Calculus · Chapter 16

Vector Calculus

This chapter studies the calculus of vector fields — functions that attach a vector to every point of space. Line integrals measure work done by a force along a curve; surface integrals measure flux across a surface. The three great theorems — Green's, Stokes', and the Divergence Theorem — are higher-dimensional cousins of the Fundamental Theorem of Calculus.

16.1

Vector Fields

Assigning a vector to each point of the plane or of space.

Vector Field

A vector field on a region \(D\) is a function \(\mathbf{F}\) that assigns to each point a vector. On a plane region \(D\subseteq\mathbb{R}^2\) it returns a 2-D vector \(\mathbf{F}(x,y)\); on a solid region \(E\subseteq\mathbb{R}^3\) it returns a 3-D vector \(\mathbf{F}(x,y,z)\). The arrow for each vector is drawn starting at the point it belongs to.

Writing \(\mathbf{F}\) in components separates it into scalar functions (scalar fields): \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}\) in the plane, or \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k}\) in space. The field is continuous exactly when each component is continuous.

Velocity & Force Fields

A velocity field attaches a flow vector to each point — wind patterns, ocean currents, or fluid moving through a pipe. A force field attaches a force vector to each point, such as the gravitational pull toward a mass or the electric force around a charge.

Gravitational field (inverse-square law)

\[ \mathbf{F}(\mathbf{x}) = -\frac{mMG}{|\mathbf{x}|^{3}}\,\mathbf{x} \]

Electric (Coulomb) field per unit charge

\[ \mathbf{F}(\mathbf{x}) = \frac{\varepsilon qQ}{|\mathbf{x}|^{3}}\,\mathbf{x}, \qquad \mathbf{E}(\mathbf{x}) = \frac{1}{q}\,\mathbf{F}(\mathbf{x}) \]

Gradient (Conservative) Fields

The gradient of a scalar function \(f\) is itself a vector field, \(\nabla f\), called a gradient field. A vector field \(\mathbf{F}\) is conservative if \(\mathbf{F}=\nabla f\) for some scalar \(f\); that \(f\) is called a potential function. Gradient vectors point in the direction of steepest increase and are perpendicular to level curves.

Gradient field in two and three variables

\[ \nabla f = f_x\,\mathbf{i}+f_y\,\mathbf{j}, \qquad \nabla f = f_x\,\mathbf{i}+f_y\,\mathbf{j}+f_z\,\mathbf{k} \]

EXAMPLE 1 Sketching a rotational field

Describe the field \(\mathbf{F}(x,y)=-y\,\mathbf{i}+x\,\mathbf{j}\).

Solution

Test the angle to the position vector \(\mathbf{x}=x\,\mathbf{i}+y\,\mathbf{j}\): the dot product \(\mathbf{x}\cdot\mathbf{F}=-xy+yx=0\), so \(\mathbf{F}\) is always perpendicular to \(\mathbf{x}\).

A vector perpendicular to the position vector is tangent to a circle centered at the origin, so the arrows wrap counterclockwise around the origin.

Its length \(|\mathbf{F}|=\sqrt{y^2+x^2}=|\mathbf{x}|\) equals the radius, so arrows grow with distance — the field describes rigid rotation about the origin.

EXAMPLE 6 A gradient field from a contour map

Find the gradient field of \(f(x,y)=x^2y-y^3\).

Solution

Differentiate: \(\nabla f = f_x\,\mathbf{i}+f_y\,\mathbf{j} = 2xy\,\mathbf{i}+(x^2-3y^2)\,\mathbf{j}.\)

Plotted over a contour map of \(f\), the gradient arrows are perpendicular to each level curve and are longest where the level curves crowd together (steep terrain).

Figure 16.1.1 Original schematic of the rotational field \(\mathbf{F}(x,y)=-y\,\mathbf{i}+x\,\mathbf{j}\): every arrow is tangent to a circle centered at the origin, with length equal to the radius.

Instructor Note

In fluid mechanics you'll meet vector fields constantly — velocity fields for flow, pressure gradients as force fields. Get comfortable reading them as "arrows at every point." When you see \(-y\,\mathbf{i}+x\,\mathbf{j}\), immediately picture rotation; that intuition saves time later in the curl and circulation sections.

16.2

Line Integrals

Integrating along a curve instead of along an interval.

