This deck covers conservative vector fields and potential functions, the cross-partial and curl tests, the Fundamental Theorem of Line Integrals and the path independence it gives, and Green's Theorem in both its circulation and flux forms, including finding area by Green's Theorem. It targets the traps of ignoring orientation and closure, claiming that a field is conservative on a domain with a hole, and dropping the extra function when rebuilding a potential.
Subject: Calculus III · 123 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
This week ties line integrals together with two big ideas: conservative fields and Green's Theorem.
1. Decide whether a vector field is conservative using the cross-partial test in the plane, or the curl test in space.
2. Rebuild a potential function by partial integration - without dropping the extra term.
3. Use the Fundamental Theorem of Line Integrals to evaluate a line integral from its endpoints only.
4. Apply Green's Theorem to turn a closed-curve line integral into a double integral - in both circulation and flux forms - and find area with it.
Warm-up
Discussion prompt
Before we open Week 14 - Conservative Fields & Green's Theorem: without looking back, what was the main idea of Week 13 - Vector Fields & Line Integrals, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers vector fields in the plane and in space, gradient fields, divergence and curl, and line integrals of both scalar fields and vector fields, the latter giving work. It targets the classic errors: dropping the speed factor in a scalar line integral, mishandling orientation, and mis-parameterizing a curve.
Concept
Start with a scalar function and take its gradient. The result is a vector field.
\[ \nabla f = \left\langle \frac{\partial f}{\partial x},\; \frac{\partial f}{\partial y} \right\rangle \]
Every scalar function gives you a field this way. The question this week: which fields come from a scalar function?
Counterexample
Discussion prompt
Start with a scalar function and take its gradient. The result is a vector field.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Every scalar function gives you a field this way. The question this week: which fields come from a scalar function?
Concept
A field is conservative when it is the gradient of some scalar function.
\[ \mathbf{F} = \nabla f \quad \text{for some scalar } f \]
conservative field — A vector field F for which there exists a scalar function f with F = grad f. That f is called a potential function for F.
Analogy
Discussion prompt
Explain What a conservative field is by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A field is conservative when it is the gradient of some scalar function.
Intuition
Picture the scalar function as the height of a landscape. The gradient field points uphill everywhere, steepest first.
A conservative field is exactly the kind of field that is 'downhill of some hill.' The hill is the potential.
In physics, the potential is potential energy and the field is force. Moving around a loop and returning home costs zero net energy - that is the whole point.
Explain it
Discussion prompt
Explain Think of it as a hill to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Picture the scalar function as the height of a landscape. The gradient field points uphill everywhere, steepest first.
Concept
A potential function is any scalar whose gradient is the field.
\[ \nabla f = \mathbf{F} \;\Longleftrightarrow\; f_x = P,\; f_y = Q \]
Here we write the plane field with components as below. Rebuilding this f is a main skill this week.
\[ \mathbf{F}(x,y) = \langle P(x,y),\, Q(x,y) \rangle \]
Ranking
Put in order
Put the moves of Worked example: confirm a given potential into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Treat y as a constant. The derivative of x^2 y is 2xy; the derivative of y^3 is 0.
Worked example
Show that the field below is conservative with the given potential.
\[ \mathbf{F} = \langle 2xy,\; x^2 + 3y^2 \rangle, \qquad f = x^2 y + y^3 \]
Compute the partial of f with respect to x
Why: Treat y as a constant. The derivative of x^2 y is 2xy; the derivative of y^3 is 0.
\[ f_x = 2xy \]
Compute the partial of f with respect to y
Why: Treat x as a constant. The derivative of x^2 y is x^2; the derivative of y^3 is 3y^2.
\[ f_y = x^2 + 3y^2 \]
Compare with the components of F
Why: The gradient of f matches P and Q exactly, so f is a potential for F.
\[ \nabla f = \langle 2xy,\; x^2+3y^2 \rangle = \mathbf{F} \]
Verify by re-reading both components
Why: Both partials reproduce F component by component, confirming F is conservative with potential f.
\[ f_x = P,\quad f_y = Q \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "confirm a given potential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both partials reproduce F component by component, confirming F is conservative with potential f.
Concept
You usually are not handed the potential. You need a quick test for whether one could exist.
On a nice (simply connected) domain, a plane field is conservative exactly when the partial of P by y equals the partial of Q by x.
\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \]
Intuition
If a potential exists, then P and Q are both partials of the same f.
\[ P = f_x, \qquad Q = f_y \]
Differentiate again and use Clairaut's theorem: the mixed second partials of a smooth f are equal.
\[ P_y = f_{xy} = f_{yx} = Q_x \]
So matching cross-partials is a necessary condition. On a simply connected domain it is also sufficient.
Step zero
Discussion prompt
Worked example: is this field conservative? — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify P and Q
Answer:
Worked example
Test the field below.
\[ \mathbf{F} = \langle 2xy + 3,\; x^2 - 2y \rangle \]
Identify P and Q
Why: P is the first component, Q the second.
\[ P = 2xy + 3, \qquad Q = x^2 - 2y \]
Differentiate P by y
Why: Hold x constant; the derivative of 2xy is 2x and of the constant 3 is 0.
\[ P_y = 2x \]
Differentiate Q by x
Why: Hold y constant; the derivative of x^2 is 2x and of -2y is 0.
\[ Q_x = 2x \]
Compare
Why: The cross-partials agree, and the domain is the whole plane (simply connected), so F is conservative.
\[ P_y = Q_x = 2x \;\Rightarrow\; \text{conservative} \]
Verify the conclusion is consistent
Why: A potential must exist; indeed f = x^2 y + 3x - y^2 has grad f equal to F, confirming the test.
\[ \nabla(x^2y + 3x - y^2) = \langle 2xy+3,\; x^2 - 2y\rangle \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "is this field conservative?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A potential must exist; indeed f = x^2 y + 3x - y^2 has grad f equal to F, confirming the test.
Fill the middle
Fill in the blanks
From Worked example: a field that fails the test — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle -y,\; x \rangle = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The derivative of -y with respect to y is -1.
