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Vector calculus

Gradient, divergence, curl and the integral theorems

Vector calculus extends differentiation and integration to fields: the gradient of a scalar field and the divergence and curl of a vector field measure how a field changes from point to point, and the theorems of Green, Stokes and Gauss convert line, surface and volume integrals into one another. The working skills are three: computing $\nabla ϕ$ , $\nabla \cdot F$ or $\nabla \times F$ at a point, parametrising a line or surface integral, and choosing the theorem (with the right orientation) that turns a hard integral into an easy one.

23 previous-year questions · 23 solved · 7 practice questions

Vector algebra: products, angles and projections

Conventions used throughout. Vectors are bold ( $A$ ). $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the $x$ , $y$ , $z$ axes of a right-handed Cartesian system; the hat on $\hat{j}$ separates it from the imaginary unit $j$ , which does not appear in this topic. A vector is written $A = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ , with magnitude $∣ A ∣ = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$ . The position vector of the point $(x, y, z)$ is $r = x \hat{i} + y \hat{j} + z \hat{k}$ , and $r = ∣ r ∣$ .

DOT (SCALAR) PRODUCT

A \cdot B = ∣ A ∣ ∣ B ∣ cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}

A scalar; $θ \in [0, π]$ is the angle between the vectors. Commutative. For non-zero vectors, $A \cdot B = 0$ means perpendicular. $\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ .

ANGLE, COMPONENT AND PROJECTION

cos θ = \frac{A \cdot B}{∣ A ∣ ∣ B ∣}, comp_{B} A = \frac{A \cdot B}{∣ B ∣}, proj_{B} A = \frac{A \cdot B}{∣ B ∣ ^{2}} B

The scalar component of $A$ along $B$ is signed (negative when $θ > 9 0^{\circ}$ ); the vector projection lies along $\pm B$ . The direction cosines of a non-zero $A$ are $A_{x} /∣ A ∣$ , $A_{y} /∣ A ∣$ , $A_{z} /∣ A ∣$ , and their squares add to 1.

CROSS (VECTOR) PRODUCT

A \times B = \hat{i} A_{x} B_{x} \hat{j} A_{y} B_{y} \hat{k} A_{z} B_{z}, ∣ A \times B ∣ = ∣ A ∣ ∣ B ∣ sin θ

A vector perpendicular to both $A$ and $B$ , direction by the right-hand rule. Anti-commutative: $B \times A = - A \times B$ . $∣ A \times B ∣$ is the area of the parallelogram on $A$ and $B$ . For non-zero vectors, $A \times B = 0$ means parallel. $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

LAGRANGE'S IDENTITY

∣ A \times B ∣^{2} = ∣ A ∣^{2} ∣ B ∣^{2} - (A \cdot B)^{2}

Follows from $sin^{2} θ = 1 - cos^{2} θ$ ; gives $∣ A \times B ∣$ when only magnitudes and the dot product are known.

SCALAR TRIPLE PRODUCT

[A B C] = A \cdot (B \times C) = A_{x} B_{x} C_{x} A_{y} B_{y} C_{y} A_{z} B_{z} C_{z}

Unchanged by a cyclic shift, $A \cdot (B \times C) = B \cdot (C \times A) = C \cdot (A \times B)$ ; changes sign when two vectors are swapped; dot and cross may be interchanged. $∣ [A B C] ∣$ is the volume of the parallelepiped on the three vectors, and it is zero exactly when they are coplanar (linearly dependent).

VECTOR TRIPLE PRODUCT

A \times (B \times C) = (A \cdot C) B - (A \cdot B) C

The result lies in the plane of $B$ and $C$ . The cross product is not associative: $(A \times B) \times C = (A \cdot C) B - (B \cdot C) A$ , which differs in general.

EXAM TIP

Geometry shortcuts: the area of the triangle $P QR$ is $\frac{1}{2} ∣ P Q \times P R ∣$ ; the volume of the tetrahedron with edges $a, b, c$ from one vertex is $\frac{1}{6} ∣ [a b c] ∣$ ; four points are coplanar exactly when the triple product of the three edge vectors drawn from one of them is zero.

Gradient, directional derivative and normals

The del operator is $\nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}$ . Applied to a differentiable scalar field $ϕ (x, y, z)$ it gives the gradient, a vector field. At each point where $\nabla ϕ \neq = 0$ , $\nabla ϕ$ points in the direction in which $ϕ$ increases fastest, its magnitude is that fastest rate, and it is perpendicular to the level surface $ϕ = c$ through the point. These three facts are all that directional-derivative and normal-vector questions need.

GRADIENT

\nabla ϕ = \frac{\partial ϕ}{\partial x} \hat{i} + \frac{\partial ϕ}{\partial y} \hat{j} + \frac{\partial ϕ}{\partial z} \hat{k}

Defined for a differentiable scalar field. Linear, with $\nabla (ϕ ψ) = ϕ \nabla ψ + ψ \nabla ϕ$ and $d ϕ = \nabla ϕ \cdot d r$ .

DIRECTIONAL DERIVATIVE

D_{\overset{a}{^}} ϕ = \nabla ϕ \cdot \overset{a}{^}, \overset{a}{^} = \frac{a}{∣ a ∣}

Rate of change of $ϕ$ per unit distance in the direction of $a$ . The direction must be a unit vector, so divide by $∣ a ∣$ first. 'From $P$ towards the point $Q$ ' means $a = P Q$ . 'Along the tangent to the curve $r (t)$ at $t = t_{0}$ ' means $a = r^{'} (t_{0})$ ; 'along the normal to the surface $ψ = c$ ' means $a = \nabla ψ$ at the point.

