WaywardHouse

44  Vector integral calculus

Green’s, Stokes’, and the Divergence theorem

A turbine blade sits in a flow of hot gas. You need to know the total heat flux through the blade’s surface — but integrating over every point on that complicated curved surface is brutal. The Divergence theorem says you only need to know the divergence of the heat flux field throughout the interior volume, which is often far simpler to compute.

This is the deepest efficiency in classical calculus: boundary integrals and interior integrals are two faces of the same coin. Green’s theorem, Stokes’ theorem, and the Divergence theorem are the exchange rate.

44.1 Line integrals

A line integral accumulates the effect of a field along a curve. There are two kinds: integrals of scalar fields and integrals of vector fields. The vector case is the one that matters physically — it measures work.

44.1.1 Scalar line integrals

If \(f\) is a scalar field and \(C\) is a curve, the scalar line integral is:

\[\int_C f \, ds\]

where \(ds\) is arc length. This answers questions like “what is the total mass of a wire with variable density \(f\)?” or “what is the average temperature along a path?”

To evaluate it, parametrise the curve: let \(\mathbf{r}(t) = (x(t), y(t), z(t))\) for \(t \in [a, b]\). Then \(ds = |\mathbf{r}'(t)| \, dt\), so:

\[\int_C f \, ds = \int_a^b f(\mathbf{r}(t)) \, |\mathbf{r}'(t)| \, dt\]

The integral is independent of how you parametrise the curve — only the geometry of \(C\) matters, not the speed at which you traverse it.

44.1.2 Vector line integrals (work)

If \(\mathbf{F}\) is a vector field and \(C\) is an oriented curve, the vector line integral is:

\[\int_C \mathbf{F} \cdot d\mathbf{r}\]

This measures the work done by the field \(\mathbf{F}\) on a particle moving along \(C\). The dot product \(\mathbf{F} \cdot d\mathbf{r}\) picks out the component of \(\mathbf{F}\) that is parallel to the direction of travel — only that component does work.

To evaluate, parametrise \(C\) as \(\mathbf{r}(t)\), so \(d\mathbf{r} = \mathbf{r}'(t) \, dt\):

\[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt\]

Writing out in components with \(\mathbf{F} = (P, Q, R)\) and \(d\mathbf{r} = (dx, dy, dz)\):

\[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C P \, dx + Q \, dy + R \, dz\]

Sign convention: reversing the orientation of \(C\) changes the sign of the integral. For a closed curve traversed anti-clockwise (positive orientation), the notation is \(\oint_C \mathbf{F} \cdot d\mathbf{r}\).

Worked example. Compute \(\int_C \mathbf{F} \cdot d\mathbf{r}\) where \(\mathbf{F} = (y, -x)\) and \(C\) is the upper semicircle of the unit circle from \((1, 0)\) to \((-1, 0)\).

Parametrise: \(\mathbf{r}(t) = (\cos t, \sin t)\), \(t \in [0, \pi]\), so \(\mathbf{r}'(t) = (-\sin t, \cos t)\).

Along \(C\): \(\mathbf{F}(\mathbf{r}(t)) = (\sin t, -\cos t)\).

\[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^\pi (\sin t, -\cos t) \cdot (-\sin t, \cos t) \, dt\]

\[= \int_0^\pi (-\sin^2 t - \cos^2 t) \, dt = \int_0^\pi -1 \, dt = -\pi\]

The field does \(-\pi\) units of work on the particle — it opposes the motion.

44.2 Path independence and conservative fields

For many fields, the line integral between two points is the same regardless of which path you take. Such fields are called conservative.

A vector field \(\mathbf{F}\) is conservative on a simply connected domain if and only if any (equivalently all) of the following hold:

  1. \(\oint_C \mathbf{F} \cdot d\mathbf{r} = 0\) for every closed curve \(C\).
  2. \(\mathbf{F} = \nabla f\) for some scalar potential function \(f\).
  3. In 3D: \(\nabla \times \mathbf{F} = \mathbf{0}\) (the field is irrotational; curl is zero).