Worked example
Test the rotation field below.
\[ \mathbf{F} = \langle -y,\; x \rangle \]
Identify P and Q
Why: Read off the components.
\[ P = -y, \qquad Q = x \]
Differentiate P by y
Why: The derivative of -y with respect to y is -1.
\[ P_y = -1 \]
Differentiate Q by x
Why: The derivative of x with respect to x is 1.
\[ Q_x = 1 \]
Compare
Why: The cross-partials disagree, so no potential can exist - the field is not conservative.
\[ P_y = -1 \ne 1 = Q_x \]
Verify with intuition
Why: This field circulates counterclockwise; going around a loop does net positive work, so it cannot come from a hill. Not conservative is consistent.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} \ne 0 \text{ for a loop } C \]
Picture it
Animation
Shows: Each line of the worked example "a field that fails the test", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: This field circulates counterclockwise; going around a loop does net positive work, so it cannot come from a hill. Not conservative is consistent.
Concept
The cross-partial test only guarantees conservative on a simply connected domain: one piece, with no holes punched in it.
simply connected — A region where every closed loop can be shrunk to a point without leaving the region. The whole plane qualifies; a plane with a point removed does not.
If the domain has a hole, matching cross-partials is not enough. The next trap shows exactly how this bites.
Definition probe
Sort into buckets
Every line below is part of the definition of conservative field or of simply connected — one or the other, never both. Put each where it belongs.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The wrong move: run the cross-partial test on the vortex field, see it pass, and declare the field conservative everywhere.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A direct quotient-rule computation gives the same result for P_y and Q_x away from the origin.
The right view: the field is undefined at the origin, so its domain is the plane with a hole - not simply connected. The test's guarantee does not apply.
Why: A direct quotient-rule computation gives the same result for P_y and Q_x away from the origin.
Trap
The wrong move: run the cross-partial test on the vortex field, see it pass, and declare the field conservative everywhere.
\[ \mathbf{F} = \left\langle \frac{-y}{x^2+y^2},\; \frac{x}{x^2+y^2} \right\rangle \]
Both cross-partials equal the same expression
Why: A direct quotient-rule computation gives the same result for P_y and Q_x away from the origin.
\[ P_y = Q_x = \frac{y^2 - x^2}{(x^2+y^2)^2} \]
Wrongly conclude conservative and expect a zero loop integral
Why: The test passed, so the tempting conclusion is a potential exists and every loop gives 0. This is false here.
\[ \text{claim: } \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 \;\; (\text{wrong}) \]
The right view: the field is undefined at the origin, so its domain is the plane with a hole - not simply connected. The test's guarantee does not apply.
Integrate around the unit circle directly
Why: Parametrize x = cos t, y = sin t. The integrand simplifies to dt, giving a nonzero loop integral.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} 1\, dt = 2\pi \ne 0 \]
Conclude: not conservative on the punctured plane
Why: A nonzero loop integral is impossible for a conservative field, so matching cross-partials was not enough - the hole matters.
\[ 2\pi \ne 0 \;\Rightarrow\; \text{not conservative} \]
Notation
Annotate
From Trap: conservative on a domain with a hole — read this one piece at a time. What is each part doing?
On: \( \mathbf{F} = \left\langle \frac{-y}{x^2+y^2},\; \frac{x}{x^2+y^2} \right\rangle \)
Concept
In space the same idea is packaged as the curl. A field on a simply connected region of space is conservative exactly when its curl is the zero vector.
\[ \operatorname{curl}\mathbf{F} = \nabla \times \mathbf{F} = \mathbf{0} \]
Writing the space field with three components, the curl is the vector below.
\[ \mathbf{F} = \langle P, Q, R\rangle \]
\[ \nabla\times\mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle \]
Intuition
Setting each curl component to zero gives three pairings - the space version of the single plane condition.
\[ P_y = Q_x, \quad P_z = R_x, \quad Q_z = R_y \]
Each pair is a mixed-partial match, just like the plane. If all three hold on a simply connected region, a potential exists.
Worked example
Is this space field conservative?
\[ \mathbf{F} = \langle 2xy,\; x^2 + z^2,\; 2yz \rangle \]
Label the components
Why: P, Q, R are the three entries.
\[ P = 2xy,\; Q = x^2+z^2,\; R = 2yz \]
First curl component: R_y minus Q_z
Why: R_y is 2z; Q_z is 2z; they cancel.
\[ R_y - Q_z = 2z - 2z = 0 \]
Second curl component: P_z minus R_x
Why: P has no z, so P_z is 0; R has no x, so R_x is 0.
\[ P_z - R_x = 0 - 0 = 0 \]
Third curl component: Q_x minus P_y
Why: Q_x is 2x; P_y is 2x; they cancel.
\[ Q_x - P_y = 2x - 2x = 0 \]
Verify all three vanish
Why: The curl is the zero vector on all of space, which is simply connected, so F is conservative.
\[ \nabla\times\mathbf{F} = \langle 0,0,0\rangle \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "curl test in space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The curl is the zero vector on all of space, which is simply connected, so F is conservative.
Concept
Once you know a field is conservative, you can recover a potential by partial integration - undoing each partial derivative one variable at a time.
Integrate P in x to get most of f, but the constant of integration can depend on the other variables. That leftover function is where students slip.
\[ f = \int P\, dx + g(y) \]
Fill the middle
Fill in the blanks
From Worked example: find a potential (plane) — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle 2xy,\; x^2 + 2y \rangle = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. P_y = 2x and Q_x = 2x agree on the plane, so a potential exists and the hunt is worthwhile.
Worked example
Find a potential for the field below.
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y \rangle \]
Confirm it is conservative first
Why: P_y = 2x and Q_x = 2x agree on the plane, so a potential exists and the hunt is worthwhile.
\[ P_y = 2x = Q_x \]
Integrate P with respect to x
Why: Undo the x-partial. The constant of integration may depend on y, so append g(y).
\[ f = \int 2xy\, dx = x^2 y + g(y) \]
Differentiate this f by y and match Q
Why: The y-partial of x^2 y is x^2; the unknown adds g'(y). Set the total equal to Q.
\[ f_y = x^2 + g'(y) = x^2 + 2y \]
Solve for g and integrate
Why: Cancel x^2 to get g'(y) = 2y, then integrate in y to recover g.
\[ g'(y) = 2y \;\Rightarrow\; g(y) = y^2 + C \]
Assemble the potential
Why: Combine the pieces into a single scalar function.
\[ f = x^2 y + y^2 + C \]
Verify by taking the gradient
Why: f_x = 2xy and f_y = x^2 + 2y reproduce F exactly, so the potential is correct.
\[ \nabla f = \langle 2xy,\; x^2+2y\rangle = \mathbf{F} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "find a potential (plane)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: f_x = 2xy and f_y = x^2 + 2y reproduce F exactly, so the potential is correct.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The wrong move: after integrating P in x, add only a plain constant instead of a function of the remaining variable.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Treating the integration constant as a number loses the y-only part of the potential.