MAXIMUM AND MINIMUM RATE OF CHANGE

\overset{a}{^} max D_{\overset{a}{^}} ϕ = ∣\nabla ϕ ∣ along \frac{\nabla ϕ}{∣\nabla ϕ ∣}, \overset{a}{^} min D_{\overset{a}{^}} ϕ = - ∣\nabla ϕ ∣

Because $D_{\overset{a}{^}} ϕ = ∣\nabla ϕ ∣ cos θ$ ( $θ$ = angle between $\overset{a}{^}$ and $\nabla ϕ$ ), the rate is largest along the gradient, most negative opposite to it, and zero along any direction tangent to the level surface.

UNIT NORMAL TO A SURFACE

\overset{n}{^} = \pm \frac{\nabla ϕ}{∣\nabla ϕ ∣} on the surface ϕ (x, y, z) = c

Valid where $\nabla ϕ \neq = 0$ . The sign picks a side; for a closed surface (a sphere, say) the outward normal points away from the enclosed region. For $z = g (x, y)$ take $ϕ = z - g (x, y)$ , so $\nabla ϕ = - g_{x} \hat{i} - g_{y} \hat{j} + \hat{k}$ , which points upward.

TANGENT PLANE AND NORMAL LINE

ϕ_{x} (x - x_{0}) + ϕ_{y} (y - y_{0}) + ϕ_{z} (z - z_{0}) = 0, \frac{x - x _{0}}{ϕ _{x}} = \frac{y - y _{0}}{ϕ _{y}} = \frac{z - z _{0}}{ϕ _{z}}

Tangent plane and normal line to $ϕ = c$ at $P (x_{0}, y_{0}, z_{0})$ , with the partial derivatives evaluated at $P$ and $\nabla ϕ (P) \neq = 0$ . The plane is $\nabla ϕ (P) \cdot (r - r_{0}) = 0$ . If one partial derivative is zero at $P$ , the matching numerator is zero along the line (for example $ϕ_{z} = 0$ gives $z = z_{0}$ ), and the other two ratios are equated.

ANGLE BETWEEN TWO SURFACES

cos θ = \frac{\nabla ϕ _{1} \cdot \nabla ϕ _{2}}{∣\nabla ϕ _{1} ∣ ∣\nabla ϕ _{2} ∣}

The angle between the surfaces $ϕ_{1} = c_{1}$ and $ϕ_{2} = c_{2}$ at a common point is the angle between their normals there, both gradients evaluated at that point. Each normal is fixed only up to sign, so the angle between the surfaces is usually quoted as the acute one: take the absolute value of the numerator. They cut orthogonally when $\nabla ϕ_{1} \cdot \nabla ϕ_{2} = 0$ .

GRADIENTS OF RADIAL FUNCTIONS

\nabla r = \frac{r}{r} = \overset{r}{^}, \nabla f (r) = f^{'} (r) \overset{r}{^}, \nabla r^{n} = n r^{n - 2} r

Valid for $r \neq = 0$ . Examples: $\nabla (r^{2}) = 2 r$ and $\nabla (1/ r) = - r / r^{3}$ .

WORKED EXAMPLE

For $ϕ (x, y, z) = x^{2} y + y z^{2}$ , find at $P (1, 2, - 1)$ : (a) the directional derivative in the direction of $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ ; (b) the maximum directional derivative and its direction; (c) the unit normal to the surface $ϕ = 4$ at $P$ .

Solution

$\nabla ϕ = 2 x y \hat{i} + (x^{2} + z^{2}) \hat{j} + 2 y z \hat{k}$ .
At $P (1, 2, - 1)$ : $\nabla ϕ = 4 \hat{i} + 2 \hat{j} - 4 \hat{k}$ , and $∣\nabla ϕ ∣ = 16 + 4 + 16 = 6$ .
(a) $∣ a ∣ = 4 + 1 + 4 = 3$ , so $\overset{a}{^} = \frac{1}{3} (2 \hat{i} - \hat{j} + 2 \hat{k})$ and $D_{\overset{a}{^}} ϕ = \frac{1}{3} (8 - 2 - 8) = - \frac{2}{3}$ . The minus sign means $ϕ$ decreases in that direction.
(b) The maximum rate is $∣\nabla ϕ ∣ = 6$ , along $\frac{1}{6} (4 \hat{i} + 2 \hat{j} - 4 \hat{k}) = \frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k})$ .
(c) $P$ lies on the surface, since $ϕ (1, 2, - 1) = 2 + 2 = 4$ . The unit normal is $\overset{n}{^} = \pm \frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k})$ .

Answer: (a) $- 2/3$ ; (b) $6$ , along $\frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k})$ ; (c) $\overset{n}{^} = \pm \frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k})$ .

EXAM TIP

A directional derivative can never exceed $∣\nabla ϕ ∣$ in magnitude, so an answer larger than that means the direction vector was not normalised. The direction of fastest decrease is $- \nabla ϕ$ ; the directions of zero change are the ones perpendicular to $\nabla ϕ$ .