The equivalence of (2) and (3) was established in Chapter 3. The equivalence of (1) and (2) follows from the Fundamental Theorem for line integrals.

44.2.1 Fundamental Theorem for line integrals

If \(\mathbf{F} = \nabla f\), then:

\[\int_C \mathbf{F} \cdot d\mathbf{r} = f(\mathbf{r}(b)) - f(\mathbf{r}(a))\]

where \(\mathbf{r}(a)\) and \(\mathbf{r}(b)\) are the start and end points of \(C\). The integral depends only on the endpoints — not on the path.

This is the vector analogue of the Fundamental Theorem of Calculus: the integral of a derivative (here, \(\nabla f\)) over a domain (here, the curve) equals the value of the original function at the boundary (here, the two endpoints, with sign).

Finding the potential function. Given \(\mathbf{F} = (P, Q, R)\) with \(\nabla \times \mathbf{F} = \mathbf{0}\), find \(f\) such that \(\nabla f = \mathbf{F}\). Integrate component by component:

\[f(x, y, z) = \int P \, dx + g(y, z)\]

then differentiate with respect to \(y\) and match against \(Q\) to determine \(g\), and so on.

Example. Is \(\mathbf{F} = (2xy + z^2,\; x^2,\; 2xz)\) conservative?

Check \(\nabla \times \mathbf{F}\): compute each component.

\((\nabla \times \mathbf{F})_x = \partial_y(2xz) - \partial_z(x^2) = 0 - 0 = 0\)

\((\nabla \times \mathbf{F})_y = \partial_z(2xy + z^2) - \partial_x(2xz) = 2z - 2z = 0\)

\((\nabla \times \mathbf{F})_z = \partial_x(x^2) - \partial_y(2xy + z^2) = 2x - 2x = 0\)

All zero, so \(\mathbf{F}\) is conservative. Find \(f\):

From \(\partial f / \partial x = 2xy + z^2\), integrate in \(x\): \(f = x^2 y + xz^2 + g(y, z)\).

From \(\partial f / \partial y = x^2\): matches \(Q = x^2\), so \(\partial g / \partial y = 0\).

From \(\partial f / \partial z = 2xz\): matches \(R = 2xz\), so \(\partial g / \partial z = 0\).

Therefore \(f(x, y, z) = x^2 y + xz^2\) (plus an arbitrary constant).

Why conservative fields matter in physics

Gravity is conservative. Electric fields (in electrostatics) are conservative. This means the work done lifting an object or moving a charge depends only on where it starts and where it ends — not on the path. This is why potential energy is a well-defined concept: the number \(f(\mathbf{r})\) at any point is the energy stored.

Friction is not conservative: a longer path dissipates more energy. The difference is that \(\nabla \times \mathbf{F} \neq \mathbf{0}\) for non-conservative forces.

44.3 Green’s theorem

Green’s theorem connects a line integral around a closed curve in the plane to a double integral over the enclosed region.

Statement. Let \(C\) be a positively oriented (anti-clockwise), piecewise smooth, simple closed curve in the \(xy\)-plane, and let \(D\) be the region it encloses. If \(P(x, y)\) and \(Q(x, y)\) have continuous partial derivatives on an open set containing \(D\), then:

\[\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA\]

44.3.1 Where it comes from: cellular decomposition

Divide \(D\) into tiny rectangles. For each rectangle, compute the line integral around its boundary (traversed anti-clockwise). When you sum all these contributions, every interior edge is traversed twice — once in each direction by adjacent rectangles — so those contributions cancel exactly. What survives is only the contribution from the outer boundary of \(D\), which is \(C\).

For each tiny rectangle with width \(\Delta x\) and height \(\Delta y\), the line integral around its boundary approximates:

\[\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \Delta x \, \Delta y\]

Summing over all rectangles and taking the limit gives the double integral on the right of Green’s theorem.