The right move: the constant of integration is really a function of every variable you did not integrate over.
Why: Treating the integration constant as a number loses the y-only part of the potential.
Trap
The wrong move: after integrating P in x, add only a plain constant instead of a function of the remaining variable.
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y\rangle \]
Write f with a bare constant
Why: Treating the integration constant as a number loses the y-only part of the potential.
\[ f = x^2 y + C \;\;(\text{wrong}) \]
Check its y-partial
Why: This f gives f_y = x^2, which does not match Q = x^2 + 2y. The 2y term is missing.
\[ f_y = x^2 \ne x^2 + 2y \]
The right move: the constant of integration is really a function of every variable you did not integrate over.
Add g(y), then solve for it
Why: Matching f_y to Q forces g'(y) = 2y, so g(y) = y^2. The leftover carries the missing term.
\[ f = x^2 y + g(y), \quad g(y) = y^2 \]
Confirm the full gradient
Why: Now f = x^2 y + y^2 gives f_y = x^2 + 2y, matching Q. The extra function was essential.
\[ \nabla(x^2 y + y^2) = \langle 2xy,\; x^2+2y\rangle \;\checkmark \]
Translation
\( \nabla(x^2 y + y^2) = \langle 2xy,\; x^2+2y\rangle \;\checkmark \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Estimation
Predict first
Find a potential for the space field from the curl test.
Commit before you compute: what does Worked example: find a potential (space) come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the gradient
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. f = x^2 y + y z^2 gives f_x = 2xy, f_y = x^2 + z^2, f_z = 2yz, matching F exactly.
Worked example
Find a potential for the space field from the curl test.
\[ \mathbf{F} = \langle 2xy,\; x^2 + z^2,\; 2yz \rangle \]
Integrate P in x
Why: Undo the x-partial; the constant may depend on y and z, so append g(y,z).
\[ f = \int 2xy\, dx = x^2 y + g(y,z) \]
Match the y-partial to Q
Why: The y-partial of f is x^2 plus g_y; set it equal to Q = x^2 + z^2.
\[ f_y = x^2 + g_y = x^2 + z^2 \;\Rightarrow\; g_y = z^2 \]
Integrate g_y in y
Why: Undo the y-partial; the leftover can still depend on z, so append h(z).
\[ g = y z^2 + h(z), \qquad f = x^2 y + y z^2 + h(z) \]
Match the z-partial to R
Why: The z-partial of f is 2yz plus h'(z); set it equal to R = 2yz, forcing h' = 0.
\[ f_z = 2yz + h'(z) = 2yz \;\Rightarrow\; h(z) = C \]
Verify the gradient
Why: f = x^2 y + y z^2 gives f_x = 2xy, f_y = x^2 + z^2, f_z = 2yz, matching F exactly.
\[ \nabla f = \langle 2xy,\; x^2+z^2,\; 2yz\rangle = \mathbf{F} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "find a potential (space)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: f = x^2 y + y z^2 gives f_x = 2xy, f_y = x^2 + z^2, f_z = 2yz, matching F exactly.
Concept
Here is the payoff. If a field is the gradient of f, a line integral along any curve depends only on the endpoints.
\[ \int_C \nabla f \cdot d\mathbf{r} = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) \]
No parametrization, no arc length, no grinding. Evaluate the potential at the finish and subtract its value at the start.
Intuition
It is the same story as a hike: your net change in altitude depends only on where you start and where you finish, not on the trail you took.
path independence — For a conservative field, the line integral between two points is the same for every path connecting them. Only the endpoints matter.
This is exactly the single-variable Fundamental Theorem of Calculus, promoted to curves: integrate a derivative, evaluate the antiderivative at the ends.
Step zero
Discussion prompt
Worked example: evaluate by the FTLI — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Recall the potential
Answer:
Worked example
Evaluate the work of the field below from the point (1,1) to (2,3), along any path.
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y \rangle \]
Recall the potential
Why: We already built it: f = x^2 y + y^2 has gradient F, so the FTLI applies.
\[ f = x^2 y + y^2 \]
Evaluate f at the endpoint
Why: Plug in (2,3): 4 times 3 is 12, plus 3^2 = 9.
\[ f(2,3) = (4)(3) + 9 = 21 \]
Evaluate f at the start point
Why: Plug in (1,1): 1 times 1 is 1, plus 1^2 = 1.
\[ f(1,1) = (1)(1) + 1 = 2 \]
Subtract
Why: The FTLI gives finish minus start.
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = 21 - 2 = 19 \]
Verify path independence
Why: Any curve from (1,1) to (2,3) gives 19, because f is single-valued; the answer cannot depend on the route. Value 19 stands.
\[ f(2,3) - f(1,1) = 19 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "evaluate by the FTLI", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Any curve from (1,1) to (2,3) gives 19, because f is single-valued; the answer cannot depend on the route. Value 19 stands.
Concept
A closed curve ends where it starts. Feed that into the FTLI and the two endpoint values are identical.
\[ \oint_C \nabla f \cdot d\mathbf{r} = f(\text{start}) - f(\text{start}) = 0 \]
Circulate a conservative field around any loop and you get exactly zero. This is the loop test for conservative fields.
Missing information
Discussion prompt
Evaluate the circulation of the conservative field around the unit circle traversed once.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Starting at the point (1,0) and going around returns to (1,0); start and finish coincide.
Worked example
Evaluate the circulation of the conservative field around the unit circle traversed once.