Line Integral with Respect to Arc Length

A line integral generalizes the single integral: instead of integrating over an interval \([a,b]\), we integrate a function over a curve \(C\). We chop \(C\) into small arcs, sample \(f\) on each, weight by the arc length \(\Delta s_i\), and take the limit of the resulting Riemann-type sum.

Line integral of a scalar field (plane)

\[ \int_C f(x,y)\,ds = \int_a^b f\big(x(t),y(t)\big)\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\;dt \]

Line integrals with respect to x and y

\[ \int_C f\,dx=\int_a^b f\,x'(t)\,dt,\qquad \int_C f\,dy=\int_a^b f\,y'(t)\,dt \]

Line integral in space (arc length)

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

Line Integral of a Vector Field (Work)

The work done by a force field \(\mathbf{F}\) in moving a particle along \(C\) is the integral of the tangential component \(\mathbf{F}\cdot\mathbf{T}\) over the curve. Unlike arc-length integrals, this depends on orientation: reversing the direction of travel flips the sign.

Work integral of a vector field

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

Component form

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

EXAMPLE 1 Scalar line integral over a semicircle

Evaluate \(\displaystyle\int_C (2+x^2y)\,ds\), where \(C\) is the upper half of the unit circle.

Solution

Parametrize: \(x=\cos t,\ y=\sin t,\ 0\le t\le\pi\), so \(\sqrt{(x')^2+(y')^2}=1\).

Substitute: \(\displaystyle\int_0^\pi (2+\cos^2 t\sin t)\,dt = \left[\,2t+\tfrac{-\cos^3 t}{3}\,\right]_0^\pi.\)

Evaluate: result is \(\displaystyle 2\pi+\tfrac{2}{3}.\)

EXAMPLE 7 Work along a quarter circle

Find the work done by \(\mathbf{F}(x,y)=x^2\,\mathbf{i}-xy\,\mathbf{j}\) along the quarter circle \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}\), \(0\le t\le \tfrac{\pi}{2}\).

Solution

Compose: \(\mathbf{F}(\mathbf{r}(t))=\cos^2 t\,\mathbf{i}-\cos t\sin t\,\mathbf{j}\), and \(\mathbf{r}'(t)=-\sin t\,\mathbf{i}+\cos t\,\mathbf{j}.\)

Dot product: \(\mathbf{F}\cdot\mathbf{r}'=-2\cos^2 t\sin t.\)

Integrate: \(\displaystyle\int_0^{\pi/2}-2\cos^2 t\sin t\,dt=\left[\tfrac{2\cos^3 t}{3}\right]_0^{\pi/2}=-\tfrac{2}{3}.\) Negative work — the field opposes the motion.

Path dependence. Two curves with the same endpoints can give different line-integral values. The value of a vector line integral generally depends on the whole path, not just where it starts and ends — until we reach conservative fields in §16.3.

Instructor Note

Watch the difference between \(ds\) (arc length, orientation-independent) and \(d\mathbf{r}\) (vector, orientation-dependent). Mass and center-of-mass problems use \(ds\); work problems use \(d\mathbf{r}\). Mixing them up is the single most common exam mistake I see.

16.3

The Fundamental Theorem for Line Integrals

When a line integral cares only about the endpoints.

Fundamental Theorem for Line Integrals

Conditions: Let \(C\) be a smooth curve given by \(\mathbf{r}(t)\), \(a\le t\le b\), and let \(f\) be a differentiable function whose gradient \(\nabla f\) is continuous on \(C\).

Conclusion: \[ \int_C \nabla f\cdot d\mathbf{r}=f\big(\mathbf{r}(b)\big)-f\big(\mathbf{r}(a)\big). \] The integral of a gradient field is just the net change in the potential between the endpoints.

Independence of Path

A line integral \(\int_C \mathbf{F}\cdot d\mathbf{r}\) is independent of path if it gives the same value for any two paths sharing the same endpoints. Line integrals of conservative fields are independent of path.

Closed-Path Criterion

Conditions: \(\mathbf{F}\) is continuous on an open region \(D\).

Conclusion: \(\int_C \mathbf{F}\cdot d\mathbf{r}\) is independent of path in \(D\) if and only if \(\oint_C \mathbf{F}\cdot d\mathbf{r}=0\) for every closed path \(C\) in \(D\).

Test for a Conservative Field

Conditions: \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}\) on an open, simply-connected region, with \(P,Q\) having continuous first-order partials.