Using Green’s theorem. The theorem converts between two kinds of integrals. Use it when:

  • The line integral \(\oint_C P \, dx + Q \, dy\) is complicated but the integrand \(\partial Q / \partial x - \partial P / \partial y\) is simple.
  • The double integral over \(D\) is complicated but the boundary \(C\) is simple.

Example. Compute \(\oint_C (-y \, dx + x \, dy)\) where \(C\) is the circle \(x^2 + y^2 = 4\) traversed anti-clockwise.

Here \(P = -y\), \(Q = x\), so \(\partial Q / \partial x - \partial P / \partial y = 1 - (-1) = 2\).

By Green’s theorem:

\[\oint_C (-y \, dx + x \, dy) = \iint_D 2 \, dA = 2 \cdot \text{Area}(D) = 2 \cdot \pi(2)^2 = 8\pi\]

The boundary integral reduced to computing an area. Notice that \(\frac{1}{2} \oint_C (-y \, dx + x \, dy) = \text{Area}(D)\) — this is a general formula for area from a line integral, directly from Green’s theorem.

Planimeters: Green’s theorem in hardware

A planimeter is a mechanical device for measuring area. You trace the boundary of a region with the pointer; the rotating wheel integrates \(-y \, dx + x \, dy\) mechanically as you go, reading off the area directly. Engineers used planimeters to measure irregular cross-sections in structural calculations before digital tools. The device is a physical implementation of Green’s theorem.

44.4 Surface integrals

Surface integrals extend integration from curves to surfaces in 3D. As with line integrals, there are scalar and vector versions.

44.4.1 Parametric surfaces

A surface \(S\) can be parametrised by \(\mathbf{r}(u, v) = (x(u,v), y(u,v), z(u,v))\) for \((u, v)\) in some region \(D\) of the parameter plane. The surface element \(d\mathbf{S}\) is the vector area of an infinitesimal patch:

\[d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \, du \, dv\]

Its direction is normal to the surface; its magnitude \(\left|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right|\) is the scalar surface area element \(dS\).

For a surface \(z = g(x, y)\) over a region \(D\) in the \(xy\)-plane:

\[dS = \sqrt{1 + \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2} \, dA\]

44.4.2 Flux integrals

The flux of a vector field \(\mathbf{F}\) through a surface \(S\) is:

\[\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \hat{\mathbf{n}} \, dS\]

where \(\hat{\mathbf{n}}\) is the outward unit normal. Physically, this is the rate at which field lines cross the surface — the volume of fluid per unit time crossing \(S\) if \(\mathbf{F}\) is a velocity field.

Example. Compute the upward flux of \(\mathbf{F} = (0, 0, z)\) through the surface \(z = 1 - x^2 - y^2\) for \(z \geq 0\) (a paraboloid cap).

On this surface, parametrise by \((x, y)\) in the disk \(D: x^2 + y^2 \leq 1\).

The normal vector pointing upward:

\[d\mathbf{S} = \left(-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1\right) dA = (2x, 2y, 1) \, dA\]

So \(\mathbf{F} \cdot d\mathbf{S} = (0)(2x) + (0)(2y) + z \cdot 1 = z = 1 - x^2 - y^2\).

\[\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D (1 - x^2 - y^2) \, dA\]

In polar coordinates (\(x = r\cos\theta\), \(y = r\sin\theta\)):

\[= \int_0^{2\pi} \int_0^1 (1 - r^2) \, r \, dr \, d\theta = 2\pi \int_0^1 (r - r^3) \, dr = 2\pi \left[\frac{1}{2} - \frac{1}{4}\right] = \frac{\pi}{2}\]

44.5 The Divergence theorem

The Divergence theorem (Gauss’s theorem) extends Green’s theorem to three dimensions, relating the flux through a closed surface to the divergence integrated over the enclosed volume.