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y \rangle, \qquad f = x^2 y + y^2 \]
Note the curve is closed
Why: Starting at the point (1,0) and going around returns to (1,0); start and finish coincide.
\[ \mathbf{r}(0) = \mathbf{r}(2\pi) = (1,0) \]
Apply the FTLI with equal endpoints
Why: Because the field is a gradient, the integral is f at the finish minus f at the start.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = f(1,0) - f(1,0) \]
Verify the result is zero
Why: The two f-values are identical, so their difference is 0 - as it must be for any loop of a conservative field.
\[ f(1,0) - f(1,0) = 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a conservative loop integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two f-values are identical, so their difference is 0 - as it must be for any loop of a conservative field.
Concept
For a field on a simply connected domain, these three statements are equivalent - each implies the others.
One: the field is conservative (it is a gradient).
Two: every line integral is path independent.
Three: every closed-loop integral is zero.
\[ \mathbf{F} = \nabla f \;\Longleftrightarrow\; \text{path indep.} \;\Longleftrightarrow\; \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 \]
Anomaly
Predict first
A student writes this, and it looks reasonable:
The wrong move: shortcut a line integral by 'just using the endpoints' before checking the field is conservative.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: For a straight path from the origin to (1,0), a start-minus-finish shortcut suggests a value that ignores the route.
The right move: test first. This field failed the cross-partial test, so it is not conservative and path independence does not hold.
Why: For a straight path from the origin to (1,0), a start-minus-finish shortcut suggests a value that ignores the route.
Trap
The wrong move: shortcut a line integral by 'just using the endpoints' before checking the field is conservative.
\[ \mathbf{F} = \langle -y,\; x \rangle \]
Pretend endpoints are enough
Why: For a straight path from the origin to (1,0), a start-minus-finish shortcut suggests a value that ignores the route.
\[ \text{guess: value depends only on ends} \;(\text{wrong}) \]
The right move: test first. This field failed the cross-partial test, so it is not conservative and path independence does not hold.
Different paths give different answers
Why: Along the straight segment from the origin to (1,0), y stays 0, so the work is 0; along a semicircular arc between the same points the work is nonzero.
\[ \int_{\text{line}} = 0, \qquad \int_{\text{arc}} \ne 0 \]
Conclude: endpoints are not enough here
Why: Two paths with the same endpoints disagree, proving the field is path dependent - you must actually parametrize and integrate.
\[ \text{not conservative} \Rightarrow \text{path matters} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Along the straight segment from the origin to (1,0), y stays 0, so the work is 0; along a semicircular arc between the same points the work is nonzero.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
For a straight path from the origin to (1,0), a start-minus-finish shortcut suggests a value that ignores the route.
Concept
What about fields that are not conservative? Green's Theorem gives a completely different shortcut: trade a closed-curve line integral for a double integral over the region it encloses.
It only works when the curve is a simple closed loop bounding a region, and it turns a one-dimensional integral into a two-dimensional one.
\[ \oint_C \; \longleftrightarrow \; \iint_D \]
Concept
Green's Theorem assumes the boundary is traced with positive orientation: counterclockwise, so the region stays on your left.
positive orientation — The counterclockwise direction around a region's outer boundary. Walking this way, the enclosed region is always on your left.
Orientation is the single most common source of sign errors here. Fix it before you compute anything.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of conservative field, simply connected, path independence, positive orientation as Week 14 - Conservative Fields & Green's Theorem uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
The circulation (tangential) form converts the loop integral of P dx plus Q dy into a double integral of one particular combination of partials.
\[ \oint_C P\, dx + Q\, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
That integrand is the two-dimensional curl, the scalar version of the k-component of the curl in space.
Intuition
Imagine tiling the region with microscopic squares. Each has its own tiny circulation, measured by the local curl.
Where two squares share an edge, their boundary contributions run opposite ways and cancel. Only the outer edges survive.
Add up all the tiny curls over the area, and what is left equals the circulation around the whole outer boundary. That is Green's Theorem.
Worked example
Compute the circulation of the field below around the circle of radius 2, counterclockwise.
\[ \oint_C -y\, dx + x\, dy, \qquad C: x^2 + y^2 = 4 \]
Identify P and Q
Why: Read them from the integrand P dx + Q dy.
\[ P = -y, \qquad Q = x \]
Form the Green integrand
Why: Compute Q_x minus P_y: the x-partial of x is 1, the y-partial of -y is -1.
\[ Q_x - P_y = 1 - (-1) = 2 \]
Integrate the constant 2 over the disk
Why: A constant integrand pulls out; the double integral of 1 over the disk is its area, pi times radius squared.
\[ \iint_D 2\, dA = 2\,(\pi\cdot 2^2) = 8\pi \]
Verify by direct parametrization
Why: With x = 2 cos t, y = 2 sin t the integrand becomes 4 dt; integrating from 0 to 2 pi gives 8 pi, matching Green.
\[ \int_0^{2\pi} 4\, dt = 8\pi \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "circulation around a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: With x = 2 cos t, y = 2 sin t the integrand becomes 4 dt; integrating from 0 to 2 pi gives 8 pi, matching Green.
Ranking
Put in order
Put the moves of Worked example: circulation over a rectangle into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Q_x is 2x; P_y is x; subtract to get the two-dimensional curl.
Worked example
Compute the circulation below around the boundary of the rectangle, counterclockwise.
\[ \oint_C xy\, dx + x^2\, dy, \qquad 0\le x\le 2,\; 0\le y\le 1 \]
Identify P and Q
Why: Read them from the integrand.
\[ P = xy, \qquad Q = x^2 \]
Form the Green integrand
Why: Q_x is 2x; P_y is x; subtract to get the two-dimensional curl.
\[ Q_x - P_y = 2x - x = x \]
Set up the double integral
Why: Integrate x over the rectangle; do the inner x-integral first, then the outer y-integral.
\[ \int_0^1 \int_0^2 x\, dx\, dy \]
Do the inner integral
Why: The x-integral of x from 0 to 2 is x^2 over 2 evaluated at 2, which is 2.
\[ \int_0^2 x\, dx = \frac{x^2}{2}\Big|_0^2 = 2 \]
Do the outer integral
Why: Integrate the constant 2 over y from 0 to 1.
\[ \int_0^1 2\, dy = 2 \]
Verify by reversing the order
Why: Integrating x over y first gives x, then x from 0 to 2 gives 2 - the same value, confirming the double integral.
\[ \int_0^2 \int_0^1 x\, dy\, dx = \int_0^2 x\, dx = 2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "circulation over a rectangle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Integrating x over y first gives x, then x from 0 to 2 gives 2 - the same value, confirming the double integral.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The wrong move: apply Green's formula directly when the curve is traced clockwise, forgetting the orientation assumption.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Plugging in as if counterclockwise gives a positive value, but the actual clockwise loop has the opposite sign.
