Vector Analysis

6.1Applications of Vector Multiplication

Dot Product (Scalar Product)

Cross Product (Vector Product)

Scalar Triple Product

This scalar represents the signed volume of the parallelepiped formed by \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\).

Vector Triple Product

This identity is commonly known as the BAC–CAB rule. The resulting vector lies in the plane of \(\vec{B}\) and \(\vec{C}\).

6.2Differentiation of Vectors

In physics, many vector quantities—such as position, velocity, and acceleration—vary with time or with spatial coordinates. We therefore need rules for differentiating and integrating vectors.

If the components of \(\vec{A}\) are differentiable functions of \(t\),

then

6.3Fields

Many physical quantities depend on position in space. Temperature, pressure, electric field, and magnetic field are examples of fields.

6.4Directional Derivative and Gradient

When a scalar field \(\phi(x,y,z)\) varies in space, we often wish to know the rate at which it changes in a given direction.

Gradient as a Normal to a Surface

Example - Tangent Plane and Normal Line to a Sphere

Find the equations of the tangent plane and the normal line to the surface of a sphere

\begin{equation} \phi(x,y,z) = x^2 + y^2 + z^2 = a^2 \end{equation}

at the point \((0,0,a)\).

Solution.

\begin{equation} \vec{\nabla}\phi = \frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k} = 2x\,\hat{i} + 2y\,\hat{j} + 2z\,\hat{k}. \end{equation}

At the point \((0,0,a)\),

\begin{equation} \boxed{\vec{\nabla}\phi(0,0,a) = 2a\,\hat{k}.} \end{equation}

Step 2: Equation of the tangent plane.

Since \(\vec{\nabla}\phi\) is normal to the surface at every point, the tangent plane at \((x_0,y_0,z_0)\) is given by

\begin{equation} (\vec{r}-\vec{r}_0)\cdot\vec{\nabla}\phi = 0. \end{equation}

At \((0,0,a)\), this becomes

\begin{equation} (\vec{r}-a\hat{k})\cdot(2a\hat{k}) = 0 \quad\Rightarrow\quad 2a(z-a)=0 \quad\Rightarrow\quad \boxed{z=a.} \end{equation}

This is the equation of the tangent plane at the top of the sphere.

Step 3: Equation of the normal line.

The line normal to the surface passes through \((0,0,a)\) and has direction \(\vec{\nabla}\phi(0,0,a) = 2a\hat{k}\). Hence, its vector equation is

\begin{equation} (\vec{r}-\vec{r}_0)\times\vec{\nabla}\phi = 0. \end{equation}

Substitute:

\begin{equation} (\vec{r}-a\hat{k})\times 2a\hat{k}=0. \end{equation}

Writing \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\):

\begin{equation} (x\hat{i}+y\hat{j})\times 2a\hat{k}=2a(y\hat{i}-x\hat{j})=0 \quad\Rightarrow\quad x=0, \; y=0. \end{equation}

Thus, the normal line is the \(z\)-axis.

Result:

\begin{equation} \boxed{ \begin{aligned} \text{Tangent plane:}&\quad z=a,\\[4pt] \text{Normal line:}&\quad x=0,\ y=0 \quad (\text{the } z\text{-axis}). \end{aligned} } \end{equation}

6.5Expressions Involving \(\vec{\nabla}\)

The differential operator

acts on scalar and vector fields in several useful ways.

Gradient of a Scalar Field

For a scalar field \(\phi(x,y,z)\),

It produces a vector field pointing in the direction of the greatest increase of \(\phi\).

Divergence of a Vector Field

Curl of a Vector Field

Example - Comparing Divergence and Curl in 2D Flows

Consider three two–dimensional vector fields in the \(xy\)–plane:

\begin{equation} \begin{aligned} (1)\;& \text{Field } \vec{A} &: &\quad \vec{A} = kx\,\hat{i} + ky\,\hat{j} \quad &\text{(Pure Divergence)} \\[4pt] (2)\;& \text{Field } \vec{B} &: &\quad \vec{B} = -\omega y\,\hat{i} + \omega x\,\hat{j} \quad &\text{(Pure Curl / Rotation)} \\[4pt] (3)\;& \text{Field } \vec{A} + \vec{B} &: &\quad (kx - \omega y)\,\hat{i} + (ky + \omega x)\,\hat{j} \quad &\text{(Combined / Spiral Flow)} \end{aligned} \end{equation}