Conclusion: \(\mathbf{F}\) is conservative if and only if \[ \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\quad\text{throughout the region.} \]

EXAMPLE 4 Finding a potential function

Show \(\mathbf{F}(x,y)=(3+2xy)\,\mathbf{i}+(x^2-3y^2)\,\mathbf{j}\) is conservative and find its potential \(f\).

Solution

Test: \(P_y=2x\) and \(Q_x=2x\) agree on all of \(\mathbb{R}^2\), so \(\mathbf{F}\) is conservative.

Partial-integrate \(f_x=3+2xy\) in \(x\): \(f=3x+x^2y+g(y).\)

Differentiate in \(y\): \(f_y=x^2+g'(y)\). Match to \(Q=x^2-3y^2\Rightarrow g'(y)=-3y^2\Rightarrow g(y)=-y^3.\)

Potential: \(f(x,y)=3x+x^2y-y^3+K.\)

Conservation of Energy

For a conservative force \(\mathbf{F}=-\nabla P\) (with \(P\) the potential energy), the work done equals the change in kinetic energy and also the drop in potential energy. Combining these gives \(P_A+K_A=P_B+K_B\): the total mechanical energy stays constant — the reason such fields are called "conservative."

Instructor Note

Before grinding through a messy work integral, always test \(P_y\stackrel{?}{=}Q_x\). If it passes, find the potential and just subtract endpoint values — minutes instead of pages. This shortcut is gold in dynamics problems involving gravity or springs.

16.4

Green's Theorem

A line integral around a closed curve equals a double integral over the region inside.

Green's Theorem

Conditions: \(C\) is a positively oriented (counterclockwise), piecewise-smooth, simple closed curve bounding a region \(D\); \(P\) and \(Q\) have continuous partials on an open region containing \(D\).

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

Area enclosed by a curve (from Green's Theorem)

\[ A=\oint_C x\,dy=-\oint_C y\,dx=\tfrac{1}{2}\oint_C x\,dy-y\,dx \]

EXAMPLE 2 Trading a hard line integral for a double integral

Evaluate \(\displaystyle\oint_C (3y-e^{\sin x})\,dx+\left(7x+\sqrt{y^4+1}\right)dy\), where \(C\) is the circle \(x^2+y^2=9\).

Solution

The direct integral is hopeless. Apply Green's Theorem with \(Q_x=7,\ P_y=3\), so \(Q_x-P_y=4.\)

\(\displaystyle\iint_D 4\,dA = 4\cdot(\text{area of disk of radius }3)=4\cdot 9\pi=36\pi.\)

EXAMPLE 3 Area of an ellipse

Find the area enclosed by \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

Solution

Parametrize: \(x=a\cos t,\ y=b\sin t,\ 0\le t\le 2\pi.\)

Use \(A=\tfrac12\oint_C x\,dy-y\,dx=\tfrac12\int_0^{2\pi} ab\,dt=\pi ab.\)

Figure 16.4.1 Original sketch: region \(D\) with its boundary \(C\) traversed counterclockwise (positive orientation) so that \(D\) stays on the left.

Instructor Note

Green's Theorem works both ways. Usually you convert a nasty line integral into an easy area integral — but the reverse trick (computing area as a boundary line integral) is exactly how planimeters and CAD area tools work in practice.

16.5

Curl and Divergence

Two derivative-like operations: one yields a vector, the other a scalar.

Curl

The curl of a vector field measures its rotational tendency. Where \(\mathbf{F}\) is a fluid velocity, a tiny paddle wheel spins fastest about the axis pointing along \(\operatorname{curl}\mathbf{F}\), and the magnitude tells how fast. A field with zero curl everywhere is called irrotational.

Curl (del cross F)

\[ \operatorname{curl}\mathbf{F}=\nabla\times\mathbf{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{k} \]

Divergence

The divergence of a vector field is a scalar measuring net outflow per unit volume at a point. For a fluid, positive divergence marks a source (fluid expanding outward) and negative divergence a sink. A field with zero divergence is incompressible.

Divergence (del dot F)

\[ \operatorname{div}\mathbf{F}=\nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} \]

Curl of a Gradient is Zero

Conditions: \(f\) has continuous second-order partials.

Conclusion: \(\operatorname{curl}(\nabla f)=\mathbf{0}\). Hence if \(\mathbf{F}\) is conservative, then \(\operatorname{curl}\mathbf{F}=\mathbf{0}\) — a quick way to show a field is not conservative.