The right move: Green's Theorem needs counterclockwise. A clockwise loop is the reverse, so its integral is the negative.
Why: Plugging in as if counterclockwise gives a positive value, but the actual clockwise loop has the opposite sign.
Trap
The wrong move: apply Green's formula directly when the curve is traced clockwise, forgetting the orientation assumption.
\[ \oint_{C_{\text{cw}}} -y\, dx + x\, dy, \quad C: x^2+y^2 = 4 \]
Blindly quote the double integral
Why: Plugging in as if counterclockwise gives a positive value, but the actual clockwise loop has the opposite sign.
\[ \text{claim } = 8\pi \;\;(\text{wrong sign}) \]
The right move: Green's Theorem needs counterclockwise. A clockwise loop is the reverse, so its integral is the negative.
Insert the sign flip
Why: Reversing orientation negates a vector line integral, so the clockwise value is minus the counterclockwise value.
\[ \oint_{C_{\text{cw}}} = -\iint_D (Q_x - P_y)\, dA \]
Get the correct answer
Why: The counterclockwise value was 8 pi, so the clockwise circulation is its negative.
\[ \oint_{C_{\text{cw}}} = -8\pi \;\checkmark \]
Notation
Annotate
From Trap: a clockwise curve — read this one piece at a time. What is each part doing?
On: \( \oint_{C_{\text{cw}}} = -8\pi \;\checkmark \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
The wrong move: use Green's Theorem on a single arc that does not close up into a loop.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: There is no enclosed region D for an open curve, so the double integral on the right has nothing to integrate over.
The right move: Green's Theorem requires a simple closed curve. Either close the path with an extra segment, or integrate the open arc directly.
Why: There is no enclosed region D for an open curve, so the double integral on the right has nothing to integrate over.
Trap
The wrong move: use Green's Theorem on a single arc that does not close up into a loop.
Try to convert an open arc
Why: There is no enclosed region D for an open curve, so the double integral on the right has nothing to integrate over.
\[ \int_{\text{arc}} P\,dx + Q\,dy \;\ne\; \iint_D (\cdots)\, dA \]
The right move: Green's Theorem requires a simple closed curve. Either close the path with an extra segment, or integrate the open arc directly.
Close the loop, then subtract
Why: Add a return segment L so the arc plus L is closed; apply Green to that loop, then subtract the integral over L to isolate the arc.
\[ \int_{\text{arc}} = \oint_{\text{arc}+L} - \int_L \]
Or just parametrize the arc
Why: When closing is awkward, fall back to a direct parametrization of the open arc - Green's Theorem simply does not apply to it.
\[ \int_{\text{arc}} \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
The second form measures flux: how much the field flows outward across the boundary. It converts the flux line integral into a double integral of the divergence.
\[ \oint_C \mathbf{F}\cdot \mathbf{n}\, ds = \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA \]
The integrand here is the divergence, adding the partials instead of subtracting them - the flux twin of the circulation form.
Intuition
Divergence at a point measures whether the field is a source (spreading out) or a sink (draining in) right there.
Add up all the little sources and sinks over the region, and the net is exactly how much fluid crosses the outer boundary. That is the flux form.
Fill the middle
Fill in the blanks
From Worked example: flux across a circle — finish the line. Write what belongs on the right of the equals sign before you look.
3 \cdot (2\pi\cdot 3) = 18\pi \;\checkmark
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The x-partial of x is 1; the y-partial of y is 1; the divergence is their sum.
Worked example
Find the outward flux of the radial field across the circle of radius 3.
\[ \mathbf{F} = \langle x,\; y\rangle, \qquad C: x^2+y^2 = 9 \]
Compute the divergence
Why: The x-partial of x is 1; the y-partial of y is 1; the divergence is their sum.
\[ P_x + Q_y = 1 + 1 = 2 \]
Integrate over the disk
Why: A constant integrand times the disk's area; the area is pi times radius squared, radius 3.
\[ \iint_D 2\, dA = 2\,(\pi\cdot 3^2) = 18\pi \]
Verify along the boundary
Why: On the circle the field points straight out with magnitude 3, so the flux is 3 times the circumference 2 pi times 3, giving 18 pi - matching the divergence integral.
\[ 3 \cdot (2\pi\cdot 3) = 18\pi \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "flux across a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: On the circle the field points straight out with magnitude 3, so the flux is 3 times the circumference 2 pi times 3, giving 18 pi - matching the divergence integral.
Concept
Here is a slick trick. Choose P and Q so the Green integrand becomes 1; then the double integral is just the area of the region.
The symmetric choice below makes Q_x minus P_y equal to 1, so the loop integral computes area directly.
\[ A = \frac{1}{2}\oint_C x\, dy - y\, dx \]
Estimation
Predict first
Use Green's Theorem to find the area enclosed by the ellipse below.
Commit before you compute: what does Worked example: area of an ellipse come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the known formula
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The area of an ellipse with semi-axes a and b is pi a b, matching our result exactly.
Worked example
Use Green's Theorem to find the area enclosed by the ellipse below.
\[ x = a\cos t, \quad y = b\sin t, \quad 0 \le t \le 2\pi \]
Differentiate the parametrization
Why: Take dx and dy in terms of dt for the area integrand.
\[ dx = -a\sin t\, dt, \qquad dy = b\cos t\, dt \]
Form x dy minus y dx
Why: Substitute and combine; the cross terms collapse using cosine squared plus sine squared.
\[ x\, dy - y\, dx = ab(\cos^2 t + \sin^2 t)\, dt = ab\, dt \]
Integrate and halve
Why: The integrand is the constant ab; integrate over a full turn and multiply by one half.
\[ A = \frac{1}{2}\int_0^{2\pi} ab\, dt = \frac{1}{2}(ab)(2\pi) = \pi ab \]
Verify against the known formula
Why: The area of an ellipse with semi-axes a and b is pi a b, matching our result exactly.
\[ A = \pi ab \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "area of an ellipse", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The area of an ellipse with semi-axes a and b is pi a b, matching our result exactly.