Statement. Let \(V\) be a bounded region in \(\mathbb{R}^3\) with piecewise smooth boundary \(\partial V\) (oriented with outward normal). If \(\mathbf{F}\) has continuous partial derivatives on an open set containing \(V\), then:

\[\oiint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV\]

44.5.1 Physical interpretation

\(\nabla \cdot \mathbf{F}\) at a point measures the net outward flow per unit volume — how much the field is “spreading out” there. The Divergence theorem says: sum up all that local spreading (the volume integral), and you get the total outward flow through the boundary (the surface integral). There is no other way out. Any flow created in the interior must eventually cross the boundary.

This is the mathematical statement of conservation laws. In fluid dynamics: if a fluid has no sources or sinks (incompressible, \(\nabla \cdot \mathbf{F} = 0\)), then the net flux through any closed surface is zero — what flows in must flow out.

44.5.2 Cellular decomposition in 3D

The argument is identical to Green’s theorem but in 3D. Subdivide \(V\) into tiny cubes. For each cube, compute the outward flux through its six faces. On each interior face, two adjacent cubes share that face — one uses it as an outward face, the other as an inward face. Their contributions cancel. What remains is the flux through the outer faces of \(V\) only.

For a cube of volume \(\Delta V\), the net outward flux through its faces approximates \((\nabla \cdot \mathbf{F}) \, \Delta V\) (this is precisely the definition of divergence). Summing over all cubes gives the Divergence theorem.

Example. Compute the flux of \(\mathbf{F} = (x, y, z)\) outward through the sphere \(x^2 + y^2 + z^2 = a^2\).

Direct computation is possible but requires parametrising the sphere. The Divergence theorem is far faster:

\(\nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3\).

Volume of the sphere: \(V = \frac{4}{3}\pi a^3\).

\[\oiint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V 3 \, dV = 3 \cdot \frac{4}{3}\pi a^3 = 4\pi a^3\]

No surface integral needed — just multiplication.

Gauss’s law (Maxwell’s equations). The electric field \(\mathbf{E}\) due to a charge distribution \(\rho\) satisfies:

\[\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}\]

This is the differential form. Applying the Divergence theorem over a closed surface \(\partial V\) enclosing volume \(V\):

\[\oiint_{\partial V} \mathbf{E} \cdot d\mathbf{S} = \iiint_V \frac{\rho}{\varepsilon_0} \, dV = \frac{Q_{\text{enc}}}{\varepsilon_0}\]

where \(Q_{\text{enc}}\) is the total enclosed charge. This is Gauss’s law in integral form — the total electric flux through any closed surface equals the enclosed charge divided by \(\varepsilon_0\). It is the Divergence theorem with \(\mathbf{F} = \mathbf{E}\).

44.6 Stokes’ theorem

Stokes’ theorem is the 3D analogue of Green’s theorem. It relates the line integral around a closed curve to the flux of the curl through any surface bounded by that curve.

Statement. Let \(S\) be an oriented, piecewise smooth surface with boundary \(\partial S\) — a closed curve. Let \(\mathbf{F}\) have continuous partial derivatives on an open set containing \(S\). Then:

\[\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}\]

The orientation of \(\partial S\) is determined by the right-hand rule: curl your right hand’s fingers in the direction you traverse the boundary, and your thumb points in the direction of the surface normal.

44.6.1 Connection to Green’s theorem

Green’s theorem is the special case where \(S\) is a flat region in the \(xy\)-plane. For \(S \subset \{z = 0\}\) with normal \(\hat{\mathbf{k}}\):

\[(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{k}} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\]

Substituting into Stokes’ theorem recovers Green’s theorem exactly. Stokes’ theorem can be understood as “un-flattening” Green’s theorem — lifting it from a 2D region to a curved surface in 3D.