Conservative Test in Space

Conditions: \(\mathbf{F}\) is defined on all of \(\mathbb{R}^3\) (or a simply-connected domain) with continuous partials, and \(\operatorname{curl}\mathbf{F}=\mathbf{0}\).

Conclusion: \(\mathbf{F}\) is conservative.

Divergence of a Curl is Zero

Conditions: \(\mathbf{F}\) has continuous second-order partials.

Conclusion: \(\operatorname{div}(\operatorname{curl}\mathbf{F})=0\). So a field with nonzero divergence cannot be written as the curl of another field.

Laplace operator

\[ \nabla^2 f=\operatorname{div}(\nabla f)=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2} \]

EXAMPLE 1 Computing a curl

For \(\mathbf{F}=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}\), find \(\operatorname{curl}\mathbf{F}\).

Solution

Set up the symbolic determinant \(\nabla\times\mathbf{F}\) with rows \(\mathbf{i},\mathbf{j},\mathbf{k}\); \(\partial_x,\partial_y,\partial_z\); \(xz,xyz,-y^2\).

Expand: \(\operatorname{curl}\mathbf{F}=(-2y-xy)\,\mathbf{i}-(0-x)\,\mathbf{j}+(yz-0)\,\mathbf{k}.\)

Result: \(\operatorname{curl}\mathbf{F}=-y(2+x)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k}.\) Nonzero, so this field is not conservative.

Instructor Note

Mnemonic that never fails me: curl is a cross product (gives a vector, measures spin); divergence is a dot product (gives a scalar, measures spreading). In electromagnetics these two operators ARE Maxwell's equations — learn them cold now.

16.6

Parametric Surfaces and Their Areas

Describing surfaces with two parameters and measuring their area.

Parametric Surface

Just as a curve is traced by a vector function of one parameter, a surface is traced by a vector function of two parameters: \(\mathbf{r}(u,v)=x(u,v)\,\mathbf{i}+y(u,v)\,\mathbf{j}+z(u,v)\,\mathbf{k}\) for \((u,v)\) in a domain \(D\). Holding one parameter fixed sweeps out grid curves that lie on the surface.

Tangent vectors and normal

\[ \mathbf{r}_u=x_u\,\mathbf{i}+y_u\,\mathbf{j}+z_u\,\mathbf{k},\quad \mathbf{r}_v=x_v\,\mathbf{i}+y_v\,\mathbf{j}+z_v\,\mathbf{k},\quad \mathbf{n}\parallel \mathbf{r}_u\times\mathbf{r}_v \]

Smooth Surface

A surface is smooth where \(\mathbf{r}_u\times\mathbf{r}_v\neq\mathbf{0}\) — it has no corners and a well-defined tangent plane. That cross product is normal to the tangent plane at each point.

Surface area of a parametric surface

\[ A(S)=\iint_D |\mathbf{r}_u\times\mathbf{r}_v|\,dA \]

Surface area of a graph z = g(x, y)

\[ A(S)=\iint_D \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}\,dA \]

EXAMPLE 4 Parametrizing a sphere

Find a parametric representation of the sphere \(x^2+y^2+z^2=a^2\).

Solution

Use spherical angles \(\phi\) (from the \(z\)-axis) and \(\theta\) (around) as parameters.

\(x=a\sin\phi\cos\theta,\quad y=a\sin\phi\sin\theta,\quad z=a\cos\phi.\)

Domain \(0\le\phi\le\pi,\ 0\le\theta\le 2\pi\). Grid curves of constant \(\phi\) are latitude circles; constant \(\theta\) are meridians.

EXAMPLE 10 Surface area of a sphere

Find the surface area of a sphere of radius \(a\).

Solution

From the parametrization, \(|\mathbf{r}_\phi\times\mathbf{r}_\theta|=a^2\sin\phi.\)

Integrate: \(\displaystyle A=\int_0^{2\pi}\!\!\int_0^{\pi} a^2\sin\phi\,d\phi\,d\theta = a^2(2\pi)(2)=4\pi a^2.\)

Instructor Note

Parametric surfaces are the language of CAD and finite-element meshing. The cross product \(\mathbf{r}_u\times\mathbf{r}_v\) you compute here is exactly the "surface normal" your CAD software draws — same math, different notation.

16.7

Surface Integrals

Integrating over a surface — including the flux of a vector field.

Surface Integral of a Scalar Field

The surface integral of \(f\) over \(S\) is the surface analogue of an arc-length line integral: chop \(S\) into patches, sample \(f\), weight by patch area, and sum. When \(f=1\) it returns the surface area; with a density it gives mass and center of mass.