Divergence, curl and conservative fields

For a vector field $F = F_{1} \hat{i} + F_{2} \hat{j} + F_{3} \hat{k}$ with differentiable components, the del operator gives two derivatives. The divergence $\nabla \cdot F$ is a scalar: the net outward flux per unit volume, in the limit of a small volume shrinking to the point; it is positive at a source and negative at a sink. The curl $\nabla \times F$ is a vector: its component along a unit vector $\overset{n}{^}$ is the circulation per unit area around a small loop in the plane perpendicular to $\overset{n}{^}$ , traversed by the right-hand rule about $\overset{n}{^}$ , in the limit as the loop shrinks to the point; so it measures local rotation. For a rigid body rotating with constant angular velocity $ω$ about an axis through the origin, the velocity is $v = ω \times r$ and $\nabla \times v = 2 ω$ .

DIVERGENCE

\nabla \cdot F = \frac{\partial F _{1}}{\partial x} + \frac{\partial F _{2}}{\partial y} + \frac{\partial F _{3}}{\partial z}

A scalar field. $\nabla \cdot r = 3$ ; a constant field has zero divergence.

CURL

\nabla \times F = \hat{i} \frac{\partial}{\partial x} F_{1} \hat{j} \frac{\partial}{\partial y} F_{2} \hat{k} \frac{\partial}{\partial z} F_{3} = (\frac{\partial F _{3}}{\partial y} - \frac{\partial F _{2}}{\partial z}) \hat{i} + (\frac{\partial F _{1}}{\partial z} - \frac{\partial F _{3}}{\partial x}) \hat{j} + (\frac{\partial F _{2}}{\partial x} - \frac{\partial F _{1}}{\partial y}) \hat{k}

A vector field. Expand the determinant along the first row; the $\hat{j}$ term, written here in its cyclic form, is where sign slips happen. $\nabla \times r = 0$ .

Field type	Test	Consequence
Solenoidal	$\nabla \cdot F = 0$ throughout the region	Zero net flux out of every closed surface whose enclosed volume lies in the region; on all of $R^{3}$ , $F = \nabla \times A$ for some vector potential $A$
Irrotational	$\nabla \times F = 0$ throughout the region	Zero circulation around every closed curve that bounds a surface lying in the region; on a simply connected region, $F = \nabla ϕ$
Conservative	$F = \nabla ϕ$ for some scalar $ϕ$	$\int_{A}^{B} F \cdot d r = ϕ (B) - ϕ (A)$ on every path; $\oint_{C} F \cdot d r = 0$
Both solenoidal and irrotational	$\nabla \cdot F = 0$ and $\nabla \times F = 0$	On a simply connected region, $F = \nabla ϕ$ with $\nabla^{2} ϕ = 0$ ( $ϕ$ harmonic)

CONSERVATIVE FIELD: EQUIVALENT CONDITIONS

F = \nabla ϕ ⟺ \oint_{C} F \cdot d r = 0 for every closed C ⟺ \int_{A}^{B} F \cdot d r depends only on A and B

Equivalent for a continuous $F$ on any connected open region. If $F$ also has continuous first partial derivatives, each implies $\nabla \times F = 0$ . The converse, curl zero implies conservative, needs $F$ to have continuous first partial derivatives on a simply connected region (every closed curve in it can be shrunk to a point without leaving it).

FUNDAMENTAL THEOREM FOR LINE INTEGRALS

\int_{A}^{B} \nabla ϕ \cdot d r = ϕ (B) - ϕ (A)

For any piecewise smooth path from $A$ to $B$ on which $ϕ$ is continuously differentiable. Sign convention: here a potential means $F = \nabla ϕ$ ; electromagnetics writes $E = - \nabla V$ , which flips the sign of the potential but not the test.

EXACTNESS TEST IN THE PLANE

P d x + Q d y = d ϕ ⟺ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

On a simply connected region with $P$ , $Q$ continuously differentiable. This is the $\hat{k}$ component of the curl test for $F = P \hat{i} + Q \hat{j}$ .

Finding a scalar potential

Check $\nabla \times F = 0$ , all three components. If any component is non-zero, no potential exists.
Integrate $F_{1}$ with respect to $x$ , holding $y$ and $z$ fixed: $ϕ = \int F_{1} d x + g (y, z)$ .
Differentiate this $ϕ$ with respect to $y$ , set it equal to $F_{2}$ , solve for $\partial g / \partial y$ and integrate: $g = \dots + h (z)$ .
Differentiate with respect to $z$ , set it equal to $F_{3}$ , and find $h (z)$ up to an additive constant.
Check that $\nabla ϕ = F$ . The work from $A$ to $B$ is $ϕ (B) - ϕ (A)$ ; the constant cancels.

WORKED EXAMPLE

Find the constants $a$ and $b$ for which $F = (2 x y + z^{2}) \hat{i} + (x^{2} + 3 z) \hat{j} + (a x z + b y) \hat{k}$ is irrotational. For these values find a scalar potential, and the work done by $F$ along any path from $(0, 0, 0)$ to $(1, 2, 1)$ .