44.6.2 The cellular argument for Stokes

Subdivide \(S\) into small surface patches. The line integral around each patch’s boundary traverses interior edges twice in opposite directions — those cancel. The sum over all patches reduces to the line integral around the outer boundary \(\partial S\). For each patch, the boundary integral approximates \((\nabla \times \mathbf{F}) \cdot d\mathbf{S}\) (this is the definition of curl). Summing gives Stokes’ theorem.

Key insight: the surface \(S\) does not need to be flat, and it is not unique. Any two surfaces spanning the same boundary curve give the same flux integral of \(\nabla \times \mathbf{F}\) — provided \(\mathbf{F}\) is smooth throughout. Engineers exploit this: choose whichever surface makes the integral easiest.

Example. Verify Stokes’ theorem for \(\mathbf{F} = (-y^2, x, z^2)\) where \(S\) is the upper hemisphere \(x^2 + y^2 + z^2 = 1\), \(z \geq 0\), and \(\partial S\) is the unit circle in the \(xy\)-plane.

Boundary integral: on \(\partial S\), parametrise \(\mathbf{r}(t) = (\cos t, \sin t, 0)\):

\[\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} (-\sin^2 t, \cos t, 0) \cdot (-\sin t, \cos t, 0) \, dt\]

\[= \int_0^{2\pi} (\sin^3 t + \cos^2 t) \, dt = 0 + \pi = \pi\]

Surface integral: compute \(\nabla \times \mathbf{F}\):

\[\nabla \times \mathbf{F} = \left(\frac{\partial z^2}{\partial y} - \frac{\partial x}{\partial z},\; \frac{\partial(-y^2)}{\partial z} - \frac{\partial z^2}{\partial x},\; \frac{\partial x}{\partial x} - \frac{\partial(-y^2)}{\partial y}\right) = (0, 0, 1 + 2y)\]

Applying Stokes’ theorem is easier via the flat disk \(D: x^2 + y^2 \leq 1\), \(z = 0\) (same boundary, much simpler surface):

\[\iint_D (0, 0, 1 + 2y) \cdot (0, 0, 1) \, dA = \iint_D (1 + 2y) \, dA = \pi \cdot 1^2 + 2 \iint_D y \, dA = \pi + 0 = \pi\]

Both sides give \(\pi\). \(\checkmark\)

Ampère’s law in integral form. The magnetic field \(\mathbf{B}\) satisfies \(\nabla \times \mathbf{B} = \mu_0 \mathbf{J}\) (Ampère’s law, differential form). Stokes’ theorem converts this to:

\[\oint_{\partial S} \mathbf{B} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{B}) \cdot d\mathbf{S} = \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{S} = \mu_0 I_{\text{enc}}\]

The circulation of the magnetic field around a closed loop equals \(\mu_0\) times the current threading through any surface bounded by that loop. This is the statement every electrical engineering student uses when computing fields around wires.

44.7 The unified picture

Green’s, the Divergence theorem, and Stokes’ are all instances of one principle, which can be stated abstractly as:

\[\int_{\partial \Omega} \omega = \int_\Omega d\omega\]

where \(\omega\) is a differential form and \(d\omega\) is its exterior derivative. The left side integrates over the boundary; the right side integrates the derivative over the interior. The Fundamental Theorem of Calculus, \(\int_a^b f'(x) \, dx = f(b) - f(a)\), is also an instance — here the “boundary” of \([a, b]\) is just the two endpoints.

Theorem Domain Boundary Integrand (interior)
FTC Interval \([a, b]\) Two endpoints \(f'(x)\)
Green’s 2D region \(D\) Closed curve \(\partial D\) \(\partial Q / \partial x - \partial P / \partial y\)
Stokes’ Surface \(S\) Closed curve \(\partial S\) \(\nabla \times \mathbf{F}\)
Divergence Volume \(V\) Closed surface \(\partial V\) \(\nabla \cdot \mathbf{F}\)

The pattern is inescapable. All of calculus, in one row of a table.