Surface integral over a parametric surface

\[ \iint_S f(x,y,z)\,dS=\iint_D f\big(\mathbf{r}(u,v)\big)\,|\mathbf{r}_u\times\mathbf{r}_v|\,dA \]

Surface integral over a graph z = g(x, y)

\[ \iint_S f\,dS=\iint_D f\big(x,y,g(x,y)\big)\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\;dA \]

Oriented Surfaces

An orientable surface has two sides; choosing a continuously-varying unit normal \(\mathbf{n}\) fixes an orientation. A closed surface is positively oriented when its normals point outward. The Möbius strip is the classic non-orientable counterexample — it has only one side.

Flux of a Vector Field

The surface integral (flux) of a continuous field \(\mathbf{F}\) across an oriented surface \(S\) with unit normal \(\mathbf{n}\) is \[ \iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_S \mathbf{F}\cdot\mathbf{n}\,dS=\iint_D \mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,dA. \] Physically it measures the rate at which fluid crosses \(S\).

Flux across a graph (upward orientation)

\[ \iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_D\left(-P\frac{\partial g}{\partial x}-Q\frac{\partial g}{\partial y}+R\right)dA \]

EXAMPLE 4 Flux across the unit sphere

Find the flux of \(\mathbf{F}(x,y,z)=z\,\mathbf{i}+y\,\mathbf{j}+x\,\mathbf{k}\) across the unit sphere.

Solution

Use the spherical parametrization; \(\mathbf{r}_\phi\times\mathbf{r}_\theta\) has known components and \(\mathbf{F}\cdot(\mathbf{r}_\phi\times\mathbf{r}_\theta)\) is built by substitution.

After integrating over \(0\le\phi\le\pi,\ 0\le\theta\le2\pi\), the cross terms vanish and the flux equals \(\dfrac{4\pi}{3}.\)

Applications: Electric Flux & Heat Flow

If \(\mathbf{E}\) is an electric field, \(\iint_S \mathbf{E}\cdot d\mathbf{S}\) is the electric flux (Gauss's Law links it to enclosed charge). If \(u\) is temperature, the heat-flow field \(\mathbf{F}=-K\nabla u\) gives the rate of heat transfer across \(S\) as a flux integral.

Instructor Note

Flux is the workhorse of thermofluids. "Mass flow rate through a duct," "heat rate through a wall," "current through a cross-section" — all are flux integrals \(\iint_S \mathbf{F}\cdot\mathbf{n}\,dS\). Learn to spot the pattern and you'll read engineering papers far faster.

16.8

Stokes' Theorem

A higher-dimensional Green's Theorem relating a surface to its boundary curve.

Stokes' Theorem

Conditions: \(S\) is an oriented, piecewise-smooth surface bounded by a simple, closed, piecewise-smooth boundary curve \(C\) with positive orientation; \(\mathbf{F}\) has continuous partials on an open region containing \(S\).

Conclusion: \[ \oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}. \] The circulation of \(\mathbf{F}\) around \(C\) equals the flux of \(\operatorname{curl}\mathbf{F}\) through any surface spanning \(C\).

Green is a special case. When \(S\) is flat in the \(xy\)-plane with upward normal \(\mathbf{k}\), Stokes' Theorem reduces exactly to the vector form of Green's Theorem.

EXAMPLE 1 Circulation via a spanning surface

Evaluate \(\oint_C \mathbf{F}\cdot d\mathbf{r}\) for \(\mathbf{F}=-y^2\,\mathbf{i}+x\,\mathbf{j}+z^2\,\mathbf{k}\), where \(C\) is the intersection of \(y+z=2\) and \(x^2+y^2=1\), oriented counterclockwise from above.

Solution

Compute \(\operatorname{curl}\mathbf{F}=(1+2y)\,\mathbf{k}.\)

Pick the flat elliptical region \(S\) in the plane \(z=2-y\) bounded by \(C\); its projection is the disk \(x^2+y^2\le1.\)

\(\displaystyle\iint_D (1+2y)\,dA=\int_0^{2\pi}\!\!\int_0^1 (1+2r\sin\theta)\,r\,dr\,d\theta=\tfrac12(2\pi)+0=\pi.\)

EXAMPLE 2 Surface integral from boundary values alone

Compute \(\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}\) for \(\mathbf{F}=xz\,\mathbf{i}+yz\,\mathbf{j}+xy\,\mathbf{k}\), where \(S\) is the part of \(x^2+y^2+z^2=4\) inside \(x^2+y^2=1\), above the \(xy\)-plane.