Solution

$\hat{i}$ component of the curl: $\frac{\partial}{\partial y} (a x z + b y) - \frac{\partial}{\partial z} (x^{2} + 3 z) = b - 3 = 0$ , so $b = 3$ .
$\hat{j}$ component: $\frac{\partial}{\partial z} (2 x y + z^{2}) - \frac{\partial}{\partial x} (a x z + b y) = 2 z - a z$ , zero for all $z$ only if $a = 2$ .
$\hat{k}$ component: $\frac{\partial}{\partial x} (x^{2} + 3 z) - \frac{\partial}{\partial y} (2 x y + z^{2}) = 2 x - 2 x = 0$ already.
Potential: $ϕ = \int (2 x y + z^{2}) d x = x^{2} y + x z^{2} + g (y, z)$ . Then $ϕ_{y} = x^{2} + g_{y} = x^{2} + 3 z$ gives $g = 3 y z + h (z)$ , and $ϕ_{z} = 2 x z + 3 y + h^{'} (z) = 2 x z + 3 y$ gives $h$ = constant.
So $ϕ = x^{2} y + x z^{2} + 3 y z + C$ . The field is smooth and irrotational on all of $R^{3}$ , which is simply connected, so the work is path independent: $W = ϕ (1, 2, 1) - ϕ (0, 0, 0) = (2 + 1 + 6) - 0 = 9$ .

Answer: $a = 2$ , $b = 3$ , $ϕ = x^{2} y + x z^{2} + 3 y z + C$ , work $= 9$ .

EXAM TIP

Zero curl is not enough on a region with a hole. $F = \frac{- y i ^ + x j ^}{x ^{2} + y ^{2}}$ has zero curl everywhere except on the $z$ -axis, where it is undefined, yet its circulation counter-clockwise around the unit circle is $\int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = 2 π \neq = 0$ . A region that encircles the $z$ -axis is not simply connected, so no single-valued potential exists on it.

Vector identities and the Laplacian

Identities replace differentiation. They hold for scalar fields $ϕ, ψ$ and vector fields $A, B, F$ with continuous second partial derivatives; the two 'zero' identities rest on the equality of mixed partial derivatives.

THE TWO ZERO IDENTITIES

\nabla \times (\nabla ϕ) = 0, \nabla \cdot (\nabla \times F) = 0

Every gradient field is irrotational and every curl field is solenoidal. So a field with non-zero divergence cannot be the curl of anything, and a field with non-zero curl cannot be a gradient.

PRODUCT RULES

\nabla \cdot (ϕ F) \nabla \times (ϕ F) = ϕ (\nabla \cdot F) + F \cdot \nabla ϕ = ϕ (\nabla \times F) + \nabla ϕ \times F

The order in $\nabla ϕ \times F$ matters: swapping the factors flips the sign.

DIVERGENCE OF A CROSS PRODUCT

\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B)

One consequence: $\nabla ϕ \times \nabla ψ$ is always solenoidal, because both curls on the right are zero.

GRADIENT OF A DOT PRODUCT, CURL OF A CROSS PRODUCT

\nabla (A \cdot B) \nabla \times (A \times B) = (A \cdot \nabla) B + (B \cdot \nabla) A + A \times (\nabla \times B) + B \times (\nabla \times A) = A (\nabla \cdot B) - B (\nabla \cdot A) + (B \cdot \nabla) A - (A \cdot \nabla) B

Here $A \cdot \nabla = A_{x} \frac{\partial}{\partial x} + A_{y} \frac{\partial}{\partial y} + A_{z} \frac{\partial}{\partial z}$ is a scalar operator, applied to each Cartesian component of the field that follows it; it is not the divergence $\nabla \cdot A$ . With $B = A$ the first identity gives $\nabla (\frac{1}{2} ∣ A ∣^{2}) = (A \cdot \nabla) A + A \times (\nabla \times A)$ .

LAPLACIAN

\nabla^{2} ϕ = \nabla \cdot (\nabla ϕ) = \frac{\partial ^{2} ϕ}{\partial x ^{2}} + \frac{\partial ^{2} ϕ}{\partial y ^{2}} + \frac{\partial ^{2} ϕ}{\partial z ^{2}}

A scalar. $ϕ$ is harmonic when $\nabla^{2} ϕ = 0$ (Laplace's equation). If $F = \nabla ϕ$ , then $\nabla \cdot F = \nabla^{2} ϕ$ .

CURL OF A CURL

\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F

Here $\nabla^{2} F = \nabla^{2} F_{1} \hat{i} + \nabla^{2} F_{2} \hat{j} + \nabla^{2} F_{3} \hat{k}$ , taking the Laplacian component by component; this componentwise rule holds for Cartesian components only. For a solenoidal field it reduces to $\nabla \times (\nabla \times F) = - \nabla^{2} F$ .

RADIAL FIELDS

\nabla \cdot (r^{n} r) = (n + 3) r^{n}, \nabla \times (f (r) r) = 0, \nabla^{2} (r^{n}) = n (n + 1) r^{n - 2}

Valid for $r \neq = 0$ . Special cases: $\nabla \cdot r = 3$ ; $\nabla \cdot (r / r^{3}) = 0$ and $\nabla^{2} (1/ r) = 0$ away from the origin.

LAPLACIAN OF A RADIAL FUNCTION

\nabla^{2} f (r) = f^{''} (r) + \frac{2}{r} f^{'} (r) = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{df}{d r})

Valid for $r \neq = 0$ , for a twice-differentiable function $f$ of the distance $r$ from the origin. With $f = r^{n}$ it gives $n (n - 1) r^{n - 2} + 2 n r^{n - 2} = n (n + 1) r^{n - 2}$ , the result above. The functions of $r$ alone that satisfy Laplace's equation for $r \neq = 0$ are exactly $c_{1} + c_{2} / r$ .