(a) Divergence

\begin{equation} \begin{aligned} \vec{\nabla}\cdot\vec{A} &= \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} = k + k = 2k, \\[4pt] \vec{\nabla}\cdot\vec{B} &= \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} = 0 + 0 = 0, \\[4pt] \vec{\nabla}\cdot(\vec{A}+\vec{B}) &= 2k. \end{aligned} \end{equation}

Thus, \(\vec{A}\) and \(\vec{A}+\vec{B}\) represent fields with a uniform outward expansion (a source flow), while \(\vec{B}\) has zero divergence — a purely rotational flow.

(b) Curl

In two dimensions, only the \(z\)–component of the curl is nonzero:

\begin{equation} (\vec{\nabla}\times\vec{A})_z = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}. \end{equation}

Hence,

\begin{equation} \begin{aligned} (\vec{\nabla}\times\vec{A})_z &= 0, \\[4pt] (\vec{\nabla}\times\vec{B})_z &= 2\omega, \\[4pt] (\vec{\nabla}\times(\vec{A}+\vec{B}))_z &= 2\omega. \end{aligned} \end{equation}

(c) Physical Interpretation

• \(\vec{A}\): Field lines radiate outward from the origin — a pure source field. \(\vec{\nabla}\cdot\vec{A} > 0\), \(\vec{\nabla}\times\vec{A} = 0.\)

• \(\vec{B}\): Field lines form closed circles — a pure rotational (vortex) field. \(\vec{\nabla}\cdot\vec{B} = 0\), \(\vec{\nabla}\times\vec{B} \neq 0.\)

• \(\vec{A}+\vec{B}\): Field lines spiral outward — a combination of source and rotation. Both divergence and curl are nonzero, producing a spiral flow.

Visualization \(k=1\) and \(\omega=1 \)

• Left: Pure source — arrows radiate outward uniformly.

• Middle: Pure rotation — arrows circle around the origin.

• Right: Spiral flow — arrows spiral outward, combining rotation and expansion.

This example clearly shows that:

• Divergence quantifies the presence of sources or sinks.

• Curl quantifies the local rotation of the field.

Useful Vector Identities

Vector Differential Operators in a General Orthogonal Curvilinear Coordinate System

In many physical problems, Cartesian coordinates \((x,y,z)\) are not always the most convenient choice. We often use other coordinate systems such as cylindrical \((\rho,\phi,z)\) or spherical \((r,\theta,\phi)\), which are examples of orthogonal curvilinear coordinates.

Vector Differential Operators in Common Coordinate Systems

In vector calculus, the four most fundamental vector differential operators are:

These operators describe spatial variation of scalar and vector fields, and their expressions depend on the coordinate system and corresponding unit vectors.

1) Cartesian Coordinates \((x, y, z)\)

2) Cylindrical Coordinates \((\rho, \phi, z)\)

The unit vectors are related to the Cartesian basis by:

3) Spherical Coordinates \((r, \theta, \phi)\)

The spherical unit vectors in terms of Cartesian components are:

6.6Maxwell’s Equations in Vacuum

In differential form, Maxwell’s equations describe the behavior of electric and magnetic fields in space and time. They connect the electric field \(\vec{E}\), magnetic field \(\vec{B}\), charge density \(\rho\), and current density \(\vec{J}\).

Theorem - Maxwell’s Equations (Differential Form in Vacuum with Sources)

\begin{equation} \begin{aligned} \text{(1) Gauss’s Law:} &\quad \vec{\nabla}\cdot\vec{E} = \frac{\rho_f}{\epsilon_0}, \\[4pt] \text{(2) Gauss’s Law for Magnetism:} &\quad \vec{\nabla}\cdot\vec{B} = 0, \\[4pt] \text{(3) Faraday’s Law of Induction:} &\quad \vec{\nabla}\times\vec{E} = -\,\frac{\partial \vec{B}}{\partial t}, \\[4pt] \text{(4) Ampère–Maxwell Law:} &\quad \vec{\nabla}\times\vec{B} = \mu_0\,\vec{J}_f + \mu_0\epsilon_0\,\frac{\partial \vec{E}}{\partial t}. \end{aligned} \end{equation}

Physical Meaning of Each Equation

1. Gauss’s Law: \(\displaystyle \vec{\nabla}\cdot\vec{E} = \frac{\rho_f}{\epsilon_0}\)

   • The divergence of the electric field gives the local charge density.