Solution

By Stokes', replace the surface integral with the line integral over the boundary circle \(C:\ x^2+y^2=1,\ z=\sqrt3.\)

Parametrize \(C\) and integrate; the two terms cancel, giving the value \(0.\)

Any other surface sharing the same boundary \(C\) gives the identical answer.

Curl as Circulation Density

Stokes' Theorem shows \(\operatorname{curl}\mathbf{F}\cdot\mathbf{n}\) is the limiting circulation per unit area about the axis \(\mathbf{n}\). A paddle wheel placed in the flow spins fastest when its axis aligns with \(\operatorname{curl}\mathbf{F}\) — the precise meaning of "rotational tendency."

Instructor Note

The killer feature of Stokes' Theorem: you can swap the surface for any easier one with the same boundary. Stuck on a weird curved surface? Replace it with the flat disk on the same rim. This trade is worth real marks under time pressure.

16.9

The Divergence Theorem

Flux out of a closed surface equals the integral of divergence inside.

The Divergence Theorem

Conditions: \(E\) is a simple solid region with boundary surface \(S\) given the outward (positive) orientation; \(\mathbf{F}\) has continuous partials on an open region containing \(E\).

Conclusion: \[ \iint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_E \operatorname{div}\mathbf{F}\,dV. \] Total outward flux equals the total divergence inside.

EXAMPLE 1 Flux out of the unit sphere

Find the flux of \(\mathbf{F}=z\,\mathbf{i}+y\,\mathbf{j}+x\,\mathbf{k}\) over the unit sphere.

Solution

Divergence: \(\operatorname{div}\mathbf{F}=0+1+0=1.\)

By the theorem, flux \(=\iiint_B 1\,dV = \text{volume of the unit ball} = \tfrac{4}{3}\pi.\) Far easier than the direct surface integral.

EXAMPLE 2 Avoiding four surface integrals

Evaluate \(\iint_S \mathbf{F}\cdot d\mathbf{S}\) for \(\mathbf{F}=xy\,\mathbf{i}+(y^2+e^{xz^2})\,\mathbf{j}+\sin(xy)\,\mathbf{k}\), with \(S\) the boundary of the region under \(z=1-x^2\) bounded by \(y=0,\ y+z=2,\ z=0\).

Solution

The boundary has four pieces — direct evaluation is brutal. Instead, \(\operatorname{div}\mathbf{F}=y+2y+0=3y.\)

Set up \(E\) as a type-3 region and integrate \(\iiint_E 3y\,dV\) in order \(dy\,dz\,dx.\)

Carrying out the iterated integral gives \(\dfrac{184}{35}.\)

Divergence as Outflow Density & Gauss's Law

Shrinking the region to a point shows \(\operatorname{div}\mathbf{F}(P)\) is the net outward flux per unit volume there — the reason for the name. Applied to the electric field, the theorem reproduces Gauss's Law: the flux through any closed surface enclosing a charge \(Q\) is \(4\pi\varepsilon Q\), independent of the surface's shape.

Instructor Note

This is the integral form of every conservation law you'll use: mass, momentum, energy, charge. "Rate of accumulation inside = net flux through the boundary." When you derive the continuity equation in fluids, the Divergence Theorem is the bridge from integral to differential form.

16.10

Summary & Connections

Every theorem in this chapter is one idea wearing different dimensions.

The unifying pattern. In each theorem, the integral of a "derivative" of \(F\) over a region equals the values of \(F\) on the boundary of that region. Raise the dimension and the same statement reappears.

Fundamental Theorem of Calculus

\[ \int_a^b F'(x)\,dx=F(b)-F(a) \]

Fundamental Theorem for Line Integrals

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

Green's Theorem

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

Stokes' Theorem

\[ \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}=\oint_C \mathbf{F}\cdot d\mathbf{r} \]

Divergence Theorem

\[ \iiint_E \operatorname{div}\mathbf{F}\,dV=\iint_S \mathbf{F}\cdot d\mathbf{S} \]

Instructor Note

If you remember just one thing from this chapter, make it this table. Once you see all five as "integrate the derivative inside = evaluate the function on the boundary," the whole chapter collapses into a single beautiful idea — and that's exactly the insight that powers continuum mechanics and field theory.