Concept
The two forms of Green's Theorem answer two different questions about the same boundary.
Circulation measures how much the field flows along the boundary - swirling with the curve.
Flux measures how much the field flows across the boundary - outward through it.
Circulation pairs with the curl (a subtraction of partials); flux pairs with the divergence (an addition). Same region, two viewpoints.
Concept
A vector line integral appears in several equivalent dresses. Recognizing them keeps Green's Theorem readable.
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_C \mathbf{F}\cdot \mathbf{T}\, ds = \int_C P\, dx + Q\, dy \]
The middle form uses the unit tangent and arc length; the right form is the one Green's circulation formula transforms.
Concept
For a plane field these two scalars drive everything this week.
\[ \text{2D curl} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \]
\[ \operatorname{div}\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \]
Conservative means the curl is zero; the divergence has nothing to do with being conservative - it drives the flux form instead.
Step zero
Discussion prompt
Worked example: curl and divergence of a field — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read off P and Q
Answer:
Worked example
Compute the two-dimensional curl and the divergence of the field below.
\[ \mathbf{F} = \langle x^2 y,\; x y^2 \rangle \]
Read off P and Q
Why: Identify the two components.
\[ P = x^2 y, \qquad Q = x y^2 \]
Compute the 2D curl
Why: Q_x is y^2; P_y is x^2; subtract for the curl.
\[ Q_x - P_y = y^2 - x^2 \]
Compute the divergence
Why: P_x is 2xy; Q_y is 2xy; add for the divergence.
\[ P_x + Q_y = 2xy + 2xy = 4xy \]
Verify the conservative reading
Why: The curl is not identically zero, so this field is not conservative - consistent with y^2 - x^2 being nonzero off the diagonal.
\[ y^2 - x^2 \ne 0 \;\Rightarrow\; \text{not conservative} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "curl and divergence of a field", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The curl is not identically zero, so this field is not conservative - consistent with y^2 - x^2 being nonzero off the diagonal.
Hypothesis
Predict first
Worked example: potential with a trig term is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Run the cross-partial test
Why: P_y is -sin y; Q_x is -sin y; they agree, so a potential exists.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find a potential for the field below.
\[ \mathbf{F} = \langle 2x + \cos y,\; -x\sin y + 3y^2 \rangle \]
Run the cross-partial test
Why: P_y is -sin y; Q_x is -sin y; they agree, so a potential exists.
\[ P_y = -\sin y = Q_x \]
Integrate P in x
Why: Undo the x-partial; append g(y) for the y-dependent constant.
\[ f = \int (2x + \cos y)\, dx = x^2 + x\cos y + g(y) \]
Match the y-partial to Q
Why: The y-partial of f is -x sin y plus g'(y); set equal to Q.
\[ -x\sin y + g'(y) = -x\sin y + 3y^2 \;\Rightarrow\; g'(y) = 3y^2 \]
Integrate to find g
Why: The antiderivative of 3y^2 is y^3.
\[ g(y) = y^3 + C, \qquad f = x^2 + x\cos y + y^3 \]
Verify the gradient
Why: f_x = 2x + cos y and f_y = -x sin y + 3y^2 reproduce F exactly.
\[ \nabla f = \langle 2x+\cos y,\; -x\sin y + 3y^2\rangle = \mathbf{F} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "potential with a trig term", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: f_x = 2x + cos y and f_y = -x sin y + 3y^2 reproduce F exactly.
Worked example
Evaluate the work of that same field from the origin to the point (2,0).
\[ f = x^2 + x\cos y + y^3 \]
Evaluate f at the endpoint
Why: At (2,0): 2^2 is 4, plus 2 times cosine of 0 which is 2, plus 0.
\[ f(2,0) = 4 + 2\cos 0 + 0 = 4 + 2 = 6 \]
Evaluate f at the start
Why: At (0,0): every term is 0.
\[ f(0,0) = 0 \]
Subtract
Why: The FTLI gives the endpoint value minus the start value.
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = 6 - 0 = 6 \]
Verify cosine of zero
Why: Cosine of 0 equals 1, so the endpoint value is 4 plus 2, confirming the work is 6 for any path between the points.
\[ \cos 0 = 1 \;\Rightarrow\; f(2,0) = 6 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "FTLI with the trig potential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Cosine of 0 equals 1, so the endpoint value is 4 plus 2, confirming the work is 6 for any path between the points.
Missing information
Discussion prompt
Confirm the conservative field gives the same work along two different paths from the origin to (1,1).
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Parametrize x = t, y = t. Then F dotted with the velocity gives 3t^2 + 2t.
Worked example
Confirm the conservative field gives the same work along two different paths from the origin to (1,1).
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y \rangle \]
Path A: the straight diagonal
Why: Parametrize x = t, y = t. Then F dotted with the velocity gives 3t^2 + 2t.
\[ \int_0^1 (3t^2 + 2t)\, dt = 1 + 1 = 2 \]
Path B: across then up
Why: Along y = 0 the integrand is 0; along x = 1 the integrand is 1 + 2y.
\[ \int_0^1 (1 + 2y)\, dy = 1 + 1 = 2 \]
Compare the two routes
Why: Both paths give 2, as path independence promises for a conservative field.
\[ \int_A = \int_B = 2 \]
Verify with the potential
Why: The potential f = x^2 y + y^2 gives f(1,1) - f(0,0) = 2 - 0 = 2, agreeing with both direct computations.
\[ f(1,1) - f(0,0) = 2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "path independence checked two ways", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The potential f = x^2 y + y^2 gives f(1,1) - f(0,0) = 2 - 0 = 2, agreeing with both direct computations.
Concept
Before you invoke Green's Theorem, confirm four things.
One: the curve is simple - it does not cross itself.
Two: the curve is closed - it returns to its start.
Three: the orientation is counterclockwise (region on the left).
Four: P and Q have continuous partials on the whole enclosed region - no singular points inside.
