The plane vector field F = (2xy, x²) is:

Conservative, with potential x²y

Conservative, with potential 2x²y

Conservative only along straight-line paths

Vector Calculus — Gradient, Divergence, Curl and the Three Theorems

Three operators and three theorems, and the entire topic is the relationship between them. Grad turns a scalar into a vector and points in the direction of steepest increase; div turns a vector into a scalar and measures how much the field is spreading out at a point; curl turns a vector into a vector and measures how much it is circulating. Getting the input and output types right is not pedantry — it eliminates wrong options instantly, because the divergence of a vector field can never be a vector and the gradient of a scalar can never be a scalar. Two identities follow from the definitions and are worth knowing cold, because they appear as one-mark questions almost every year: curl(grad f) = 0 for any scalar f, and div(curl F) = 0 for any vector F. Both are consequences of mixed partial derivatives commuting, and both mean that the operators are not independent of each other. On top of that sit Green, Stokes and Gauss, which are the same statement in different dimensions: an integral over a region equals an integral over its boundary. Green relates a double integral over a plane area to a line integral around the curve enclosing it, Stokes relates a surface integral of curl to the line integral around the surface's rim, and Gauss relates a triple integral of divergence to the flux through the closed surface bounding the volume. So read the geometry, not the name: a closed surface with a volume inside means Gauss, a closed curve bounding a surface means Stokes or Green. And whenever a line integral is asked, check first whether curl F = 0 — a conservative field has a potential function, the integral becomes a difference of two potential values, and the path stops mattering entirely.

The Three Operators and What They Return

Input and output types, which decide half the options
Operator	Takes → returns	What it means, and a worked value
grad f = ∇f	scalar → vector	Direction of steepest increase. For f = x² + y² + z², ∇f = (2x, 2y, 2z), which at (1, 1, 1) is (2, 2, 2).
div F = ∇·F	vector → scalar	Net outflow per unit volume. For F = (x²y, y²z, z²x), div F = 2xy + 2yz + 2zx, which is 6 at (1, 1, 1).
curl F = ∇×F	vector → vector	Circulation per unit area. In the plane only the k-component survives: ∂F₂/∂x − ∂F₁/∂y.

🧠 The directional derivative is a dot product with a unit vector

The rate of change of f in the direction of a vector a is ∇f · (a/|a|). The unit-vector step is the whole difficulty: for f = x² + y² + z² at (1, 1, 1) along (1, 1, 1), ∇f = (2, 2, 2) and the unit vector is (1, 1, 1)/√3, so the answer is 6/√3 = 2√3 ≈ 3.464. Dotting with the un-normalised (1, 1, 1) gives 6, which is a distractor in every question of this shape. The maximum possible directional derivative at a point is |∇f| itself, achieved along ∇f, and that fact answers a whole family of questions on its own.

curl(grad f) = 0 for every scalar field f. A gradient field never circulates, which is the same as saying that a field with a potential is conservative.
div(curl F) = 0 for every vector field F. A curl field never has sources or sinks, so a question that offers a non-zero answer for div(curl F) can be settled without computing anything.
div(grad f) = ∇²f, the Laplacian. A function with ∇²f = 0 is called harmonic, and that is the same condition that appears in the complex-variables chapter and in the Laplace equation — the three topics meet here.

Line Integrals and the Conservative Shortcut

A line integral ∫_C F·dr is in general a parametrisation exercise: write x, y, z as functions of a parameter, substitute, and integrate over the parameter. But before doing any of that, test whether the field is conservative, because if it is, the work collapses to arithmetic.

Compute curl F. In the plane that is one subtraction: for F = (2xy, x²) it is ∂(x²)/∂x − ∂(2xy)/∂y = 2x − 2x = 0, so the field is conservative.
Find the potential φ with ∇φ = F. Here φ = x²y, since ∂φ/∂x = 2xy and ∂φ/∂y = x².
The integral is now φ(end) − φ(start), and the path is irrelevant. For φ = xyz from (0, 0, 0) to (1, 2, 3) the answer is 6 − 0 = 6, whatever curve joins the two points.

🎯 Why a closed path in a conservative field gives zero

If the integral is a difference of potentials at the two endpoints, then a closed path — where the endpoints coincide — must give zero. That single observation answers any question of the form "the work done by this field around this closed loop", provided curl F = 0, and it does so without a parametrisation. It also explains why a non-zero answer around a closed loop is a proof that the field is not conservative, which is sometimes the actual question.