   • Positive charges are sources of electric field lines, negative charges are sinks.

2. Gauss’s Law for Magnetism: \(\displaystyle \vec{\nabla}\cdot\vec{B} = 0\)

   • Magnetic field lines are always closed loops; there are no magnetic monopoles.

   • The field has no divergence—no “beginning” or “end.”

3. Faraday’s Law: \(\displaystyle \vec{\nabla}\times\vec{E} = -\frac{\partial \vec{B}}{\partial t}\)

   • A time-varying magnetic field induces a circulating electric field.

   • The curl of \(\vec{E}\) represents the tendency of \(\vec{E}\) to form closed loops.

   • This is the principle behind electromagnetic induction.

4. Ampère–Maxwell Law: \(\displaystyle \vec{\nabla}\times\vec{B} = \mu_0\vec{J}_f + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}\)

   • A circulating magnetic field is produced by electric currents \(\vec{J}\) and by changing electric fields \(\frac{\partial \vec{E}}{\partial t}\).

   • The second term (\(\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}\)) is the displacement current density, introduced by Maxwell to ensure charge conservation.

Derivation of the Electromagnetic Wave Equation

We now derive how Maxwell’s equations predict that changing electric and magnetic fields propagate as waves in vacuum. In Vacuum (no charges or currents): \(\rho_f = 0\), \(\vec{J}_f = 0.\) Then Maxwell’s equations become:

Derivation of the wave equation.

Take the curl of Faraday’s law:

Substitute \(\vec{\nabla}\times\vec{B} = \mu_0\epsilon_0\,\frac{\partial \vec{E}}{\partial t}\):

Using the vector identity

and since \(\vec{\nabla}\cdot\vec{E}=0\) in vacuum, we obtain:

Similarly, taking the curl of \(\vec{\nabla}\times\vec{B} = \mu_0\epsilon_0\,\frac{\partial \vec{E}}{\partial t}\) gives:

These are wave equations for \(\vec{E}\) and \(\vec{B}\), describing electromagnetic waves propagating through space.

The wave speed.

Comparing with the standard wave equation \(\nabla^2 u = \dfrac{1}{v^{2}}\frac{\partial^{2} u}{\partial t^{2}}\), whose solutions travel at speed \(v\), we can read off the speed of an electromagnetic wave:

Substituting the measured constants \(\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}\) and \(\epsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/(N\,m^2)}\) gives \(c \approx 3.00\times10^{8}\ \mathrm{m/s}\) — the measured speed of light. This is the historical significance of the calculation: two constants obtained from benchtop experiments on static electricity and on magnets combine to give the speed of light, which is what identified light itself as an electromagnetic wave.

6.7Vector Integrals

In vector calculus, integrals of scalar or vector functions can be defined over curves, surfaces, or volumes. These are classified as:

6.7.1Line Integrals

We may encounter three types of line integrals along a curve \( C \):

Example

Evaluate

\begin{equation} I = \int_C \vec{F}\cdot d\vec{r}, \qquad \text{where } \vec{F} = (x+y)\,\hat{i} + (y-x)\,\hat{j}, \end{equation}

along each of the following paths.

(i) Parabolic path \(y^2 = x\) from \((1,1)\) to \((4,2)\).

Step 1. Express \(d\vec{r}\).

Since \(x = y^2\), we have

\begin{equation} d\vec{r} = dx\,\hat{i} + dy\,\hat{j} = (2y\,dy)\,\hat{i} + dy\,\hat{j}. \end{equation}

Step 2. Compute \(\vec{F}\cdot d\vec{r}\).

\begin{equation} \vec{F}\cdot d\vec{r} = (x+y)\,dx + (y-x)\,dy = (y^2+y)(2y\,dy) + (y - y^2)\,dy = (2y^3 + y^2 + y)\,dy. \end{equation}

Step 3. Integrate from \(y=1\) to \(y=2\).

\begin{equation} I = \int_1^2 (2y^3 + y^2 + y)\,dy = \left[\frac{y^4}{2} + \frac{y^3}{3} + \frac{y^2}{2}\right]_1^2 = \frac{34}{3}. \end{equation}

(ii) Parametric curve \(x=2u^2+u+1\), \(y=1+u^2\) from \((1,1)\) to \((4,2)\).