Estimation
Predict first
Compute the circulation below around the triangle with vertices at the origin, (1,0), and (0,1), counterclockwise.
Commit before you compute: what does Worked example: circulation over a triangle come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the area another way
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Setting up the double integral, the inner y-integral runs from 0 to 1 - x, and integrating 1 - x from 0 to 1 gives one half.
Worked example
Compute the circulation below around the triangle with vertices at the origin, (1,0), and (0,1), counterclockwise.
\[ \oint_C y\, dx + 2x\, dy \]
Identify P and Q
Why: Read them off the integrand.
\[ P = y, \qquad Q = 2x \]
Form the Green integrand
Why: Q_x is 2; P_y is 1; their difference is a constant.
\[ Q_x - P_y = 2 - 1 = 1 \]
Integrate 1 over the triangle
Why: The double integral of 1 over any region is its area; this right triangle has legs of length 1.
\[ \iint_D 1\, dA = \text{area} = \tfrac{1}{2}(1)(1) = \tfrac{1}{2} \]
Verify the area another way
Why: Setting up the double integral, the inner y-integral runs from 0 to 1 - x, and integrating 1 - x from 0 to 1 gives one half.
\[ \int_0^1 (1-x)\, dx = 1 - \tfrac{1}{2} = \tfrac{1}{2} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "circulation over a triangle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting up the double integral, the inner y-integral runs from 0 to 1 - x, and integrating 1 - x from 0 to 1 gives one half.
Step zero
Discussion prompt
Worked example: circulation over a square — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify P and Q
Answer:
Worked example
Compute the circulation below around the unit square, counterclockwise.
\[ \oint_C (x + y^2)\, dx + (2x - y)\, dy, \qquad 0\le x\le 1,\; 0\le y\le 1 \]
Identify P and Q
Why: Read them off the integrand.
\[ P = x + y^2, \qquad Q = 2x - y \]
Form the Green integrand
Why: Q_x is 2; P_y is 2y; subtract.
\[ Q_x - P_y = 2 - 2y \]
Integrate over the square
Why: Integrate 2 - 2y over x from 0 to 1 (no x-dependence), then over y.
\[ \int_0^1 \int_0^1 (2 - 2y)\, dx\, dy = \int_0^1 (2 - 2y)\, dy \]
Finish the y-integral
Why: The antiderivative is 2y - y^2, evaluated from 0 to 1 gives 2 - 1.
\[ \big[\,2y - y^2\,\big]_0^1 = 2 - 1 = 1 \]
Verify by reversing the order
Why: Integrating 2 - 2y over y first gives 1 for each x, and integrating 1 over x from 0 to 1 gives 1 - the same total.
\[ \int_0^1 \Big(\int_0^1 (2-2y)\, dy\Big) dx = \int_0^1 1\, dx = 1 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "circulation over a square", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Integrating 2 - 2y over y first gives 1 for each x, and integrating 1 over x from 0 to 1 gives 1 - the same total.
Intuition
Notice the shared shape: the Fundamental Theorem of Line Integrals and Green's Theorem both say the same thing.
Integrate a derivative of something over a region, and you can instead evaluate the original thing on the boundary of that region.
For the FTLI the region is a curve and the boundary is its two endpoints. For Green's Theorem the region is an area and the boundary is its enclosing loop. Later, Stokes and the Divergence Theorem extend the very same pattern.
Concept
Three shortcuts, three triggers.
If the field is conservative and you know the endpoints, use the Fundamental Theorem of Line Integrals.
If the curve is a closed loop and the field is not conservative, use Green's Theorem to drop into a double integral.
If neither fits - an open, non-conservative arc - parametrize and integrate directly.
Explain it
Discussion prompt
Explain Which tool for which job to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
If the field is conservative and you know the endpoints, use the Fundamental Theorem of Line Integrals.
Intuition
Think of the field as a force and the line integral as the work it does on a particle you push along the curve.
For a conservative force - like gravity - carrying an object around any loop and back to the start returns all the energy you spent. Net work is zero.
For a non-conservative field - like a whirlpool - a loop can do net work, which is exactly why its closed integral is nonzero and why Green's Theorem is worth having.
Analogy
Discussion prompt
Explain Work around a loop, physically by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of the field as a force and the line integral as the work it does on a particle you push along the curve.
Concept
The definition runs one way for free: if a field is written as the gradient of some f, it is conservative by definition, no test needed.
\[ \mathbf{F} = \nabla f \;\Rightarrow\; \mathbf{F} \text{ is conservative} \]
The test earns its keep in the other direction: given raw components, it tells you whether such an f could exist before you spend effort hunting for it.
Counterexample
Discussion prompt
The definition runs one way for free: if a field is written as the gradient of some f, it is conservative by definition, no test needed.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The test earns its keep in the other direction: given raw components, it tells you whether such an f could exist before you spend effort hunting for it.
Ranking
Put in order
Put the moves of Worked example: total differential viewpoint into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The combination f_x dx + f_y dy is exactly df, the total change in f along the path.
Worked example
See why the FTLI works by writing the integrand as a total differential.
\[ \mathbf{F} = \nabla f, \qquad \mathbf{F}\cdot d\mathbf{r} = f_x\, dx + f_y\, dy \]
Recognize the total differential
Why: The combination f_x dx + f_y dy is exactly df, the total change in f along the path.
\[ \mathbf{F}\cdot d\mathbf{r} = df \]
Integrate df along the curve
Why: Integrating a total differential from start to finish gives the net change in f - the single-variable Fundamental Theorem, applied along a curve.
\[ \int_C df = f(\text{end}) - f(\text{start}) \]
Verify it reduces to the FTLI
Why: This is precisely the Fundamental Theorem of Line Integrals, so the endpoint-only rule is not a coincidence - it is the chain rule in reverse.
\[ \int_C \nabla f\cdot d\mathbf{r} = f(\text{end}) - f(\text{start}) \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "total differential viewpoint", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: This is precisely the Fundamental Theorem of Line Integrals, so the endpoint-only rule is not a coincidence - it is the chain rule in reverse.
Pattern
1. Run the appropriate test
Why: In the plane check whether the partial of P by y equals the partial of Q by x; in space check that the curl is zero.
