of its square. That is the whole proof, and the same picture returns for Stokes and for the divergence theorem.
Another way: steps
To use the theorem:
The theorem is an equality, so it is always legitimate to compute either side. In practice one of them is far easier, and recognising which is most of the skill.
Go from the line integral to the double integral when the boundary is made of several pieces — a triangle needs three parametrisations, a square four — and especially when $Q_x - P_y$ turns out to be constant. Three line integrals collapse into one multiplication.
Go the other way when the region is awkward and the boundary is simple, and above all when computing an area. The whole point of the area formulas above is that they let a machine tracing an outline report the area enclosed without knowing anything about the interior.
And notice what the theorem says about conservative fields. If $P_y = Q_x$ everywhere on a region with no holes, the right side is zero for every loop in it, so the circulation vanishes round every loop — which is the cross-partials test of the last lesson, now with a proof rather than a plausible argument. Green's theorem is where that hypothesis about holes finally earns its place.
Ignoring the orientation. A clockwise loop gives the negative. The answer's sign is part of the answer, and a problem that says counterclockwise has told you what it wants.
Subtracting in the wrong order. The integrand is $Q_x - P_y$: the second component differentiated in the first variable, minus the first in the second. Reversing it negates every answer.
Checking the field only along the curve. This is the one that produces confidently wrong answers rather than obviously wrong ones. The vortex field is perfectly well behaved on a circle round the origin, and the theorem still does not apply.
Forgetting the inner boundary. On a region with a hole the boundary has two pieces and both appear. Green, Stokes and the divergence theorem each hold on a region of a particular shape, with a field defined everywhere inside it and a boundary oriented in a particular way. A theorem quoted where one of those fails has not been applied; it has been guessed with, and the answer it gives can be wrong by exactly the amount the missing hypothesis was carrying.
The theorem's field hypothesis is about $D$, the inside, and every instinct pulls the other way: the integral being computed is written along $C$, the parametrisation is of $C$, and the field is evaluated at points of $C$. It is entirely natural to check that the field behaves there and conclude that all is well.
The vortex field is the standing counterexample, and it is not rare. Any field of the form something divided by distance from a point has this shape: the electric field of a line charge, the magnetic field round a current, the velocity field of an idealised tornado. Each is beautifully smooth on any circle that misses the axis, and each breaks Green's theorem on any loop that encloses it.
The habit that prevents it: before writing the double integral, ask where the integrand $Q_x - P_y$ fails to exist, and then ask whether any of those points is inside the curve. If one is, the region-with-a-hole version is the version to use.
Find $\oint_C (y\,dx + 3x\,dy)$ counterclockwise round the triangle with corners $(0,0)$, $(4,0)$, $(0,4)$. Here $P = y$ and $Q = 3x$.
A closed curve made of three segments.
$Q_x - P_y = 3 - 1 = 2$, a constant, so the double integral is $2$ times the area of the triangle.
A constant integrand needs no limits.
The area is $\tfrac12 \cdot 4 \cdot 4 = 8$, so the circulation is $16$. Directly it would have been three parametrisations and three integrals, with three chances to lose a sign.
The theorem traded three integrals for one multiplication.
Take the rectangle $0 \le x \le 5$, $0 \le y \le 3$ and compute $\oint_C x\,dy$ counterclockwise.
Choosing $P = 0$, $Q = x$ makes the integrand one.
On the bottom and top, $dy = 0$, so those sides contribute nothing. On the right side $x = 5$ and $y$ runs from $0$ to $3$, contributing $15$; on the left $x = 0$, contributing nothing.
Only one side of the four contributes.
So the integral is $15$, which is the area of the rectangle. The interior was never visited: tracing the outline was enough, and that is what makes this formula the basis of a planimeter and of the shoelace formula for polygons.
An area computed from its boundary alone.
Find $\oint_C (x^2\,dx + xy\,dy)$ counterclockwise round the square $0 \le x \le 1$, $0 \le y \le 1$. Here $P = x^2$ and $Q = xy$.
Check first: simple, closed, counterclockwise, field smooth everywhere.
$Q_x = y$ and $P_y = 0$, so the integrand is $y$ — not constant this time, so the double integral has to be set up properly.
$\int_0^1 \int_0^1 y\,dy\,dx = \int_0^1 \tfrac12\,dx = \tfrac12$. The answer is the average height of the integrand times the area, which is what a double integral always is — and the four line integrals it replaced would each have needed its own substitution.
Put the stages of applying Green's theorem in order, for a loop of length $2$.
Number the steps in order (write the number in the box):
For $\mathbf{F} = \langle 6y, 7x \rangle$ round the rectangle $0 \le x \le 6$, $0 \le y \le 5$ counterclockwise, fill in the table.
| Value | |
|---|---|
| The partial of the second component in the first variable | |
| The partial of the first component in the second variable | |
| The difference of those two | |
| The area of the rectangle | |
| The circulation round the boundary |
Find the counterclockwise circulation of $\mathbf{F} = \langle 6y, 3x \rangle$ round the triangle with corners $(0,0)$, $(6, 0)$ and $(0, 6)$.
Answer:
A student applies Green's theorem to the vortex field round a circle of radius $8$ centred at the origin, gets $0$, and is told the circulation is $2\pi$. What went wrong?
Put the stages of applying Green's theorem in order, for a loop of length $4$.
Number the steps in order (write the number in the box):
On the region between an outer square of side $4$ and an inner square of area $4$ removed from inside it, $Q_x - P_y = 4$ everywhere. The counterclockwise circulation round the inner square is $4$. What is it round the outer square?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $\mathbf{F} = \langle 6y, 3x \rangle$ round the rectangle $0 \le x \le 5$, $0 \le y \le 3$ counterclockwise, fill in the table.
| Value | |
|---|---|
| The partial of the second component in the first variable | |
| The partial of the first component in the second variable | |
| The difference of those two | |
| The area of the rectangle | |
| The circulation round the boundary |
You can convert between a loop integral and a double integral in either direction, choose whichever side is easier, and name the hypothesis that fails when a field has a singularity inside. Next: the same idea one dimension up, where the loop bounds a surface rather than a region of the plane.
10. Your turn: a circulation with a non-constant integrand, step 3