Strategy: which way to go

Given a problem, decide which direction the theorem should run:

  • If the boundary integral looks hard but the interior is simple: Green’s / Stokes’ / Divergence converts boundary → interior.
  • If the interior integral looks hard but the boundary is simple: run it the other way.
  • If \(\nabla \cdot \mathbf{F} = 0\) (solenoidal): the flux through any closed surface is zero. No computation needed.
  • If \(\nabla \times \mathbf{F} = \mathbf{0}\) (irrotational): the line integral around any closed curve is zero. Use the potential function.

The theorems are not formulas to plug into — they are a choice of which representation of the problem is simpler.

44.8 Applications

44.8.1 Electric flux and Gauss’s law

An isolated point charge \(q\) sits at the origin. Compute the electric flux through a sphere of radius \(r\).

The electric field is \(\mathbf{E} = \frac{q}{4\pi\varepsilon_0 r^2} \hat{\mathbf{r}}\). On the sphere of radius \(r\), the field is radial with constant magnitude \(E = q / (4\pi\varepsilon_0 r^2)\), and the outward normal \(\hat{\mathbf{n}} = \hat{\mathbf{r}}\), so \(\mathbf{E} \cdot \hat{\mathbf{n}} = E\) everywhere.

\[\oiint_S \mathbf{E} \cdot d\mathbf{S} = E \cdot 4\pi r^2 = \frac{q}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}\]

This equals the enclosed charge divided by \(\varepsilon_0\), confirming Gauss’s law. Notice the \(r\) cancelled — the flux is the same through any sphere centred on the charge. This is a consequence of the Divergence theorem and the fact that \(\nabla \cdot \mathbf{E} = 0\) everywhere except at the charge itself.

44.8.2 Mass flow through a surface (fluid mechanics)

A velocity field \(\mathbf{v} = (2x, -y, z)\) describes fluid flow in a region. Is mass accumulating inside the unit cube \([0,1]^3\)?

\(\nabla \cdot \mathbf{v} = 2 - 1 + 1 = 2 > 0\).

By the Divergence theorem, the net outward mass flux is:

\[\oiint_{\partial V} \mathbf{v} \cdot d\mathbf{S} = \iiint_V 2 \, dV = 2\]

Two units of mass per unit time are flowing out of the cube. The divergence is positive throughout, so the fluid is expanding — it acts like a source term. In an incompressible flow (water, to a first approximation), this is impossible: \(\nabla \cdot \mathbf{v}\) must be zero everywhere.

44.8.3 Work done against gravity (conservative field)

Gravity is \(\mathbf{g} = (0, 0, -g)\) with potential \(\phi = gz\) (so \(\mathbf{g} = -\nabla\phi\)). A mass \(m\) is moved along a complicated path from the ground level \((0, 0, 0)\) to a height \((x_1, y_1, h)\).

Since \(\mathbf{F}_{\text{grav}} = m\mathbf{g} = -m\nabla(gz)\), it is conservative. Work done by gravity:

\[W = \int_C \mathbf{F}_{\text{grav}} \cdot d\mathbf{r} = -m g z \Big|_{z=0}^{z=h} = -mgh\]

The negative sign indicates gravity does negative work (you are lifting against it). Work done against gravity is \(+mgh\) — the familiar result from introductory physics, now derived precisely from path independence.

44.8.4 Stokes’ theorem and Faraday’s law

Faraday’s law of electromagnetic induction states \(\nabla \times \mathbf{E} = -\partial \mathbf{B} / \partial t\). Apply Stokes’ theorem to a surface \(S\) bounded by a circuit loop \(C\):

\[\oint_C \mathbf{E} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{E}) \cdot d\mathbf{S} = -\frac{d}{dt} \iint_S \mathbf{B} \cdot d\mathbf{S} = -\frac{d\Phi_B}{dt}\]

The EMF (voltage) around the circuit equals the negative rate of change of magnetic flux through it. This is the principle behind every transformer and generator ever built.