DIVERGENCE OF A RADIAL FIELD (SPHERICAL COORDINATES)

\nabla \cdot (F (r) \overset{r}{^}) = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} F (r))

For a field that points radially and depends only on $r$ (the distance from the origin), $r \neq = 0$ . The cylindrical analogue for $F (ρ) \overset{ρ}{^}$ , with $ρ$ the distance from the $z$ -axis, is $\frac{1}{ρ} \frac{d}{d ρ} (ρF (ρ))$ for $ρ \neq = 0$ .

GRADIENT AND DIVERGENCE IN CYLINDRICAL AND SPHERICAL COORDINATES

cylindrical (ρ, φ, z) : spherical (r, θ, φ) : \nabla f = \frac{\partial f}{\partial ρ} \overset{ρ}{^} + \frac{1}{ρ} \frac{\partial f}{\partial φ} \overset{φ}{^} + \frac{\partial f}{\partial z} \hat{k} \nabla \cdot F = \frac{1}{ρ} \frac{\partial ( ρ F _{ρ} )}{\partial ρ} + \frac{1}{ρ} \frac{\partial F _{φ}}{\partial φ} + \frac{\partial F _{z}}{\partial z} \nabla f = \frac{\partial f}{\partial r} \overset{r}{^} + \frac{1}{r} \frac{\partial f}{\partial θ} \hat{θ} + \frac{1}{r sin θ} \frac{\partial f}{\partial φ} \overset{φ}{^} \nabla \cdot F = \frac{1}{r ^{2}} \frac{\partial ( r ^{2} F _{r} )}{\partial r} + \frac{1}{r sin θ} \frac{\partial ( sin θ F _{θ} )}{\partial θ} + \frac{1}{r sin θ} \frac{\partial F _{φ}}{\partial φ}

$ρ$ is the distance from the $z$ -axis, $θ$ the angle from the $+ z$ axis ( $0 \leq θ \leq π$ ) and $φ$ the angle around the $z$ -axis measured from the $+ x$ axis. $F_{ρ}, F_{φ}, F_{z}$ (or $F_{r}, F_{θ}, F_{φ}$ ) are the components along the local unit vectors, which change direction from point to point; that is why the factors $ρ$ , $r$ and $r sin θ$ appear. Valid away from the $z$ -axis ( $ρ \neq = 0$ , $sin θ \neq = 0$ ). The previous block is the special case in which only $F_{ρ}$ or $F_{r}$ is non-zero.

WORKED EXAMPLE

(a) With $ϕ = x y z$ and $r = x \hat{i} + y \hat{j} + z \hat{k}$ , find $\nabla \cdot (ϕ r)$ and evaluate it at $(1, 2, - 1)$ . (b) For which $n$ is $r^{n} r$ solenoidal for $r \neq = 0$ ?

Solution

(a) $\nabla \cdot (ϕ r) = ϕ (\nabla \cdot r) + r \cdot \nabla ϕ$ , with $\nabla \cdot r = 3$ and $\nabla ϕ = y z \hat{i} + x z \hat{j} + x y \hat{k}$ .
$r \cdot \nabla ϕ = x y z + x y z + x y z = 3 x y z$ , so $\nabla \cdot (ϕ r) = 3 x y z + 3 x y z = 6 x y z$ .
Direct check: $ϕ r = x^{2} y z \hat{i} + x y^{2} z \hat{j} + x y z^{2} \hat{k}$ has divergence $2 x y z + 2 x y z + 2 x y z = 6 x y z$ . At $(1, 2, - 1)$ this is $6 (1) (2) (- 1) = - 12$ .
(b) $\nabla \cdot (r^{n} r) = (n + 3) r^{n}$ , which vanishes for $r \neq = 0$ only when $n = - 3$ : the inverse-square field $r / r^{3}$ .

Answer: (a) $6 x y z$ , equal to $- 12$ at $(1, 2, - 1)$ ; (b) $n = - 3$ .

EXAM TIP

Look for an identity before differentiating. $\nabla \cdot (\nabla \times F)$ and $\nabla \times (\nabla ϕ)$ are zero for every $F$ and $ϕ$ with continuous second partial derivatives, however complicated; so is the flux of $\nabla \times F$ out of any closed surface and the circulation of $\nabla ϕ$ around any closed curve.

Line, surface and volume integrals

A line integral adds up a field along a curve $C$ . Parametrise the curve as $r (t) = x (t) \hat{i} + y (t) \hat{j} + z (t) \hat{k}$ , $a \leq t \leq b$ , with $t$ increasing in the direction of travel; then $d r = r^{'} (t) d t$ and the integral becomes an ordinary integral in $t$ . If $F$ is a force, $\int_{C} F \cdot d r$ is the work done along $C$ ; around a closed curve, $\oint_{C} F \cdot d r$ is the circulation.

LINE INTEGRAL OF A VECTOR FIELD

\int_{C} F \cdot d r = \int_{C} (F_{1} d x + F_{2} d y + F_{3} d z) = \int_{a}^{b} F (r (t)) \cdot r^{'} (t) d t

Depends on orientation: reversing the direction of travel changes the sign. It does not depend on which parametrisation is used, as long as the direction is the same.