Green, Stokes and Gauss — Choosing by Geometry

One statement, three dimensions
Theorem	Trades	Use it when the question shows
Green	A double integral over a plane region ↔ a line integral round its boundary curve	A closed curve in the xy-plane. The special case ∮(−y dx + x dy) = 2·Area is worth memorising: round the unit circle it is 2π.
Stokes	A surface integral of curl F ↔ a line integral of F round the rim of the surface	An open surface with a boundary curve — a hemisphere, a disc, a paraboloid cap. Any two surfaces sharing the same rim give the same answer.
Gauss (divergence)	A triple integral of div F ↔ the flux of F through the closed surface bounding the volume	A closed surface with a volume inside. For F = (x, y, z) over the unit sphere: div F = 3, so the flux is 3·(4π/3) = 4π.

⚠️ Gauss needs the surface closed, Stokes needs it open

The divergence theorem is false for an open surface — a hemisphere on its own is not the boundary of anything, and applying Gauss to it silently omits the flat disc that would close it. If a question gives a hemisphere and asks for flux, either close it with the disc and subtract the disc contribution, or integrate directly. Conversely Stokes needs a rim to integrate around, so it says nothing about a closed surface: on a closed surface the rim is empty and the surface integral of curl F is therefore zero, which is a neat result in its own right and occasionally the answer.

Key takeaways

Grad takes a scalar to a vector, div takes a vector to a scalar, curl takes a vector to a vector. Type-checking the options removes the wrong ones before any computation.
curl(grad f) = 0 and div(curl F) = 0 always. Both are one-mark questions and neither needs a calculation.
A directional derivative needs a unit vector: ∇f·(a/|a|). Forgetting to normalise is the standard error, and the un-normalised value is always among the options.
Test curl F before parametrising a line integral. If it vanishes, find the potential and take a difference of two numbers; a closed loop then gives zero.
Green, Stokes and Gauss all trade a region integral for a boundary integral. Pick by geometry: closed surface with volume means Gauss, curve bounding a surface means Stokes or Green.
The divergence theorem requires a closed surface. Applied to a hemisphere alone it silently drops the closing disc.
Two worth carrying: ∮(−y dx + x dy) = 2 × area, and the flux of (x, y, z) out of any closed surface is 3 × the enclosed volume.