Step 1. Express \(d\vec{r}\).

\begin{equation} d\vec{r} = dx\,\hat{i} + dy\,\hat{j} = (4u+1)\,du\,\hat{i} + (2u)\,du\,\hat{j}. \end{equation}

Step 2. Compute \(\vec{F}\cdot d\vec{r}\).

\begin{equation} \vec{F}\cdot d\vec{r} = (x+y)\,dx + (y-x)\,dy. \end{equation}

Substitute \(x=2u^2+u+1\), \(y=1+u^2\):

\begin{equation} x+y = 3u^2 + u + 2, \qquad y-x = -u^2 - u. \end{equation}

Thus,

\begin{equation} \vec{F}\cdot d\vec{r} = (3u^2 + u + 2)(4u+1)\,du - (u^2 + u)(2u)\,du . \end{equation}

Expanding each product separately,

\begin{equation} (3u^2 + u + 2)(4u+1) = 12u^3 + 7u^2 + 9u + 2, \qquad (u^2+u)(2u) = 2u^3 + 2u^2, \end{equation}

so that

\begin{equation} \vec{F}\cdot d\vec{r} = \big(12u^3 + 7u^2 + 9u + 2 - 2u^3 - 2u^2\big)\,du = (10u^3 + 5u^2 + 9u + 2)\,du. \end{equation}

Step 3. Integrate from \(u=0\) to \(u=1\).

\begin{equation} I = \int_0^1 (10u^3 + 5u^2 + 9u + 2)\,du = \left[\frac{10u^4}{4} + \frac{5u^3}{3} + \frac{9u^2}{2} + 2u\right]_0^1 = \frac{5}{2} + \frac{5}{3} + \frac{9}{2} + 2 = \frac{32}{3}. \end{equation}

(iii) Piecewise linear path.

Segment 1: along \(y=1\) from \((1,1)\) to \((4,1)\).

\begin{equation} d\vec{r} = dx\,\hat{i} \quad (dy=0), \quad \vec{F}\cdot d\vec{r} = (x+y)\,dx = (x+1)\,dx. \end{equation}

\begin{equation} I_1 = \int_1^4 (x+1)\,dx = \left[\frac{x^2}{2}+x\right]_1^4 = \frac{21}{2}. \end{equation}

Segment 2: along \(x=4\) from \((4,1)\) to \((4,2)\).

\begin{equation} d\vec{r} = dy\,\hat{j} \quad (dx=0), \quad \vec{F}\cdot d\vec{r} = (y-x)\,dy = (y-4)\,dy. \end{equation}

\begin{equation} I_2 = \int_1^2 (y-4)\,dy = \left[\frac{y^2}{2}-4y\right]_1^2 = -\frac{5}{2}. \end{equation}

Total:

\begin{equation} I = I_1 + I_2 = \frac{21}{2} - \frac{5}{2} = 8. \end{equation}

Example

Evaluate the line integral

\begin{equation} I = \oint_C x\,dy, \end{equation}

where \(C\) is the circle in the \(xy\)–plane defined by

\begin{equation} x^2 + y^2 = a^2, \qquad z = 0. \end{equation}

Solution.

This integral is very simple in polar coordinates.

\begin{equation} x = a\cos\theta, \qquad y = a\sin\theta, \qquad dy = a\cos\theta\,d\theta. \end{equation}

\begin{equation} I = \oint_C x\,dy = \int_0^{2\pi} a\cos\theta\,(a\cos\theta\,d\theta) = a^2\int_0^{2\pi}\cos^2\theta\,d\theta. \end{equation}

Using the trigonometric identity \(\cos^2\theta = \dfrac{1+\cos(2\theta)}{2}\),

\begin{equation} I = a^2\int_0^{2\pi}\frac{1+\cos(2\theta)}{2}\,d\theta = \frac{a^2}{2}\Big[\theta + \frac{\sin(2\theta)}{2}\Big]_0^{2\pi}. \end{equation}

Since \(\sin(2\pi)=0\),

\begin{equation} \boxed{\,I = \pi a^2.\,} \end{equation}

Example - Compute the line integral

\begin{equation} I=\int_C \frac{x\,dy - y\,dx}{x^2+y^2} \end{equation}

from \((-1,0)\) to \((1,0)\) along the two paths shown below.