LINE INTEGRAL WITH RESPECT TO ARC LENGTH

\int_{C} f d s = \int_{a}^{b} f (r (t)) ∣ r^{'} (t) ∣ d t

Independent of orientation, because $d s > 0$ . With $f = 1$ it gives the length of $C$ ; with $f$ a line density it gives the mass of a wire.

STANDARD PARAMETRISATIONS

segment P_{0} \to P_{1} : circle x^{2} + y^{2} = a^{2} : curve y = g (x) : r (t) = r_{0} + t (r_{1} - r_{0}), 0 \leq t \leq 1 x = a cos t, y = a sin t, 0 \leq t \leq 2 π x = t, y = g (t)

The circle parametrisation runs counter-clockwise, which keeps the enclosed disc on the left of the direction of travel. For the ellipse $x^{2} / a^{2} + y^{2} / b^{2} = 1$ use $x = a cos t$ , $y = b sin t$ .

A surface integral of a vector field is its flux: $\iint_{S} F \cdot \overset{n}{^} d S$ adds the normal component of $F$ over the surface. An open surface has two sides and the question must say which normal is meant (for example 'upward'); a closed surface is taken with the outward normal unless stated otherwise. A volume integral $∭_{V} f d V$ is an ordinary triple integral, usually set up in the coordinates that match the solid.

FLUX THROUGH A PARAMETRISED SURFACE

\iint_{S} F \cdot \overset{n}{^} d S = \iint_{D} F (r (u, v)) \cdot (r_{u} \times r_{v}) d u d v

$r_{u} \times r_{v}$ is a normal whose length is the area scale factor, so $d S = ∣ r_{u} \times r_{v} ∣ d u d v$ . If it points to the wrong side, use $r_{v} \times r_{u}$ , which flips the sign. For a scalar $f$ , $\iint_{S} f d S = \iint_{D} f (r (u, v)) ∣ r_{u} \times r_{v} ∣ d u d v$ ; with $f = 1$ this is the area of $S$ , and it does not depend on the side chosen.

SURFACE Z = G(X, Y) OVER A REGION R OF THE XY-PLANE

\iint_{S} F \cdot \overset{n}{^} d S = \iint_{R} (- F_{1} g_{x} - F_{2} g_{y} + F_{3}) d x d y, d S = 1 + g_{x}^{2} + g_{y}^{2} d x d y

The flux formula is for the upward normal ( $\overset{n}{^} \cdot \hat{k} > 0$ ); negate it for the downward normal. In general, projecting onto the $x y$ -plane uses $d S = \frac{d x d y}{∣ n ^ \cdot k ^ ∣}$ , valid where $\overset{n}{^} \cdot \hat{k} \neq = 0$ .

SPHERE AND CYLINDER SURFACE ELEMENTS

sphere r = a : cylinder ρ = a : d S = a^{2} sin θ d θ d φ, \overset{n}{^} = \frac{r}{a} d S = a d φ d z, \overset{n}{^} = \frac{x i ^ + y j ^}{a}

Outward normals. $θ$ is measured from the $+ z$ axis ( $0 \leq θ \leq π$ ) and $φ$ around it ( $0 \leq φ < 2 π$ ). Areas: $4 π a^{2}$ for the sphere, $2 π ah$ for the curved side of a cylinder of height $h$ .

VOLUME ELEMENTS

d V = d x d y d z = ρ d ρ d φ d z = r^{2} sin θ d r d θ d φ

Cartesian, cylindrical and spherical. Volumes: sphere $\frac{4}{3} π a^{3}$ , cylinder $π a^{2} h$ , cone $\frac{1}{3} π a^{2} h$ .

WORKED EXAMPLE

Evaluate $\int_{C} F \cdot d r$ for $F = y^{2} \hat{i} + 2 x \hat{j}$ from $(0, 0)$ to $(1, 1)$ (a) along the parabola $y = x^{2}$ and (b) along the straight line $y = x$ . Is $F$ conservative?

Solution

(a) $r (t) = t \hat{i} + t^{2} \hat{j}$ , $0 \leq t \leq 1$ , so $r^{'} (t) = \hat{i} + 2 t \hat{j}$ and $F = t^{4} \hat{i} + 2 t \hat{j}$ on the curve.
$F \cdot r^{'} = t^{4} + 4 t^{2}$ , so the integral is $\int_{0}^{1} (t^{4} + 4 t^{2}) d t = \frac{1}{5} + \frac{4}{3} = \frac{23}{15}$ .
(b) $r (t) = t \hat{i} + t \hat{j}$ , $r^{'} = \hat{i} + \hat{j}$ , $F = t^{2} \hat{i} + 2 t \hat{j}$ , so the integral is $\int_{0}^{1} (t^{2} + 2 t) d t = \frac{1}{3} + 1 = \frac{4}{3} = \frac{20}{15}$ .
Two paths with the same end points give different values, so $F$ is not conservative. The curl test agrees: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 2 y \neq = 0$ .

Answer: (a) $23/15$ ; (b) $4/3$ ; $F$ is not conservative.

EXAM TIP

Split a piecewise path (the sides of a triangle or square) into segments, parametrise each with its own $t$ in the direction of travel, and add. On a segment where $x$ is constant, $d x = 0$ and the $F_{1} d x$ term drops out; likewise for $y$ and $z$ .