Practice questions (9)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For F = (x²y, y²z, z²x), the divergence of F at the point (1, 1, 1) is:
1. 3
2. 6
3. 0
4. (2, 2, 2)
Show answer
Answer: B — 6
div F = ∂(x²y)/∂x + ∂(y²z)/∂y + ∂(z²x)/∂z = 2xy + 2yz + 2zx, which at (1, 1, 1) is 2 + 2 + 2 = 6. Option D can be discarded on type alone: the divergence of a vector field is a scalar, never a vector, and an option that is a triple is answering a question about the gradient or the curl instead. Option A is what you get by differentiating each component with respect to its own variable but forgetting the factor from the other variable — a slip worth catching by checking one term by hand.
For any twice-differentiable scalar field f, the value of curl(grad f) is:
1. The zero vector
2. The scalar zero
3. ∇²f
4. It depends on f
Show answer
Answer: A — The zero vector
The curl of any gradient is identically the zero vector, because each component of the result is a difference of two mixed second partial derivatives taken in opposite orders, and those are equal for a twice-differentiable function. Options A and B differ only in type, and the distinction is the point: curl returns a vector, so the answer is the zero vector, and an examiner who offers both is checking that you know what curl produces. Option C is div(grad f), the Laplacian — a different composition of the same two operators, and the one that is generally not zero.
The directional derivative of f = x² + y² + z² at (1, 1, 1) in the direction of the vector (1, 1, 1) is:
1. 6
2. 2√3
3. 2
4. 6√3
Show answer
Answer: B — 2√3
∇f = (2x, 2y, 2z) is (2, 2, 2) at the point. The direction must be a unit vector, so divide (1, 1, 1) by its magnitude √3. The directional derivative is then (2, 2, 2)·(1, 1, 1)/√3 = 6/√3 = 2√3 ≈ 3.464. Option A is precisely the answer obtained by dotting with the un-normalised direction vector, and it appears in every question of this shape — normalising is the step being tested. As a check, the largest directional derivative available at this point is |∇f| = 2√3, and the given direction happens to be exactly the direction of ∇f, so the answer had to be the maximum.
The flux of F = (x, y, z) outward through the closed surface of the unit sphere x² + y² + z² = 1 is:
1. 4π
2. 4π/3
3. 3
4. 12π
Show answer
Answer: A — 4π
Use the divergence theorem rather than a surface integral. div F = 1 + 1 + 1 = 3, a constant, so the flux is 3 × (volume enclosed) = 3 × (4π/3) = 4π. Option B is the volume of the unit sphere on its own, which is what you get by forgetting to multiply by the divergence; option C is the divergence on its own, forgetting the volume. The general result is worth carrying into the exam: the outward flux of the position field (x, y, z) through any closed surface is three times the volume it encloses, whatever the shape.
The line integral ∮_C (−y dx + x dy) taken counter-clockwise around the unit circle equals:
1. 0
2. π
3. 2π
4. 4π
Show answer
Answer: C — 2π
By Green’s theorem the integral equals ∫∫ (∂x/∂x − ∂(−y)/∂y) dA = ∫∫ (1 + 1) dA = 2 × area = 2π for the unit circle, whose area is π. So the answer is 2π. Option A is the answer for a conservative field, and the check is quick: here the plane curl is 1 − (−1) = 2 ≠ 0, so the field is not conservative and a closed loop need not give zero. Option B is the area itself, obtained by forgetting the factor 2. The identity ∮(−y dx + x dy) = 2 × area is worth memorising in its own right — it turns any "find the area enclosed by this curve" question into a line integral and back.
The plane vector field F = (2xy, x²) is:
1. Conservative, with potential x²y
2. Conservative, with potential 2x²y
3. Not conservative
4. Conservative only along straight-line paths
Show answer
Answer: A — Conservative, with potential x²y
The plane curl is ∂F₂/∂x − ∂F₁/∂y = ∂(x²)/∂x − ∂(2xy)/∂y = 2x − 2x = 0, so the field is conservative. Its potential must satisfy ∂φ/∂x = 2xy and ∂φ/∂y = x², and φ = x²y does both — checking both partial derivatives is the verification step, since option B satisfies neither. Option D describes something that cannot happen: conservativeness is a property of the field, not of a class of paths, and once it holds the line integral is path-independent for every path joining the same two points.
For any vector field F with continuous second derivatives, div(curl F) is:
1. Always zero
2. Zero only if F is conservative
3. Equal to ∇²F
4. Non-zero in general
Show answer
Answer: A — Always zero
div(curl F) = 0 always, for the same structural reason as curl(grad f) = 0: expanding it produces pairs of mixed third-order partial derivatives that cancel term by term. No property of F beyond smoothness is needed, so option B adds a condition that does no work — and note that a conservative field has curl F = 0 to begin with, which makes the divergence of it trivially zero, a special case rather than the reason. This identity is the mathematical statement behind the physical fact that a magnetic field, being a curl, has no monopoles.
F = ∇φ with φ = xyz. The value of ∫_C F·dr along any path from (0, 0, 0) to (1, 2, 3) is _____.
Numerical answer — type the value.
Show answer
Answer: 6
F is a gradient, so it is conservative and the integral is simply φ(end) − φ(start) = (1·2·3) − 0 = 6. The phrase "along any path" in the stem is the examiner confirming that path-independence applies, and it is also a hint: if the answer depended on the path, the question could not have been asked that way. No parametrisation, no curve, no integration — a two-line answer to a question that looks like a page of work. As a numerical-answer question it carries no negative marking, so it should never be left blank.
For any sufficiently smooth scalar field f and vector field F, which of the following are identically zero? (More than one option may be correct.)
1. grad(div F)
2. curl(grad f)
3. div(grad f)
4. div(curl F)
Show answer
Answer: B — curl(grad f); D — div(curl **F**)
The two vanishing identities are curl(grad f) = 0 and div(curl F) = 0, both because mixed second partial derivatives commute — options B and D. Option C is the Laplacian ∇²f, which is zero only for a harmonic function and not in general; option A is a perfectly ordinary vector field with no reason to vanish. The four options are deliberately the four compositions that type-check, and knowing which two are the identities is a one-mark fact that appears almost every year — here it is asked as an MSQ, where getting three of the four judgements right still scores nothing.