2. Confirm the domain is simply connected
Why: The test only guarantees conservative when there are no holes; watch for points where the field blows up.
3. Integrate one component to start f
Why: Integrate P in x, and always append a function of the remaining variables as the constant of integration.
4. Differentiate and match the next component
Why: Take the partial of your f in the next variable, set it equal to that component, and solve for the leftover function.
5. Verify by taking the full gradient
Why: The gradient of your f must reproduce every component of F before you trust it.
Pattern
Ask: is the field conservative?
Why: Run the cross-partial or curl test first - the answer decides everything downstream.
Conservative and given endpoints: use the FTLI
Why: Build a potential and subtract its endpoint values; the path is irrelevant.
Closed loop, not conservative: use Green's Theorem
Why: Convert to a double integral of the curl (circulation) or divergence (flux) over the enclosed region.
Neither: integrate directly
Why: Parametrize the curve and evaluate; no shortcut applies to an open, non-conservative path.
Pattern
1. Check simple, closed, counterclockwise
Why: If the loop is clockwise, attach a minus sign; if it is open, Green's Theorem does not apply.
2. Pick the form
Why: Use circulation (curl, a subtraction) for flow along the curve; use flux (divergence, an addition) for flow across it.
3. Build the integrand
Why: Compute Q_x minus P_y for circulation, or P_x plus Q_y for flux.
4. Set up limits from the region
Why: Sketch D and choose the order of integration that makes the limits simplest.
5. Evaluate the double integral
Why: The double integral is the answer; a constant integrand is just that constant times the area.
Real world
Discussion prompt
Outside this lesson: where does Week 14 - Conservative Fields & Green's Theorem actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: applying Green's Theorem is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers conservative vector fields and potential functions, the cross-partial and curl tests, the Fundamental Theorem of Line Integrals and the path independence it gives, and Green's Theorem in both its circulation and flux forms, including finding area by Green's Theorem. It targets the traps of ignoring orientation and closure, claiming that a field is conservative on a domain with a hole, and dropping the extra function when rebuilding a potential.
Check
Apply the cross-partial test.
\[ \mathbf{F} = \langle 3x^2 y,\; x^3 \rangle \]
Check your understanding
Is F conservative on the plane?
Answer: A
Why: The partial of P = 3x^2 y with respect to y is 3x^2, and the partial of Q = x^3 with respect to x is 3x^2. They match on the whole plane, which is simply connected, so F is conservative.
Elimination
Eliminate the wrong options
What is the value of the line integral from (0,0) to (2,1)?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: By the FTLI the value is f at the endpoint minus f at the start. f(2,1) = (4)(1) + 1 = 5 and f(0,0) = 0, so the integral is 5 - 0 = 5.
Check
The field below has potential f as shown. Evaluate the line integral from the origin to (2,1).
\[ \mathbf{F} = \langle 2xy,\; x^2 + 2y\rangle, \qquad f = x^2 y + y^2 \]
Check your understanding
What is the value of the line integral from (0,0) to (2,1)?
Answer: A
Why: By the FTLI the value is f at the endpoint minus f at the start. f(2,1) = (4)(1) + 1 = 5 and f(0,0) = 0, so the integral is 5 - 0 = 5.
Check
Use Green's Theorem on the unit square, traversed counterclockwise.
\[ \oint_C x^2\, dx + xy\, dy, \qquad 0\le x\le 1,\; 0\le y\le 1 \]
Check your understanding
What is the circulation around the square?
Answer: A
Why: Here Q_x - P_y = y - 0 = y. The double integral of y over the unit square is the integral of y from 0 to 1, which equals one half.
Prediction
Predict first
Why does this loop integral compute the enclosed area?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Because the Green integrand for P = -y, Q = x is Q_x - P_y = 2, and one half of that is 1, so the double integral is the area.
Why: With P = -y and Q = x, the circulation integrand Q_x - P_y equals 1 - (-1) = 2. The one-half factor turns the double integral of 2 into the double integral of 1, which is exactly the area of the region.
Check
Recall the area formula from Green's Theorem.
\[ A = \frac{1}{2}\oint_C x\, dy - y\, dx \]
Check your understanding
Why does this loop integral compute the enclosed area?
Answer: A
Why: With P = -y and Q = x, the circulation integrand Q_x - P_y equals 1 - (-1) = 2. The one-half factor turns the double integral of 2 into the double integral of 1, which is exactly the area of the region.
Elimination
Eliminate the wrong options
What is the value of the same integral taken clockwise?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Reversing the orientation of a vector line integral negates its value. Green's Theorem assumes counterclockwise, so the clockwise version is minus the counterclockwise value, giving negative 10.
Check
A counterclockwise loop integral over a region equals 10. Now traverse the same curve clockwise.
Check your understanding
What is the value of the same integral taken clockwise?
Answer: A
Why: Reversing the orientation of a vector line integral negates its value. Green's Theorem assumes counterclockwise, so the clockwise version is minus the counterclockwise value, giving negative 10.
Check
You must evaluate the work of a conservative field along an open curve between two known points.
Check your understanding
Which method is the right first choice?
Answer: A
Why: The field is conservative and the endpoints are known, so the FTLI applies: find a potential f and compute f at the finish minus f at the start, ignoring the path entirely.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: test and build a potential · Pattern: choosing your shortcut · Pattern: applying Green's Theorem · Recall: a gradient field · What a conservative field is. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now decide whether a field is conservative, rebuild its potential, and pick the fastest way to evaluate a line integral.
Conservative test: in the plane, check that the partial of P by y equals the partial of Q by x; in space, check that the curl is zero - and only on a simply connected domain.
Potential: integrate one component, then differentiate and match the rest, never forgetting the extra function of the other variables.
FTLI: for a conservative field, the line integral is just the potential evaluated at the endpoints; around any loop it is zero.
Green's Theorem: turn a counterclockwise closed-curve integral into a double integral - curl for circulation, divergence for flux - and use it to find area.
| Situation | Best tool |
|---|---|
| Conservative field, known endpoints | Fundamental Theorem of Line Integrals |
| Closed loop, not conservative | Green's Theorem (circulation or flux) |
| Need an enclosed area | Green's area formula |
| Open, non-conservative arc | Direct parametrization |
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