We can write this integral as

\begin{equation} I=\int_C \vec{F}\cdot d\vec{r}, \qquad \text{where } \vec{F} = \frac{-y\,\hat{i} + x\,\hat{j}}{x^2 + y^2}, \end{equation}

along each of the following paths.

Path 1: upper semicircle of the unit circle

Use polar coordinates on the circle \(r=1\): \(x=\cos\theta,\ y=\sin\theta,\ dx=-\sin\theta\,d\theta,\ dy=\cos\theta\,d\theta,\ x^2+y^2=1.\)

\begin{equation} \frac{x\,dy-y\,dx}{x^2+y^2} =\cos\theta(\cos\theta\,d\theta)-\sin\theta(-\sin\theta\,d\theta) =\big(\cos^2\theta+\sin^2\theta\big)d\theta=d\theta. \end{equation}

Along the upper semicircle from \((-1,0)\) to \((1,0)\) we have \(\theta:\ \pi\to 0\). Hence

\begin{equation} \boxed{\,I_1=\int_{\pi}^{0} d\theta=-\pi.\,} \end{equation}

Path 2: two straight-line segments

Segment A.

From \((-1,0)\) to \((0,1)\) along \(y=x+1\). Then \(dy=dx\) and

\begin{equation} x\,dy-y\,dx = \big(x-(x+1)\big)\,dx = -\,dx,\qquad x^2+y^2 = x^2+(x+1)^2=2x^2+2x+1. \end{equation}

Thus

\begin{equation} I_A=\int_{-1}^{0}\frac{-\,dx}{2x^2+2x+1} =\int_{-1}^{0}\frac{-2\,dx}{(2x+1)^2+1} =-\Big[\arctan(2x+1)\Big]_{-1}^{0} =-\frac{\pi}{2}. \end{equation}

Segment B.

From \((0,1)\) to \((1,0)\) along \(y=1-x\). Then \(dy=-dx\) and

\begin{equation} x\,dy-y\,dx = \big(-x-(1-x)\big)\,dx = -\,dx,\qquad x^2+y^2 = x^2+(1-x)^2=2x^2-2x+1. \end{equation}

Hence

\begin{equation} I_B=\int_{0}^{1}\frac{-\,dx}{2x^2-2x+1} =\int_{0}^{1}\frac{-2\,dx}{(2x-1)^2+1} =-\Big[\arctan(2x-1)\Big]_{0}^{1} =-\frac{\pi}{2}. \end{equation}

\begin{equation} \boxed{\,I_2=I_A+I_B=-\frac{\pi}{2}-\frac{\pi}{2}=-\pi.\,} \end{equation}

6.7.2Conservative Fields

A force field \(\vec{F}\) is said to be conservative if the work done by it is independent of the path between two points:

If \(W\) depends on the path, then \(\vec{F}\) is non-conservative, and energy is being dissipated (e.g., by friction).

Condition for a Conservative Field

A vector field \(\vec{F}\) is conservative if and only if

In this case, \(\vec{F}\) can be expressed as the gradient of a scalar potential \(W(x,y,z)\):

Taking the curl of both sides:

which confirms that \(\vec{F}\) is indeed conservative.

Work in Terms of Potential Function

If \(\vec{F} = \vec{\nabla}W\), then

Hence,

The work done depends only on the values of \(W\) at the endpoints.

Potential Function

For a conservative field we have \(\vec{F} = \vec{\nabla}W\), where \(W\) is the work function. In physics it is conventional to work instead with the potential energy \(\Phi\), defined as the negative of the work function:

and therefore,

The sign is not arbitrary: with this definition the force points in the direction of decreasing potential, so a ball rolls downhill and a positive charge moves away from another positive charge. Work done by the field lowers the potential energy, \(W(B)-W(A) = -[\Phi(B)-\Phi(A)]\), which is what makes \(\Phi\) the quantity that adds to kinetic energy to give a conserved total. Both conventions appear in the literature; the sign of the gradient must be checked whenever a potential is quoted.