44.9 Exercises

These exercises build from evaluation to application. The first two establish mechanical fluency; the later ones require choosing which theorem to apply and why.

1. Line integral along a parametric curve. A particle moves along the helix \(\mathbf{r}(t) = (\cos t, \sin t, t)\) for \(t \in [0, 2\pi]\) in the force field \(\mathbf{F} = (-y, x, z)\). Compute the work done \(W = \int_C \mathbf{F} \cdot d\mathbf{r}\).

2. Conservative field — find the potential, evaluate by endpoints. Show that \(\mathbf{F} = (2xy + z,\; x^2,\; x + 2z)\) is conservative, find its potential function \(f\), and evaluate \(\int_C \mathbf{F} \cdot d\mathbf{r}\) from \((0, 0, 0)\) to \((1, 2, 3)\) along any path.

3. Green’s theorem — line integral to double integral. Compute \(\oint_C (x^2 - y) \, dx + (x + y^2) \, dy\) where \(C\) is the boundary of the triangle with vertices \((0,0)\), \((2,0)\), \((0,2)\), traversed anti-clockwise.

4. Electric flux through a hemisphere. Compute the upward flux of \(\mathbf{E} = (x, y, 2z)\) through the upper hemisphere \(S: x^2 + y^2 + z^2 = 9\), \(z \geq 0\) (with outward normal pointing away from the origin).

5. Divergence theorem — convert surface integral to volume integral. Let \(\mathbf{F} = (x^3, y^3, z^3)\). Compute the outward flux of \(\mathbf{F}\) through the closed surface of the cylinder \(x^2 + y^2 \leq 4\), \(0 \leq z \leq 3\).

6. Stokes’ theorem — Ampère’s law application. A long straight wire carries current \(I\) along the \(z\)-axis. The magnetic field is \(\mathbf{B} = \frac{\mu_0 I}{2\pi(x^2 + y^2)}(-y, x, 0)\). Use Stokes’ theorem to compute \(\oint_C \mathbf{B} \cdot d\mathbf{r}\) where \(C\) is the circle \(x^2 + y^2 = R^2\) in the \(xy\)-plane.

44.10 Where this goes

The three theorems in this chapter are the operational heart of classical field theory. The next chapter, Fourier series and PDEs, puts them to work in a new setting: instead of integrating over a domain once, you decompose a function into sinusoidal modes and work with each mode independently. The Divergence theorem reappears immediately in the derivation of the heat equation and the wave equation — those PDEs arise precisely from applying it to a tiny control volume and letting the volume shrink to a point.

Complex analysis (the chapter after that) makes this picture even stranger and more powerful: in the complex plane, the Cauchy integral theorem is Green’s theorem in disguise, and the residue theorem for computing integrals is a direct consequence of Stokes’ theorem applied to a deformed contour.

The unifying thread is the generalised Stokes’ theorem \(\int_{\partial \Omega} \omega = \int_\Omega d\omega\). Once you have seen that formulation, you cannot unsee it — every integration theorem you encounter from here on is a special case.

Where this shows up

  • A mechanical engineer computing the torque on a rotating shaft uses the surface integral of the stress tensor — a generalisation of flux.
  • A geophysicist inverting gravity survey data uses Gauss’s theorem to relate surface measurements to subsurface mass distribution.
  • A climate scientist computing net CO₂ flux into the ocean integrates the divergence of a diffusion field over a volume — the Divergence theorem makes global estimates tractable.
  • An electrical engineer designing an antenna housing uses Gauss’s law (Divergence theorem) to verify shielding effectiveness from bulk material properties rather than surface-by-surface calculation.

The theorems do not just simplify calculation. In each case, they are the reason the physical law can be stated locally (as a PDE or a differential form) and globally (as a conservation statement). That equivalence is not obvious — it is exactly what Green, Stokes, and Gauss proved.