Green's theorem in the plane

Each integral theorem equates an integral over a region with an integral over its boundary, with a derivative of the field on the region side. Green's theorem links a closed curve in the $x y$ -plane to the area it encloses. Its orientation convention is counter-clockwise positive: walking along $C$ , the region $R$ stays on the left. Traverse $C$ the other way and the answer changes sign.

GREEN'S THEOREM

\oint_{C} (P d x + Q d y) = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y

$C$ is a piecewise smooth simple closed curve bounding $R$ , traversed counter-clockwise; $P$ , $Q$ and their first partial derivatives are continuous on an open region containing $R$ . Clockwise traversal changes the sign. For a region with holes, the outer boundary runs counter-clockwise and each inner boundary clockwise.

AREA AS A LINE INTEGRAL

A = \oint_{C} x d y = - \oint_{C} y d x = \frac{1}{2} \oint_{C} (x d y - y d x)

Green's theorem with $\partial Q / \partial x - \partial P / \partial y = 1$ , $C$ counter-clockwise. For the ellipse $x = a cos t$ , $y = b sin t$ it gives $π ab$ .

GREEN'S THEOREM, FLUX (NORMAL) FORM

\oint_{C} F \cdot \overset{n}{^} d s = \oint_{C} (P d y - Q d x) = \iint_{R} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d x d y

For $F = P \hat{i} + Q \hat{j}$ , with $\overset{n}{^}$ the outward normal to $C$ and $C$ counter-clockwise: the divergence theorem in two dimensions.

WORKED EXAMPLE

Verify Green's theorem for $F = y^{2} \hat{i} + 2 x \hat{j}$ on the region $R$ between $y = x^{2}$ and $y = x$ , $0 \leq x \leq 1$ , with the boundary traversed counter-clockwise.

Solution

Counter-clockwise keeps $R$ on the left: go out along the lower curve $y = x^{2}$ from $(0, 0)$ to $(1, 1)$ , then back along $y = x$ from $(1, 1)$ to $(0, 0)$ .
From the line-integral example above: the parabola gives $\frac{23}{15}$ , and the line traversed backwards gives $- \frac{4}{3} = - \frac{20}{15}$ . So $\oint_{C} F \cdot d r = \frac{3}{15} = \frac{1}{5}$ .
Region side: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 2 y$ , and $\iint_{R} (2 - 2 y) d A = \int_{0}^{1} \int_{x^{2}}^{x} (2 - 2 y) d y d x = \int_{0}^{1} [2 y - y^{2}]_{x^{2}}^{x} d x$ .
$= \int_{0}^{1} (2 x - x^{2} - 2 x^{2} + x^{4}) d x = \int_{0}^{1} (2 x - 3 x^{2} + x^{4}) d x = 1 - 1 + \frac{1}{5} = \frac{1}{5}$ . Both sides agree.

Answer: $\oint_{C} F \cdot d r = \iint_{R} (2 - 2 y) d A = 1/5$ .

EXAM TIP

Green's theorem pays off when $\partial Q / \partial x - \partial P / \partial y$ is simple even though $P$ and $Q$ are not: if it is a constant $c$ , the integral is $c$ times the enclosed area. Terms such as $f (x) d x$ or $g (y) d y$ (with $f$ , $g$ continuous) are exact and contribute nothing around a closed curve. To use it on an open path, close the path with a simple segment, apply the theorem to the loop, and subtract the segment's own integral.

EXAM TIP

Common slips: writing $\partial P / \partial x - \partial Q / \partial y$ instead of $\partial Q / \partial x - \partial P / \partial y$ ; forgetting the sign change for a clockwise curve; applying the theorem when $P$ or $Q$ blows up inside $R$ (as $\frac{- y}{x ^{2} + y ^{2}}$ , $\frac{x}{x ^{2} + y ^{2}}$ do at the origin).

Stokes' and Gauss's theorems

Stokes' theorem links the circulation around a closed curve in space to the flux of the curl through any surface spanning that curve; Green's theorem is its special case in the $x y$ -plane with $\overset{n}{^} = \hat{k}$ . Gauss's divergence theorem links the outward flux through a closed surface to the volume integral of the divergence inside it. In both, the orientation is part of the statement.

STOKES' THEOREM

\oint_{C} F \cdot d r = \iint_{S} (\nabla \times F) \cdot \overset{n}{^} d S

$S$ is a piecewise smooth oriented surface whose boundary is the closed curve $C$ ; $F$ has continuous first partial derivatives on a region containing $S$ . Right-hand rule: with the thumb along $\overset{n}{^}$ , the fingers curl in the direction of $C$ (for $\overset{n}{^} = \hat{k}$ , $C$ runs counter-clockwise seen from above). Every surface that meets these conditions and has the same boundary and orientation gives the same value.

GAUSS'S DIVERGENCE THEOREM

\iint_{S} F \cdot \overset{n}{^} d S = ∭_{V} (\nabla \cdot F) d V

$S$ is a closed piecewise smooth surface enclosing the volume $V$ , $\overset{n}{^}$ is the outward unit normal, and $F$ has continuous first partial derivatives throughout $V$ (no singularity inside).