Physical Examples of Line Integrals

Line integrals appear frequently in physics, where they often represent quantities such as work, circulation, or magnetic flux linkage.

Below are some important physical contexts where line integrals are used.

1. Work Done by a Force Field

This represents the work done by a force \(\vec{F}\) in moving a particle along a path \(C\).

2. Electrostatic Potential Energy

The electrostatic potential energy gained by moving a charge \(q\) through an electric field \(\vec{E}\) along a path \(C\) is

The negative sign indicates that the electric field does work against the potential difference.

3. Ampère’s Law (Magnetostatics)

The circulation of the magnetic field around a closed loop \(C\) is proportional to the current \(I\) enclosed by that loop:

This integral form of Ampère’s law relates magnetic fields to the steady currents that produce them.

4. Magnetic Force on a Current Loop

A current-carrying wire placed in a magnetic field \(\vec{B}\) experiences a magnetic force given by

This expression gives the total force on a loop of current \(I\) in a magnetic field. Each element of the wire experiences an infinitesimal force \(d\vec{F} = I\,d\vec{r}\times\vec{B}\).

6.7.11Surface and Volume Integrals

Surface Integrals

A surface integral generalizes the concept of a line integral to a two-dimensional surface. For a surface \(\sigma\) with unit normal vector \(\hat{n}\) and differential area element \(d\vec{\sigma} = \hat{n}\,d\sigma\), we can define three common types of surface integrals:

The most important and frequently encountered is

which represents the flux of the vector field \(\vec{F}\) through the surface \(S\). Physically, this measures how much of the field “flows” through \(S\). Only the normal component of \(\vec{F}\) contributes to the flux; tangential components make no contribution.

Volume Integrals

A volume integral extends the concept to three dimensions. For a scalar field \(\phi(x,y,z)\),

represents the total accumulation of \(\phi\) over a volume \(V\).

For a vector field \(\vec{A} = A_x\,\hat{i} + A_y\,\hat{j} + A_z\,\hat{k}\),

reducing the vector integral to a sum of scalar volume integrals over the components.

6.7.14Green’s Theorem in the Plane

The one–dimensional Fundamental Theorem of Calculus states:

relating the integral of a derivative over an interval to the function’s change at its endpoints. In two dimensions, a similar idea holds: the integral of derivatives over an area \(A\) corresponds to function values along its boundary \(C\). This reasoning leads directly to Green’s Theorem.

Proof

Part I: the term involving \(\frac{\partial P}{\partial y}\).

Consider

Assume the region can be described by the equations

so that

Along a vertical line (\(x=\text{const}\)), \(dx=0\) and \(dP = \frac{\partial P}{\partial y}\,dy\). Then

Hence,

Part II: the term involving \(\frac{\partial Q}{\partial x}\).

Consider

Assume the region can also be described by

so that

Along a horizontal line (\(y=\text{const}\)), \(dy=0\) and \(dQ = \frac{\partial Q}{\partial x}\,dx\). Then

Hence,

The two parts must now be combined with care over signs. From Part I, \(I = -\oint_{\partial A} P\,dx\), so that \(\oint_{\partial A} P\,dx = -I\); from Part II, \(\oint_{\partial A} Q\,dy = II\). The boundary integral we want is therefore the difference \(II - I\), not the sum:

Writing each term as its area integral,

and hence

which is Green's theorem. The minus sign in front of \(\frac{\partial P}{\partial y}\) traces back to Part I, where traversing the boundary counterclockwise means the upper curve \(y_2(x)\) is crossed from right to left.

Physical Interpretation

• The left-hand side represents the circulation of the vector field \(\vec{F}(x,y) = P(x,y)\,\hat{i} + Q(x,y)\,\hat{j}\) around the closed curve \(C\).

• The right-hand side measures the total curl flux of \(\vec{F}\) through the enclosed region \(A\).

• Hence, Green’s theorem states that the local rotation (curl) of a field sums to the total circulation along its boundary.

Example - Method One: Use Green’s Theorem

Let

\begin{equation} \vec{F}(x,y)= P\,\hat{i}+Q\,\hat{j} = \big(xy\big)\,\hat{i} + \big(-y^{2}\big)\,\hat{j}. \end{equation}

Compute the work \(W=\displaystyle\oint_C \vec{F}\cdot d\vec{r}\).