Theorem	Converts	Orientation	Use it when
Green	closed line integral in the plane $\leftrightarrow$ double integral over the enclosed region	$C$ counter-clockwise, region on the left	the path has several pieces, or $Q_{x} - P_{y}$ is simple (often a constant)
Stokes	circulation around a closed space curve $\leftrightarrow$ flux of the curl through any spanning surface	right-hand rule linking $\overset{n}{^}$ and $C$	the curl is simple, or a curved spanning surface can be swapped for a flat one (or for the boundary curve)
Gauss	flux through a closed surface $\leftrightarrow$ volume integral of the divergence	outward normal	the surface has several faces (cube, cylinder) or the divergence is simple; for an open surface, close it with flat caps and subtract their flux

WORKED EXAMPLE

Find the outward flux of $F = x \hat{i} + y \hat{j} + z \hat{k}$ (a) through the whole closed surface of the solid cylinder $x^{2} + y^{2} \leq 4$ , $0 \leq z \leq 3$ , and (b) through its curved side only.

Solution

(a) $\nabla \cdot F = 3$ , a constant, so by Gauss's theorem the flux is $3 \times volume = 3 \times π (2^{2}) (3) = 36 π$ .
(b) Subtract the flat caps from the closed-surface total. Top ( $z = 3$ , $\overset{n}{^} = \hat{k}$ ): $F \cdot \overset{n}{^} = z = 3$ over area $4 π$ , flux $12 π$ . Bottom ( $z = 0$ , $\overset{n}{^} = - \hat{k}$ ): $F \cdot \overset{n}{^} = - z = 0$ , flux $0$ .
Curved side: $36 π - 12 π - 0 = 24 π$ .
Direct check: on $ρ = 2$ , $\overset{n}{^} = (x \hat{i} + y \hat{j}) /2$ , so $F \cdot \overset{n}{^} = (x^{2} + y^{2}) /2 = 2$ over area $2 π (2) (3) = 12 π$ , giving $24 π$ .

Answer: (a) $36 π$ ; (b) $24 π$ .

WORKED EXAMPLE

Let $F = z \hat{i} + x \hat{j} + y \hat{k}$ and let $S$ be the paraboloid $z = 4 - x^{2} - y^{2}$ , $z \geq 0$ , with upward normal. Find $\iint_{S} (\nabla \times F) \cdot \overset{n}{^} d S$ .

Solution

$\nabla \times F = (\frac{\partial y}{\partial y} - \frac{\partial x}{\partial z}) \hat{i} + (\frac{\partial z}{\partial z} - \frac{\partial y}{\partial x}) \hat{j} + (\frac{\partial x}{\partial x} - \frac{\partial z}{\partial y}) \hat{k} = \hat{i} + \hat{j} + \hat{k}$ .
The boundary of $S$ is the circle $x^{2} + y^{2} = 4$ , $z = 0$ , and the upward normal makes it counter-clockwise seen from above. $F$ is smooth everywhere, so by Stokes' theorem every surface with this boundary and orientation carries the same flux of the curl: replace the paraboloid with the flat disc $x^{2} + y^{2} \leq 4$ , $z = 0$ , $\overset{n}{^} = \hat{k}$ .
On the disc $(\nabla \times F) \cdot \hat{k} = 1$ , so the flux equals the disc's area, $4 π$ .
Check with the circulation: $r = 2 cos t \hat{i} + 2 sin t \hat{j}$ gives $F = 2 cos t \hat{j} + 2 sin t \hat{k}$ and $r^{'} = - 2 sin t \hat{i} + 2 cos t \hat{j}$ , so $\oint_{C} F \cdot d r = \int_{0}^{2 π} 4 cos^{2} t d t = 4 π$ .

Answer: $4 π$ .

EXAM TIP

Results with no computation: the flux of a constant vector field out of any closed surface is 0; the flux of $r$ out of a closed surface is $3 V$ (so $4 π a^{3}$ for a sphere of radius $a$ ); the flux of $\nabla \times F$ out of any closed surface is 0; and a constant divergence $c$ gives outward flux $c \times$ volume.

EXAM TIP

Check for singularities before applying Gauss's theorem. For $F = r / r^{3}$ , $\nabla \cdot F = 0$ everywhere except the origin, yet the outward flux through any sphere centred at the origin is $4 π$ (on $r = a$ , $F \cdot \overset{n}{^} = 1/ a^{2}$ over area $4 π a^{2}$ ). The theorem does not apply to a volume containing the origin; through a closed surface that does not enclose the origin, the flux is 0.

PREVIOUS-YEAR QUESTIONS

23 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

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GATE 2026 EC Q.12

GATE 2026 · ECQ.12 · 1 mark · MCQ

A surface is given by $z^{2} = 2 x^{2} - y^{2}$ and $n$ and $- n$ are unit normal vectors to the surface at the point $P = \overset{}{^} + 2 \hat{k}$ .

Which of the following vectors can be $n$ , where $\overset{}{^}, \overset{}{^}$ and $\hat{k}$ are the unit vectors along x, y and z axes, respectively?

(A)
$\overset{}{^} - 2 \hat{k}$
(B)
$\frac{2}{3} \overset{}{^} - \frac{1}{3} \hat{k}$
(C)
$2 \overset{}{^} - 3 \hat{k}$
(D)
$\frac{2  ^ - k ^}{3}$

Answer and solution

GATE 2026 EC Q.12: answer and solution

See the question in the paper

Vector calculus

Vector algebra: products, angles and projections

Gradient, directional derivative and normals

Divergence, curl and conservative fields

Vector identities and the Laplacian

Line, surface and volume integrals

Green's theorem in the plane

Stokes' and Gauss's theorems

23 questions from the EC, EE and IN papers

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