Using Green’s theorem. Since

\begin{equation} \frac{\partial Q}{\partial x}=0,\qquad \frac{\partial P}{\partial y}=x, \end{equation}

we have

\begin{equation} W=\iint_A\!\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy =\iint_A (-x)\,dx\,dy =-\int_{y=0}^{1}\!\!\int_{x=0}^{2\sqrt{y}}x\,dx\,dy =-1. \end{equation}

Method Two: Direct Evaluation

\begin{equation} \vec{F}\cdot d\vec{r}=P\,dx+Q\,dy=xy\,dx-y^{2}\,dy. \end{equation}

The boundary is traversed counterclockwise as:

\begin{equation} \begin{aligned} C_1&:\ x=2\sqrt{y},\ (0,0)\to(2,1),\\ C_2&:\ y=1,\ (2,1)\to(0,1),\\ C_3&:\ x=0,\ (0,1)\to(0,0). \end{aligned} \end{equation}

Path \(C_1\): The curve is \(x = 2\sqrt{y}\), or equivalently \(y = \dfrac{x^{2}}{4}\), with \(x:0\to2.\)

\begin{equation} d\vec{r} = \hat{i}\,dx + \hat{j}\,dy, \qquad dy = \frac{x}{2}\,dx. \end{equation}

Then

\begin{equation} \begin{aligned} W_1 &= \int_{0}^{2} \big(xy\,dx - y^{2}\,dy\big) = \int_{0}^{2} \left[x\!\left(\frac{x^{2}}{4}\right) - \left(\frac{x^{2}}{4}\right)^{2}\!\left(\frac{x}{2}\right)\right] dx \\[6pt] &= \int_{0}^{2} \left(\frac{x^{3}}{4} - \frac{x^{5}}{32}\right) dx = \left[\frac{x^{4}}{16} - \frac{x^{6}}{192}\right]_{0}^{2} \\[6pt] &= \left(\frac{16}{16} - \frac{64}{192}\right) = 1 - \frac{1}{3} = \boxed{\frac{2}{3}}. \end{aligned} \end{equation}

Path \(C_2\): \(y=1,\ dy=0,\ x:2\to0\).

\begin{equation} d\vec{r} = \hat{i}\,dx, \qquad \vec{F} = (x, -1). \end{equation}

Thus

\begin{equation} \begin{aligned} W_2 &=\int_{2}^{0} (xy\,dx - y^{2}\,dy) = \int_{2}^{0} x(1)\,dx = \left[\frac{x^{2}}{2}\right]_{2}^{0} =\boxed{-2}. \end{aligned} \end{equation}

Path \(C_3\): \(x=0,\ dx=0,\ y:1\to0\).

\begin{equation} d\vec{r} = \hat{j}\,dy, \qquad \vec{F} = (0, -y^{2}). \end{equation}

Then

\begin{equation} \begin{aligned} W_3 &=\int_{1}^{0} (xy\,dx - y^{2}\,dy) =\int_{1}^{0} (-y^{2})\,dy =\left[-\frac{y^{3}}{3}\right]_{1}^{0} =\boxed{\frac{1}{3}}. \end{aligned} \end{equation}

Total Work (Counterclockwise):

\begin{equation} W = W_1 + W_2 + W_3 = \frac{2}{3} - 2 + \frac{1}{3} = \boxed{-1}. \end{equation}

6.8The Divergence and Stokes’s Theorems

6.8.1The Divergence Theorem in the Plane

Let

be a continuously differentiable vector field defined on a region \(A\) of the \(xy\)-plane, bounded by a positively oriented (counterclockwise) closed curve \(\partial A\).

To connect this with Green’s theorem, define

Then the combination appearing in Green’s theorem,

becomes

Geometric Relation to the Line Integral

Along the curve \(\partial A\), the differential displacement vector is

which is tangent to the boundary. The outward normal vector (perpendicular to the tangent) is

where \(ds = \sqrt{(dx)^2 + (dy)^2}\) is the differential arc length.

Now, using \(P=-F_y\) and \(Q=F_x\),

Substituting into Green’s theorem gives

This states that the total flux of \(\vec{F}\) through the boundary \(\partial A\) equals the integral of the divergence of \(\vec{F}\) over the enclosed region \(A\).

Extension to Three Dimensions (Gauss’s Theorem)

6.8.4Stokes’s Theorem in the Plane

Let

Then the combination in Green’s theorem,

is the \(z\)-component of the curl of \(\vec{F}\), which represents the local rotation or vorticity.

Using

we can write

Substituting this into Green’s theorem gives

This expresses that the circulation of \(\vec{F}\) around a closed boundary equals the total curl flux through the enclosed region.

Extension to Three Dimensions (General Stokes’s Theorem)

6.9Maxwell’s Equations and the Continuity Equation

1. Differential Form

Taking the divergence of the Ampère–Maxwell law and using \(\vec{\nabla}\cdot(\vec{\nabla}\times\vec{B})=0\) gives

which is the continuity equation in differential form, expressing local charge conservation.

2. Integral Form via Divergence and Stokes’ Theorems

We now apply the Divergence Theorem

and the Stokes’ Theorem

Applying these to each of Maxwell’s equations:

3. Continuity Equation in Integral Form

Starting from the differential form

integrate over a fixed volume \(\tau\):

Apply the Divergence Theorem to the first term:

Define

so that

This expresses charge conservation: the rate of decrease of enclosed charge within the volume \(\tau\) equals the net outward current through its bounding closed surface \(\sigma\).

4. Physical Interpretation

• Gauss’s law for \(\vec{E}\): electric flux through a closed surface equals the enclosed charge.

• Gauss’s law for \(\vec{B}\): magnetic field lines form closed loops — no magnetic monopoles.

• Faraday’s law: a changing magnetic flux induces an electric field (emf).

• Ampère–Maxwell law: magnetic fields arise from currents and changing electric fields.

• Continuity equation: charge is conserved — current outflow equals the decrease of enclosed charge.

Summary of Key Vector Identities

6.11Where Vector Analysis Is Used

Maxwell's equations, treated twice in this chapter, are the clearest demonstration of what vector analysis is for: four statements about \(\vec{\nabla}\cdot\) and \(\vec{\nabla}\times\) contain the whole of classical electromagnetism, including the existence of light. The same operators do similar work elsewhere.

• Electromagnetism. Gauss's law is a statement about divergence, Faraday's and Ampère's about curl. The integral and differential forms are connected by exactly the divergence and Stokes's theorems proved here, which is why those theorems are not an optional extra.

• Conservative forces and potentials. \(\vec{\nabla}\times\vec{F}=\vec 0\) is the test for whether a potential energy exists at all. Gravity and electrostatics pass; friction does not. This is what makes energy conservation usable.

• Fluid dynamics. The divergence of the velocity field measures whether fluid is being compressed, and its curl measures local rotation — literally the rate at which a small paddle wheel would spin, as the rigid-rotation example showed.

• Heat and diffusion. The Laplacian \(\nabla^2 T\) measures how much a point differs from the average of its neighbours, which is why it governs diffusion and why \(\nabla^2 T = 0\) describes steady state (Chapter 10).

• Curvilinear coordinates. The general expressions for gradient, divergence and curl in orthogonal coordinates let the same physics be written in whatever coordinates match the symmetry of the problem, as in Chapter 5.

For the Interested Reader

Vector analysis is where the geometry and the algebra can easily come apart: it is possible to compute a curl correctly for a whole course without ever picturing what it measures. The first item below is the strongest remedy for that. Everything listed is free.

Seeing divergence and curl

3Blue1Brown builds both operators from what they do to a flowing fluid, and then shows why Maxwell's equations say what they say. Twenty minutes here is worth a great deal of symbol-pushing:

Video

Divergence and curl: the language of Maxwell's equations — 3Blue1Brown

Watch on YouTube

Open video on YouTube

The operators and the integral theorems

• OpenStax, Calculus Volume 3, Chapter 6. 6.5 Divergence and Curl, 6.4 Green's Theorem, 6.7 Stokes' Theorem and 6.8 The Divergence Theorem — the same three theorems, with more examples and with the orientation conventions spelled out more slowly than we do.

• Paul's Online Math Notes. Line Integrals and Stokes' Theorem — practice at parametrising curves and surfaces, which is where most of the difficulty in these integrals actually lies.